Important Questions for CBSE Class 10 Maths Chapter 8 Introduction to Trigonometry

1. If \[x\cos \theta y\sin \theta =a,\,x\sin \theta +y\cos \theta =b\], Prove that \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}+{{b}^{2}}\].

Ans: 

Given:

\[x\cos \theta y\sin \theta =a\]        …… (1)

\[x\sin \theta +y\cos \theta =b\]        …… (2)

Squaring and adding the equation (1) and (2) on both sides.

\[{{x}^{2}}{{\cos }^{2}}\theta +{{y}^{2}}{{\sin }^{2}}\theta -2xy\cos \theta \sin \theta +{{x}^{2}}{{\sin }^{2}}\theta +{{y}^{2}}{{\cos }^{2}}\theta +2xy\cos \theta \sin \theta ={{a}^{2}}+{{b}^{2}}\]

\[\Rightarrow {{x}^{2}}\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta  \right)+{{y}^{2}}\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta  \right)={{a}^{2}}+{{b}^{2}}\]

\[\Rightarrow {{x}^{2}}+{{y}^{2}}={{a}^{2}}+{{b}^{2}}\]

\[\therefore {{x}^{2}}+{{y}^{2}}={{a}^{2}}+{{b}^{2}}\]

Hence proved.

2. Prove that \[{{\sec }^{2}}\theta +\cos e{{c}^{2}}\theta \] Can Never Be Less Than \[2\].

Ans: 

Given: \[{{\sec }^{2}}\theta +\cos e{{c}^{2}}\theta \] 

We know that, ${{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta $ and ${{\operatorname{cosec}}^{2}}\theta =1+{{\cot }^{2}}\theta $.

\[\Rightarrow {{\sec }^{2}}\theta +\cos e{{c}^{2}}\theta =1+{{\tan }^{2}}\theta +1+{{\cot }^{2}}\theta \]

\[\Rightarrow {{\sec }^{2}}\theta +\cos e{{c}^{2}}\theta =2+{{\tan }^{2}}\theta +{{\cot }^{2}}\theta \]

Therefore, \[{{\sec }^{2}}\theta +\cos e{{c}^{2}}\theta \] can never be less than \[2\].

Hence proved.

3. If \[\sin \varphi =\dfrac{1}{2}\], show that \[3\cos \varphi -4{{\cos }^{3}}\varphi =0\].

Ans: 

Given: \[\sin \varphi =\dfrac{1}{2}\]

We know that $\sin 30{}^\circ =\dfrac{1}{2}$.

While comparing the angles of $\sin $, we get 

\[\Rightarrow \varphi ={{30}^{\circ }}\]

Substitute \[\varphi ={{30}^{\circ }}\] to get 

\[3\cos \varphi -4{{\cos }^{3}}\varphi =3\cos \left( 30{}^\circ  \right)-4{{\cos }^{3}}\left( 30{}^\circ  \right)\]

\[\Rightarrow 3\left( \dfrac{\sqrt{3}}{2} \right)-4\left( \dfrac{3\sqrt{3}}{8} \right)\Rightarrow 0\]

Therefore, \[3\cos \varphi -4{{\cos }^{3}}\varphi =0\].

Hence proved.

4. If \[7{{\sin }^{2}}\varphi +3{{\cos }^{2}}\varphi =4\], then Show that \[\tan \varphi =\dfrac{1}{\sqrt{3}}\].

Ans: 

Given: \[7{{\sin }^{2}}\varphi +3{{\cos }^{2}}\varphi =4\]

We know that, ${{\sin }^{2}}\varphi +{{\cos }^{2}}\varphi =1$ and \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\]

Then, \[7{{\sin }^{2}}\varphi +3{{\cos }^{2}}\varphi =4({{\sin }^{2}}\varphi +{{\cos }^{2}}\varphi )\]

\[\Rightarrow 7{{\sin }^{2}}\varphi -4{{\sin }^{2}}\varphi =4{{\cos }^{2}}\varphi -3{{\cos }^{2}}\varphi \]

\[\Rightarrow 3{{\sin }^{2}}\varphi ={{\cos }^{2}}\varphi \]

\[\Rightarrow \dfrac{{{\sin }^{2}}\varphi }{{{\cos }^{2}}\varphi }=\dfrac{1}{3}\]

\[\Rightarrow {{\tan }^{2}}\varphi =\dfrac{1}{3}\]

\[\Rightarrow \tan \varphi =\dfrac{1}{\sqrt{3}}\]

\[\therefore \tan \varphi =\dfrac{1}{\sqrt{3}}\]

Hence proved.

5. If \[\cos \varphi +\sin \varphi =\sqrt{2}\cos \varphi \], Prove that \[\cos \varphi -\sin \varphi =\sqrt{2}\sin \varphi \].

Ans: 

Given: \[\cos \varphi +\sin \varphi =\sqrt{2}\cos \varphi \]

Squaring on both sides, we get

\[\Rightarrow {{\left( \cos \varphi +\sin \varphi  \right)}^{2}}=2{{\cos }^{2}}\varphi \]

\[\Rightarrow {{\cos }^{2}}\varphi +{{\sin }^{2}}\varphi +2\cos \varphi \sin \varphi =2{{\cos }^{2}}\varphi \]

$\Rightarrow {{\sin }^{2}}\varphi =2{{\cos }^{2}}\varphi -{{\cos }^{2}}\varphi -2\cos \varphi \sin \varphi $

$\Rightarrow {{\sin }^{2}}\varphi ={{\cos }^{2}}\varphi -2\cos \varphi \sin \varphi $

Add ${{\sin }^{2}}\varphi $ on both sides

\[\Rightarrow 2{{\sin }^{2}}\varphi ={{\cos }^{2}}\varphi -2\cos \varphi \sin \varphi +{{\sin }^{2}}\varphi \]

\[\Rightarrow 2{{\sin }^{2}}\varphi ={{\left( \cos \varphi -\sin \varphi  \right)}^{2}}\] 

\[\therefore \cos \varphi -\sin \varphi =\sqrt{2}\sin \varphi \]

Hence proved.

6. If \[\tan A+\sin A=m\] and \[\tan A-\sin A=n\], then Show that \[{{m}^{2}}-{{n}^{2}}=4\sqrt{mn}\].

Ans: 

Given: 

\[\tan A+\sin A=m\]           …… (1)

\[\tan A-\sin A=n\]            …… (2)

Now to prove \[{{m}^{2}}-{{n}^{2}}=4\sqrt{mn}\].

Take left-hand side

\[{{m}^{2}}-{{n}^{2}}={{\left( \tan A+\sin A \right)}^{2}}-{{\left( \tan A-\sin A \right)}^{2}}\]

\[\Rightarrow {{\tan }^{2}}A+{{\sin }^{2}}A+2\tan A\sin A-{{\tan }^{2}}A-{{\sin }^{2}}A+2\tan A\sin A\]

\[\Rightarrow 4\tan A\sin A\]         

$\therefore {{m}^{2}}-{{n}^{2}}=4\tan A\sin A$      …… (3)

Now take right-hand side

 \[4\sqrt{mn}=4\sqrt{\left( \tan A+\sin A \right)\left( \tan A-\sin A \right)}\]

\[\Rightarrow 4\sqrt{{{\tan }^{2}}A-{{\sin }^{2}}A}\Rightarrow 4\sqrt{\dfrac{{{\sin }^{2}}A}{{{\cos }^{2}}A}-{{\sin }^{2}}A}\]

\[\Rightarrow 4\sqrt{{{\sin }^{2}}A\left( \dfrac{1}{{{\cos }^{2}}A}-1 \right)}\Rightarrow 4\sin A\sqrt{{{\sec }^{2}}A-1}\]

\[\Rightarrow 4\sin A\sqrt{{{\tan }^{2}}A}\Rightarrow 4\sin A\tan A\]

Hence, $4\sqrt{mn}=4\tan A\sin A$

\[\therefore {{m}^{2}}-{{n}^{2}}=4\sqrt{mn}\]

Hence proved. 

7. If \[\sec A=x+\dfrac{1}{4x}\], then prove that \[\sec A+\tan A=2x\] or \[\dfrac{1}{2x}\].

Ans: 

Given: \[\sec A=x+\dfrac{1}{4x}\]

Squaring on both sides.

\[\Rightarrow {{\sec }^{2}}A={{\left( x+\dfrac{1}{4x} \right)}^{2}}\]

We know that, \[{{\sec }^{2}}A=1+{{\tan }^{2}}A\]

\[\Rightarrow 1+{{\tan }^{2}}A={{\left( x+\dfrac{1}{4x} \right)}^{2}}\]

\[\Rightarrow {{\tan }^{2}}A={{\left( x+\dfrac{1}{4x} \right)}^{2}}-1\]

\[\Rightarrow {{\tan }^{2}}A={{x}^{2}}+\dfrac{1}{16{{x}^{2}}}+\dfrac{1}{2}-1\Rightarrow {{x}^{2}}+\dfrac{1}{16{{x}^{2}}}-\dfrac{1}{2}\Rightarrow {{\left( x-\dfrac{1}{4x} \right)}^{2}}\]

Taking square root on both sides,

\[\Rightarrow \tan A=\pm \left( x-\dfrac{1}{4x} \right)\].

Now, find $\sec A+\tan A$

If $\tan A=x-\dfrac{1}{4x}$ means

$\sec A+\tan A=x+\dfrac{1}{4x}+x-\dfrac{1}{4x}\Rightarrow 2x$

$\therefore \sec A+\tan A=2x$

And if $\tan A=-x+\dfrac{1}{4x}$ means

$\sec A+\tan A=x+\dfrac{1}{4x}-x+\dfrac{1}{4x}\Rightarrow \dfrac{2}{4x}\Rightarrow \dfrac{1}{2x}$

$\therefore \sec A+\tan A=\dfrac{1}{2x}$

Hence proved.

8. If \[A,B\]are Acute Angles and \[\sin A=\cos B\], then Find the Value of \[A+B\].

Ans: 

Given: $\sin A=\cos B$

We know that $\sin A=\cos \left( 90{}^\circ -A \right)$

While comparing the values to get

$\cos B=\cos \left( 90{}^\circ -A \right)$

$\Rightarrow B=90{}^\circ -A\Rightarrow A+B=90{}^\circ $

\[\therefore A+B={{90}^{\circ }}\].

9. Evaluate the Following Questions:

a. Solve for \[\phi \], if \[\tan 5\phi =1\].

Ans: 

Given: \[\tan 5\phi =1\]

We know that, ${{\tan }^{-1}}\left( 1 \right)=45{}^\circ $

$5\phi ={{\tan }^{-1}}\left( 1 \right)\Rightarrow 45{}^\circ $

$5\phi =45{}^\circ $

$\phi =\dfrac{45{}^\circ }{5}\Rightarrow 9{}^\circ $

\[\because \phi ={{9}^{\circ }}\]

b. Solve for \[\varphi \], if \[\dfrac{\sin \varphi }{1+\cos \varphi }+\dfrac{1+\cos \varphi }{\sin \varphi }=4\].

Ans: 

Given: \[\dfrac{\sin \varphi }{1+\cos \varphi }+\dfrac{1+\cos \varphi }{\sin \varphi }=4\]

\[\dfrac{{{\sin }^{2}}\varphi +1+{{\cos }^{2}}\varphi +2\cos \varphi }{\sin \varphi \left( 1+\cos \varphi  \right)}=4\]

\[\dfrac{{{\sin }^{2}}\varphi +{{(1+\cos \varphi )}^{2}}}{\sin \varphi (1+\cos \varphi )}=4\]

\[\dfrac{2+2\cos \varphi }{\sin \varphi \left( 1+\cos \varphi  \right)}=4\]

\[\dfrac{2(1+\cos \varphi )}{\sin \varphi (1+\cos \varphi )}=4\]

\[\dfrac{2}{\sin \varphi }=4\]

\[\sin \varphi =\dfrac{1}{2}\]

We know that, $\sin 30{}^\circ =\dfrac{1}{2}$

\[\sin \varphi =\sin {{30}^{\circ }}\Rightarrow \varphi ={{30}^{\circ }}\]

\[\therefore \varphi ={{30}^{\circ }}\].

10. If \[\dfrac{\cos \alpha }{\cos \beta }=m\] and \[\dfrac{\cos \alpha }{\sin \beta }=n\], show that \[({{m}^{2}}+{{n}^{2}}){{\cos }^{2}}\beta ={{n}^{2}}\].

Ans: 

Given: \[\dfrac{\cos \alpha }{\cos \beta }=m\]         …… (1)

\[\dfrac{\cos \alpha }{\sin \beta }=n\]             …… (2)

Squaring equation (1) and (2). We get,

\[{{m}^{2}}=\dfrac{{{\cos }^{2}}\alpha }{{{\cos }^{2}}\beta }\]

\[{{n}^{2}}=\dfrac{{{\cos }^{2}}\alpha }{{{\sin }^{2}}\beta }\]

Now to prove \[({{m}^{2}}+{{n}^{2}}){{\cos }^{2}}\beta ={{n}^{2}}\],

Take left-hand side, 

\[({{m}^{2}}+{{n}^{2}}){{\cos }^{2}}\beta =\left( \dfrac{{{\cos }^{2}}\alpha }{{{\cos }^{2}}\beta }+\dfrac{{{\cos }^{2}}\alpha }{{{\sin }^{2}}\beta } \right){{\cos }^{2}}\beta \]

\[=\left( \dfrac{{{\cos }^{2}}\alpha {{\sin }^{2}}\beta +{{\cos }^{2}}\alpha {{\cos }^{2}}\beta }{{{\cos }^{2}}\beta {{\sin }^{2}}\beta } \right){{\cos }^{2}}\beta \]

\[={{\cos }^{2}}\alpha \left( \dfrac{1}{{{\cos }^{2}}\beta {{\sin }^{2}}\beta } \right){{\cos }^{2}}\beta \]

\[=\dfrac{{{\cos }^{2}}\alpha }{{{\sin }^{2}}\beta }\]

\[={{n}^{2}}\]

\[\therefore ({{m}^{2}}+{{n}^{2}}){{\cos }^{2}}\beta ={{n}^{2}}\]

Hence proved.

11. If \[7\cos ec\varphi -3\cot \varphi =7\], then prove that \[7\cot \varphi -3\cos ec\varphi =3\].

Ans: 

Given: \[7\cos ec\varphi -3\cot \varphi =7\]

Then prove that, \[7\cot \varphi -3\cos ec\varphi =3\]

\[7\cos ec\varphi -3\cot \varphi =7\]

Squaring on both sides, we get

\[49{{\operatorname{cosec}}^{2}}\varphi +9{{\cot }^{2}}\varphi -42\operatorname{cosec}\varphi \cot \varphi =49\]

We know that, ${{\operatorname{cosec}}^{2}}\varphi =1+{{\cot }^{2}}\varphi $ and ${{\cot }^{2}}\varphi ={{\operatorname{cosec}}^{2}}\varphi -1$.

\[49\left( {{\cot }^{2}}\varphi +1 \right)+9\left( {{\operatorname{cosec}}^{2}}\varphi -1 \right)-42\operatorname{cosec}\varphi \cot \varphi =49\]

\[49{{\cot }^{2}}\varphi +49+9{{\operatorname{cosec}}^{2}}\varphi -9-2\left( 3\operatorname{cosec}\varphi \cdot 7\cot \varphi  \right)=49\]

\[{{\left( 7\cot \varphi -3\operatorname{cosec}\varphi  \right)}^{2}}=49-49+9\]

\[{{\left( 7\cot \varphi -3\operatorname{cosec}\varphi  \right)}^{2}}=9\]

Take square root on both sides, we get

\[\therefore 7\cot \varphi -3\operatorname{cosec}\varphi =3\]

Hence proved.

12. Prove that \[2\left( {{\sin }^{6}}\varphi +{{\cos }^{6}}\varphi  \right)\text{ }3\left( {{\sin }^{4}}\varphi +{{\cos }^{4}}\varphi  \right)+1=0\].

Ans: 

Given: \[2\left( {{\sin }^{6}}\varphi +{{\cos }^{6}}\varphi  \right)\text{ }3\left( {{\sin }^{4}}\varphi +{{\cos }^{4}}\varphi  \right)+1=0\]

Let us take left-hand side,

\[\begin{align}&2\left( {{\sin }^{6}}\varphi +{{\cos }^{6}}\varphi  \right)\text{ }3\left( {{\sin }^{4}}\varphi +{{\cos }^{4}}\varphi  \right)+1 \\ & =2\left( {{\left( {{\sin }^{2}}\varphi  \right)}^{3}}+{{\left( {{\cos }^{2}}\varphi  \right)}^{3}} \right)-3\left( {{\left( {{\sin }^{2}}\varphi  \right)}^{2}}+{{\left( {{\cos}^{2}}\varphi  \right)}^{2}} \right)+1 \\ \end{align}\]

\[=2\left[ {{\left( {{\sin }^{2}}\varphi +{{\cos }^{2}}\varphi  \right)}^{3}}-3{{\sin }^{2}}\varphi {{\cos }^{2}}\varphi \left( {{\sin }^{2}}\varphi +{{\cos }^{2}}\varphi  \right) \right]-3\left[ {{\left( {{\sin }^{2}}\varphi +{{\cos }^{2}}\varphi  \right)}^{2}}-2{{\sin }^{2}}\varphi {{\cos }^{2}}\varphi  \right]+1\]

\[=2\left[ 1-3{{\sin }^{2}}\varphi {{\cos }^{2}}\varphi  \right]-3\left[ 1-2{{\sin }^{2}}\varphi {{\cos }^{2}}\varphi  \right]+1\]

\[=2-6{{\sin }^{2}}\varphi {{\cos }^{2}}\varphi -3+6{{\sin }^{2}}\varphi {{\cos }^{2}}\varphi +1\]

\[=-1+1\]

\[=0\]

Therefore, \[2\left( {{\sin }^{6}}\varphi +{{\cos }^{6}}\varphi  \right)\text{ }3\left( {{\sin }^{4}}\varphi +{{\cos }^{4}}\varphi  \right)+1=0\].

Hence proved.

13. If \[\tan \theta =\dfrac{5}{6}\] and \[\theta =\phi ={{90}^{\circ }}\]. What is the value of \[\cot \phi \].

Ans: 

Given: \[\tan \theta =\dfrac{5}{6}\] and \[\theta =\phi ={{90}^{\circ }}\]

We know that, $\tan \theta =\dfrac{1}{\cot \theta }$.

$\cot \phi =\dfrac{1}{\tan \phi }$

$=\dfrac{1}{{5}/{6}\;}$

$=\dfrac{6}{5}$ 

$\therefore \cot \phi =\dfrac{6}{5}$.

14. What is the Value of \[\tan \varphi \] in terms of \[\sin \varphi \] ?

Ans: 

Given: \[\tan \varphi \]

We know that, \[\tan \varphi =\dfrac{\sin \varphi }{\cos \varphi }\] and \[{{\cos }^{2}}\varphi +{{\sin }^{2}}\varphi =1\]

\[\tan \varphi =\dfrac{\sin \varphi }{\cos \varphi }\]

\[\therefore \tan \varphi =\dfrac{\sin \varphi }{\sqrt{1-{{\sin }^{2}}\varphi }}\]

15. If \[\sec \varphi +\tan \varphi =4\], Find the Value of \[\sin \varphi \], \[\cos \varphi \].

Ans: 

Given: \[\sec \varphi +\tan \varphi =4\]

\[\dfrac{1}{\cos \varphi }+\dfrac{\sin \varphi }{\cos \varphi }=4\]

\[\dfrac{1+\sin \varphi }{\cos \varphi }=4\]

$1+\sin \varphi =4\cos \varphi $

Squaring on both sides.

${{\left( 1+\sin \varphi  \right)}^{2}}={{\left( 4\cos \varphi  \right)}^{2}}$

$1+2\sin \varphi +{{\sin }^{2}}\varphi =16{{\cos }^{2}}\varphi $

$1+2\sin \varphi +{{\sin }^{2}}\varphi =16\left( 1-{{\sin }^{2}}\varphi  \right)$

$1+2\sin \varphi +{{\sin }^{2}}\varphi =16-16{{\sin }^{2}}\varphi $

$17{{\sin }^{2}}\varphi +2\sin \varphi -15=0$

$17{{\sin }^{2}}\varphi +17\sin \varphi -15\sin \varphi -15=0$

$17\sin \varphi \left( \sin \varphi +1 \right)-15\left( \sin \varphi +1 \right)=0$

$\left( \sin \varphi +1 \right)\left( 17\sin \varphi -15 \right)=0$

If $\sin \varphi +1=0$

Hence, $\sin \varphi =-1$ is not possible.

Then, $17\sin \varphi -15=0$

$\therefore \sin \varphi =\dfrac{15}{17}$

Now find $\cos \varphi $

\[\dfrac{1+\sin \varphi }{\cos \varphi }=4\]

Substitute the value of $\sin \varphi $

\[1+\dfrac{15}{17}=4\cos \varphi \]

\[\dfrac{32}{17}=4\cos \varphi \]

\[\Rightarrow \cos \varphi =\dfrac{32}{17\left( 4 \right)}\Rightarrow \dfrac{8}{17}\]

\[\therefore \cos \varphi =\dfrac{8}{17}\]

Short Answer Questions (2 Marks)

1. In \[\Delta ABC\] , Right Angled at \[B,AB=24cm,BC=7cm\].

Determine the Following Equations:    

(i) \[\sin A,\cos A\]

Ans: 

Let us draw a right-angled triangle \[ABC\], right angled at \[B\].

Using Pythagoras theorem, find $AC$.

\[A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}\]

\[={{(24)}^{2}}+{{(7)}^{2}}\]

\[=576+49\]

\[=625\]

\[\therefore AC=25cm\]

Then,

\[\sin A=\dfrac{BC}{AC}=\dfrac{7}{25}\]

\[\therefore \sin A=\dfrac{7}{25}\]

 \[\cos A=\dfrac{AB}{AC}=\dfrac{24}{25}\]

\[\therefore \cos A=\dfrac{24}{25}\]

(ii) \[\sin C,\cos C\]

Ans: 

Let us draw a right-angled triangle \[ABC\], right angled at \[B\].

Using Pythagoras theorem, find $AC$.

\[A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}\]

\[={{(24)}^{2}}+{{(7)}^{2}}\]

\[=576+49\]

\[=625\]

\[\therefore AC=25cm\]

Then,

\[\sin C=\dfrac{AB}{AC}=\dfrac{24}{25}\]       

\[\therefore \sin C=\dfrac{24}{25}\]

\[\cos C=\dfrac{BC}{AC}=\dfrac{7}{25}\]

\[\therefore \cos C=\dfrac{7}{25}\]

2. In Adjoining Figure, Find the Value of \[\tan P-\cot R\].     

Ans: 

Using Pythagoras theorem,

$P{{R}^{2}}=P{{Q}^{2}}+Q{{R}^{2}}$

\[{{(13)}^{2}}={{(12)}^{2}}+Q{{R}^{2}}\]

\[Q{{R}^{2}}=169-144\Rightarrow 25\]

\[\therefore QR=5cm\]

Then find \[\tan P-\cot R\] ,

First find the value of $\tan P$

$\tan P=\dfrac{side\,opp.\,to\,\angle P}{side\,adj.\,to\,\angle P}=\dfrac{QR}{PQ}=\dfrac{5}{12}$

$\therefore \tan P=\dfrac{5}{12}$

Now find the value of $\cot R$

We know that, $\tan R=\dfrac{1}{\cot R}$

For that we need to first find the value of $\tan R$

$\tan R=\dfrac{side\,opp.\,to\,\angle R}{side\,adj.\,to\,\angle R}=\dfrac{PQ}{QR}=\dfrac{12}{5}$

$\therefore \cot R=\dfrac{5}{12}$

Then, 

\[\tan P-\cot R=\dfrac{QR}{PQ}-\dfrac{QR}{PQ}\]

\[\Rightarrow \dfrac{5}{12}-\dfrac{5}{12}\Rightarrow 0\]

\[\therefore \tan P-\cot R=0\].

3. If \[\sin A=\dfrac{3}{4}\], Calculate the Value of \[\cos A\] and \[\tan A\].                                                         

Ans: 

Given that the triangle \[ABC\] in which \[\angle B={{90}^{\circ }}\]

Let us take \[BC=3k\] and \[AC=4k\]

Then using Pythagoras theorem,

\[AB=\sqrt{{{(AC)}^{2}}-{{(BC)}^{2}}}\]

 \[\Rightarrow \sqrt{{{(4k)}^{2}}-{{(3k)}^{2}}}\]

 \[\Rightarrow \sqrt{16k-9k}\]

\[\Rightarrow k\sqrt{7}\]

$\therefore AB=k\sqrt{7}$

Calculate the value of \[\cos A\] 

$\cos A=\dfrac{AB}{AC}=\dfrac{k\sqrt{7}}{4k}=\dfrac{\sqrt{7}}{4}$

$\therefore \cos A=\dfrac{\sqrt{7}}{4}$

And calculate the value of \[\tan A\]

$\tan A=\dfrac{BC}{AB}=\dfrac{3k}{k\sqrt{7}}=\dfrac{3}{\sqrt{7}}$

$\therefore \tan A=\dfrac{3}{\sqrt{7}}$

4. Given \[15\cot A=8\], Find the Values of \[\sin A\] and \[\sec A\].                                                           

Ans: 

Given: \[15\cot A=8\]

Let us assume a triangle \[ABC\] in which \[\angle B={{90}^{\circ }}\]. 

Then,

\[15\cot A=8\]

\[\Rightarrow \cot A=\dfrac{8}{15}\]

Since $\cot A=\dfrac{adj}{hyp}=\dfrac{AB}{BC}$.

Let us draw the triangle.

Now, \[AB=8k\] and \[BC=15k\].

Using Pythagoras theorem, find the value of $AC$.

\[AC=\sqrt{{{(AB)}^{2}}+{{(BC)}^{2}}}\]

\[\Rightarrow \sqrt{{{(8k)}^{2}}+{{(15k)}^{2}}}\]

$\Rightarrow \sqrt{64{{k}^{2}}+225{{k}^{2}}}$

$\Rightarrow \sqrt{289{{k}^{2}}}$

\[\Rightarrow 17k\]

$\therefore AC=17k$

Now, find the values of \[\sin A\] and \[\sec A\].

\[\sin A=\dfrac{BC}{AC}=\dfrac{15k}{17k}=\dfrac{15}{17}\]

$\therefore \sin A=\dfrac{15}{17}$

\[\sec A=\dfrac{AC}{AB}=\dfrac{17k}{8k}=\dfrac{17}{8}\]

 \[\therefore \sec A=\dfrac{17}{8}\]

5. If \[\angle A\] and \[\angle \,B\] are Acute Angles Such That \[\cos A=\cos B\], then show that \[\angle A=\angle \,B\]

Ans: 

Given: \[\cos A=\cos B\]

In right triangle \[ABC\],

\[\cos A=\dfrac{side\,adj.\,A}{hyp.}=\dfrac{AC}{AB}\]  …… (1)

And, \[\cos B=\dfrac{side\,adj.\,B}{hyp.}=\dfrac{BC}{AB}\] …… (2)

Then, \[\cos A=\cos B\]

Now, equate equation (1) and (2).

\[\Rightarrow \dfrac{AC}{AB}=\dfrac{BC}{AB}\]

$\Rightarrow AC=BC$

\[\Rightarrow \angle A=\angle B\].

Therefore, Angles opposite to equal sides are equal.

Hence proved.

6. State Whether the Following are True or False. Justify Your Answer.           

1.  The Value of \[\tan A\] is Always Less than \[1\].

Ans: 

False because sides of a right triangle may have any length, so \[\tan A\] may have any value. For example, $\tan A=\dfrac{BC}{AB}=\dfrac{15}{10}=\dfrac{3}{2}=1.5$. 

2.  \[\sec A=\dfrac{12}{5}\] for Some Value of Angle \[A\].

Ans:

True as \[\sec A\] is always greater than \[1\]. For example, $\sec A=\dfrac{hyp.}{side\,adj.\,A}$ . As hypotenuse will be the largest side. So, it is true.

3.  \[\cos A\] is the Abbreviation Used for the Cosecant of Angle \[A\].

Ans:

False as \[\cos A\] is the abbreviation of \[\operatorname{cosineA}\]. Because $\cos A$ means cosine of angle $A$ and $\operatorname{co}\sec A$ means cosecant of angle $A$.

4.  \[\cot A\] is the Product of \[\cot \]and \[A\].

Ans:

False as \[\cot A\] is not the product of \[cot\] and \[A\]. \[cot\] without \[A\] doesn’t have meaning. 

5.  \[\sin \theta =\dfrac{4}{3}\] for Some Angle \[\theta \].

Ans:  

False as \[\sin \theta \] cannot be greater than \[1\]. For example, $\sin \theta =\dfrac{side\,opp.\,\theta }{hyp}$. Since the hypotenuse is the largest side. So, \[\sin \theta \] will be less than $1$.

7. Evaluate the Following Equations: 

i. \[\dfrac{\sin {{18}^{\circ }}}{\cos {{72}^{\circ }}}\]

Ans:

Given: \[\dfrac{\sin {{18}^{\circ }}}{\cos {{72}^{\circ }}}\]

We know that, $\sin \left( 90{}^\circ -\theta  \right)=\cos \theta $

Then, 

\[\dfrac{\sin {{18}^{\circ }}}{\cos {{72}^{\circ }}}=\dfrac{\sin ({{90}^{\circ }}-{{72}^{\circ }})}{\cos {{72}^{\circ }}}=\dfrac{\cos {{72}^{\circ }}}{\cos {{72}^{\circ }}}=1\]                   

$\therefore \dfrac{\sin 18{}^\circ }{\cos 72{}^\circ }=1$

ii. \[\dfrac{\tan {{26}^{\circ }}}{\cot {{64}^{\circ }}}\]

Ans:

Given: \[\dfrac{\tan {{26}^{\circ }}}{\cot {{64}^{\circ }}}\]

We know that, $\tan \left( 90{}^\circ -\theta  \right)=\cot \theta $

Then,

\[\dfrac{\tan {{26}^{\circ }}}{\cot {{64}^{\circ }}}=\dfrac{\tan ({{90}^{\circ }}-{{64}^{\circ }})}{\cot {{64}^{\circ }}}=\dfrac{\cot {{64}^{\circ }}}{\cot {{64}^{\circ }}}=1\]

$\therefore \dfrac{\tan 26{}^\circ }{\cot 64{}^\circ }=1$

iii. \[\cos {{48}^{\circ }}-\sin {{42}^{\circ }}\]

Ans:

Given: \[\cos {{48}^{\circ }}-\sin {{42}^{\circ }}\]

We know that, $\cos \left( 90{}^\circ -\theta  \right)=\sin \theta $

Then, 

\[\Rightarrow \cos ({{90}^{\circ }}-{{42}^{\circ }})-\sin {{42}^{\circ }}\]

\[\Rightarrow \sin {{42}^{\circ }}-\sin {{42}^{\circ }}\Rightarrow 0\]

 \[\therefore \cos 48{}^\circ -\sin 42{}^\circ =0\]

iv. \[\operatorname{cosec}31{}^\circ -\sec 59{}^\circ \]

Ans:  

Given: \[\operatorname{cosec}31{}^\circ -\sec 59{}^\circ \]

We know that, $\operatorname{cosec}\left( 90{}^\circ -\theta  \right)=\sec \theta $

Then,

\[\Rightarrow \operatorname{cosec}(({{90}^{\circ }}-{{59}^{\circ }})-\sec {{59}^{\circ }}\]

\[\Rightarrow \sec {{59}^{\circ }}-\sec {{59}^{\circ }}\Rightarrow 0\]

\[\therefore \operatorname{cosec}31{}^\circ -\sec 59{}^\circ =0\]

1. Show that the Following Equations:                       

1. \[\tan {{48}^{\circ }}\tan {{23}^{\circ }}\tan {{42}^{\circ }}\tan {{67}^{\circ }}=1\]

Ans:

Given: \[\tan {{48}^{\circ }}\tan {{23}^{\circ }}\tan {{42}^{\circ }}\tan {{67}^{\circ }}=1\]

We know that, $\tan \left( 90{}^\circ -\theta  \right)=\cot \theta $.

Now let us take left-hand side,

\[\tan {{48}^{\circ }}\tan {{23}^{\circ }}\tan {{42}^{\circ }}\tan {{67}^{\circ }}\]

\[\Rightarrow \tan ({{90}^{\circ }}-{{42}^{\circ }})\tan ({{90}^{\circ }}-{{67}^{\circ }})\tan {{42}^{\circ }}\tan {{67}^{\circ }}\]

\[\Rightarrow \cot {{42}^{\circ }}\cot {{67}^{\circ }}\tan {{42}^{\circ }}\tan {{67}^{\circ }}\]

\[\Rightarrow \dfrac{1}{\tan {{42}^{\circ }}}.\dfrac{1}{\tan {{67}^{\circ }}}.\tan {{42}^{\circ }}\tan {{67}^{\circ }}\]

\[\Rightarrow 1\] is equal to R.H.S

\[\therefore \tan {{48}^{\circ }}\tan {{23}^{\circ }}\tan {{42}^{\circ }}\tan {{67}^{\circ }}=1\]

Hence proved.

(ii) \[\cos {{38}^{\circ }}\cos {{52}^{\circ }}-\sin {{38}^{\circ }}\sin {{52}^{\circ }}=0\]

Ans: 

Given: \[\cos {{38}^{\circ }}\cos {{52}^{\circ }}-\sin {{38}^{\circ }}\sin {{52}^{\circ }}=0\]

We know that, $\cos \left( 90{}^\circ -\theta  \right)=\sin \theta $

Now let us take left-hand side, 

\[\cos {{38}^{\circ }}cos{{52}^{\circ }}-\sin {{38}^{\circ }}\sin {{52}^{\circ }}\]

\[\Rightarrow \cos ({{90}^{\circ }}-{{52}^{\circ }})\cos ({{90}^{\circ }}-{{38}^{\circ }})-\sin {{38}^{\circ }}\sin {{52}^{\circ }}\]

\[\Rightarrow \sin {{52}^{\circ }}\sin {{38}^{\circ }}-\sin {{38}^{\circ }}\sin {{52}^{\circ }}\]

\[\Rightarrow 0\] is equal to R.H.S

$\therefore \cos {{38}^{\circ }}cos{{52}^{\circ }}-\sin {{38}^{\circ }}\sin {{52}^{\circ }}=0$

Hence proved.

2. If  \[\tan 2A=\cot (A-{{18}^{\circ }})\] where \[2A\] is an Acute Angle, Find the Value of \[A\]. 

Ans: 

Given: \[\tan 2A=\cot (A-{{18}^{\circ }})\]

We know that, $\cot \left( 90{}^\circ -\theta  \right)=\tan \theta $

Then,

\[\Rightarrow \cot ({{90}^{\circ }}-2A)=\cot (A-{{18}^{\circ }})\]

Now equalise the angles,

\[{{90}^{\circ }}-2A=A-{{18}^{\circ }}\]

\[-2A-A=-{{18}^{\circ }}-{{90}^{\circ }}\]

\[-3A=-{{108}^{\circ }}\]

\[A=\dfrac{108{}^\circ }{3}\]

$\therefore A=36{}^\circ $

3. If \[\tan A=\cot B\], then Prove That \[A+B={{90}^{\circ }}\].                                                            

Ans: 

Given: \[\tan A=\cot B\]

We know that, $\cot \left( 90{}^\circ -\theta  \right)=\tan \theta $

Then,

\[\cot ({{90}^{\circ }}-A)=\cot B\]

Now equalise the angles,

\[{{90}^{\circ }}-A=B\]

\[\Rightarrow A+B={{90}^{\circ }}\]

$\therefore A+B=90{}^\circ $

Hence proved.

4. If \[\sec 4A=\operatorname{cosec}(A-{{20}^{\circ }})\], Where \[4A\] is an Acute Angle, Then Find the Value of $A$.

Ans:  

Given: \[\sec 4A=\operatorname{cosec}(A-{{20}^{\circ }})\]

We know that, $\operatorname{cosec}\left( 90{}^\circ -\theta  \right)=\sec \theta $

Then,

\[\Rightarrow \operatorname{cosec}({{90}^{\circ }}-4A)=\operatorname{cosec}(A-{{20}^{\circ }})\]

Now equalise the angles,

\[{{90}^{\circ }}-4A=A-{{20}^{\circ }}\]

\[-4A-A=-{{20}^{\circ }}-{{90}^{\circ }}\]

\[-5A=-{{110}^{\circ }}\]

\[A=\dfrac{{{110}^{\circ }}}{5}\]

\[\therefore A={{22}^{\circ }}\]

Hence proved.

5. If \[A,B\] and \[C\] are Interior Angles of a \[\Delta ABC\], then Show That \[\sin \left( \dfrac{B+C}{2} \right)=\cos \dfrac{A}{2}\].

Ans: 

Given: \[A,B\] and \[C\] are interior angles of a \[\Delta ABC\].

We know that, \[A+B+C={{180}^{\circ }}\].

Let us consider, 

\[\dfrac{A+B+C}{2}={{90}^{\circ }}\]

\[\Rightarrow \dfrac{B+C}{2}={{90}^{\circ }}-\dfrac{A}{2}\]

Multiply $\sin $ on both sides,

\[\sin \left( \dfrac{B+C}{2} \right)=\sin \left( {{90}^{\circ }}-\dfrac{A}{2} \right)\]

\[\therefore \sin \left( \dfrac{B+C}{2} \right)=\cos \dfrac{A}{2}\]

Hence proved.

6. Express \[\sin {{67}^{\circ }}+\cos {{75}^{\circ }}\] in terms of trigonometric ratios of angles between  \[{{0}^{\circ }}\] and \[{{45}^{\circ }}\].                                                   

Ans: 

Given: \[\sin {{67}^{\circ }}+\cos {{75}^{\circ }}\].

We know that, $\sin \left( 90{}^\circ -\theta  \right)=\cos \theta $ and $\cos \left( 90{}^\circ -\theta  \right)=\sin \theta $.

Then, 

\[\sin {{67}^{\circ }}+\cos {{75}^{\circ }}=\sin ({{90}^{\circ }}-{{23}^{\circ }})+\cos ({{90}^{\circ }}-{{15}^{\circ }})\]

\[=\cos {{23}^{\circ }}+\sin {{15}^{\circ }}\]

\[\therefore \cos {{23}^{\circ }}+\sin {{15}^{\circ }}\] is the required value.

7. Express the Trigonometric Ratios \[\sin A,\sec A\] and \[\tan A\] in Terms of \[\cot A\]. 

Ans: 

Find the value of \[\sin A\] in terms of \[\cot A\].

By using identity \[{{\operatorname{cosec}}^{2}}A-{{\cot }^{2}}A=1\].

Then, use $\operatorname{cosec}A=\dfrac{1}{\sin A}$

\[\Rightarrow {{\operatorname{cosec}}^{2}}A=1+{{\cot }^{2}}A\]

\[\Rightarrow \dfrac{1}{{{\sin }^{2}}A}=1+{{\cot }^{2}}A\]

\[\Rightarrow {{\sin }^{2}}A=\dfrac{1}{1+{{\cot }^{2}}A}\]

\[\Rightarrow \sin A=\dfrac{1}{\sqrt{1+{{\cot }^{2}}A}}\]

\[\therefore \sin A=\dfrac{1}{\sqrt{1+{{\cot }^{2}}A}}\]

Now find the value for \[\sec A\] in terms of \[\cot A\].

Using identity \[{{\sec }^{2}}A-{{\tan }^{2}}A=1\]

\[\Rightarrow {{\sec }^{2}}A=1+{{\tan }^{2}}A\]

\[\Rightarrow {{\sec }^{2}}A=1+\dfrac{1}{{{\cot }^{2}}A}\]

\[\Rightarrow {{\sec }^{2}}A=\dfrac{{{\cot }^{2}}A+1}{{{\cot }^{2}}A}\]

\[\Rightarrow \sec A=\dfrac{\sqrt{1+{{\cot }^{2}}A}}{\cot A}\] 

\[\therefore \sec A=\dfrac{\sqrt{1+{{\cot }^{2}}A}}{\cot A}\]

And find the value for \[\tan A\] in terms of \[\cot A\]

By trigonometric ratio property, $\tan A=\dfrac{1}{\cot A}$

Hence, \[\tan A=\dfrac{1}{\cot A}\]

Therefore, \[\sin A,\sec A\] and \[\tan A\] are founded in terms of \[\cot A\]. 

8. Write the Other Trigonometric Ratios of $A$ in Terms of \[\sec A\].                        

Ans: 

Find the value of \[\sin A\] in terms of \[\sec A\]

By using identity,\[{{\sin }^{2}}A+{{\cos }^{2}}A=1\]

\[\Rightarrow {{\sin }^{2}}A=1-{{\cos }^{2}}A\]

\[\Rightarrow {{\sin }^{2}}A=1-\dfrac{1}{{{\sec }^{2}}A}\]

\[\Rightarrow {{\sin }^{2}}A=\dfrac{{{\sec }^{2}}A-1}{{{\sec }^{2}}A}\]

\[\Rightarrow \sin A=\dfrac{\sqrt{{{\sec }^{2}}A-1}}{\sec A}\]

\[\therefore \sin A=\dfrac{\sqrt{{{\sec }^{2}}A-1}}{\sec A}\]

Now find the value for \[\cos A\] in terms of \[\sec A\],

By trigonometric ratio property, $\cos A=\dfrac{1}{\sec A}$

$\therefore \cos A=\dfrac{1}{\sec A}$

Find the value for $\tan A$ in terms of \[\sec A\],

By using identity, \[{{\sec }^{2}}A-{{\tan }^{2}}A=1\]

\[\Rightarrow {{\tan }^{2}}A={{\sec }^{2}}A-1\]

\[\Rightarrow \tan A=\sqrt{{{\sec }^{2}}A-1}\]

\[\therefore \tan A=\sqrt{{{\sec }^{2}}A-1}\]

Find the value for \[\operatorname{cosec}A\] in terms of \[\sec A\]

By trigonometric ratio property, \[\operatorname{cosec}A=\dfrac{1}{\sin A}\]

\[\Rightarrow \operatorname{cosec}A=\dfrac{1}{\sin A}\]

Substitute the value of \[\sin A=\dfrac{\sqrt{{{\sec }^{2}}A-1}}{\sec A}\].

\[\Rightarrow \operatorname{cosec}A=\dfrac{1}{\dfrac{\sqrt{{{\sec }^{2}}A-1}}{\sec A}}\]

\[\Rightarrow \operatorname{cosec}A=\dfrac{\sec A}{\sqrt{{{\sec }^{2}}A-1}}\]

\[\therefore \operatorname{cosec}A=\dfrac{\sec A}{\sqrt{{{\sec }^{2}}A-1}}\]

Finally, find the value for \[\cot A\] in terms of \[\sec A\]

By trigonometric ratio property, \[\cot A=\dfrac{1}{\tan A}\]

$\Rightarrow \cot A=\dfrac{1}{\tan A}$

Substitute the value of \[\tan A=\sqrt{{{\sec }^{2}}A-1}\]

\[\Rightarrow \cot A=\dfrac{1}{\sqrt{{{\sec }^{2}}A-1}}\]

$\therefore \cot A=\dfrac{1}{\sqrt{{{\sec }^{2}}A-1}}$.

9. Evaluate the Following Equations:                                    

(i) \[\dfrac{{{\sin }^{2}}{{63}^{\circ }}+{{\sin }^{2}}{{27}^{\circ }}}{{{\cos }^{2}}{{17}^{\circ }}+{{\cos }^{2}}{{73}^{\circ }}}\]

Ans:

Given: \[\dfrac{{{\sin }^{2}}{{63}^{\circ }}+{{\sin }^{2}}{{27}^{\circ }}}{{{\cos }^{2}}{{17}^{\circ }}+{{\cos }^{2}}{{73}^{\circ }}}\]

We know that, \[\sin ({{90}^{\circ }}-\theta )=\cos \theta ,\cos ({{90}^{\circ }}-\theta )=\sin \theta \] and  \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]

Then,

\[\Rightarrow \dfrac{{{\sin }^{2}}{{63}^{\circ }}+{{\sin }^{2}}({{90}^{\circ }}-{{63}^{\circ }})}{{{\cos }^{2}}({{90}^{\circ }}-{{73}^{\circ }})+{{\cos }^{2}}{{73}^{\circ }}}\]

\[\Rightarrow \dfrac{{{\sin }^{2}}{{63}^{\circ }}+{{\cos }^{2}}{{63}^{\circ }}}{{{\sin }^{2}}{{73}^{\circ }}+{{\cos }^{2}}{{73}^{\circ }}}\]         

\[\Rightarrow \dfrac{1}{1}\Rightarrow 1\]                      

$\therefore \dfrac{{{\sin }^{2}}{{63}^{\circ }}+{{\sin }^{2}}{{27}^{\circ }}}{{{\cos }^{2}}{{17}^{\circ }}+{{\cos }^{2}}{{73}^{\circ }}}=1$         

(ii) \[\sin {{25}^{\circ }}\cos {{65}^{\circ }}+\cos {{25}^{\circ }}\sin {{65}^{\circ }}\]

Ans:  

Given: \[\sin {{25}^{\circ }}\cos {{65}^{\circ }}+\cos {{25}^{\circ }}\sin {{65}^{\circ }}\]

We know that, $\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$

Then,

\[\Rightarrow \sin \left( 25{}^\circ +65{}^\circ  \right)\]

\[\Rightarrow \sin 90{}^\circ \]                    

\[\Rightarrow 1\]

$\therefore \sin {{25}^{\circ }}\cos {{65}^{\circ }}+\cos {{25}^{\circ }}\sin {{65}^{\circ }}=1$

10. Show that Any Positive Odd Integer Is of the Form \[\mathbf{6q}\text{ }+\text{ }\mathbf{1}\], or \[\mathbf{6q}\text{ }+\text{ }\mathbf{3},\]or \[\mathbf{6q}\text{ }+\text{ }\mathbf{5}\], where \[q\] is some integer.                                                      

Ans: 

Let \[a\] be any positive integer and \[b=\text{ }6\]. 

Then, by Euclid’s algorithm,

\[a=6q+r\] for some integer \[q\ge 0\], and \[r=0,1,2,3,4,5\] because \[0\le r<\text{ }6\].

Therefore, \[a=6q\] or \[6q+\text{ }1\] or \[6q+\text{ }2\]or \[6q\text{ }+3\]or \[6q+\text{ }4\]or \[6q+\text{ }5\]

Also, \[6q+1=2\times 3q+1=2{{k}_{1}}+1,\] where \[{{k}_{1}}\] is a positive integer

\[6q+3=(6q+2)+1=2(3q+1)+1=2{{k}_{2}}+\text{ }1,\]Where \[{{k}_{2}}\]is an integer

\[6q+5=(6q+4)+1=2(3q+2)+1=2{{k}_{3}}+1\], where \[{{k}_{3}}\] is an integer

Clearly, \[6q+1,6q+3,6q+5\] are of the form \[2k+\text{ }1,\]where \[k\]an integer is.

Therefore, \[6q+1,6q+3,6q+5\] are not exactly divisible by \[2\].

Hence, these expressions of numbers are odd numbers.

And therefore, any odd integer can be expressed in the form \[6q+1\], or \[6q+3\], or\[6q+5.\]

11. An Army Contingent of \[\mathbf{616}\] Members are to March Behind an Army Band of \[\mathbf{32}\] Members in a Parade. The Two Groups Are to March in the Same Number of Columns. What Is the Maximum Number of Columns in Which They Can March?

Ans: 

We have to find the \[HCF\left( 616,\text{ }32 \right)\] to find the maximum number of columns in which they can march. 

To find the HCF, we can use Euclid’s algorithm.

\[616=32\times 19+8\]

\[\Rightarrow 32=8\times 4+0\]

Hence,  \[HCF\left( 616,\text{ }32 \right)\] is \[8\].

Therefore, they can march in \[8\] columns each.

12. Use Euclid’s Division Lemma to Show That the Square of Any Positive Integer Is Either of Form \[\mathbf{3}m\]or \[\mathbf{3}m+\mathbf{1}\] for some integer \[m\]. 

[Hint: Let \[x\] be any positive integer then it is of the form\[\mathbf{3}q,\mathbf{3}q+\mathbf{1}\] or \[\mathbf{3}q+\mathbf{2}\text{ }\]. Now square each of these and show that they can be rewritten in the form \[\mathbf{3}m\]or \[\mathbf{3}m+\mathbf{1}\].]

Ans: 

Let\[\text{ }a\] be any positive integer and \[b=3\].

Then \[a=3q+r\] for some integer \[q\ge 0\]

And \[r=0,1,2\] because \[0\le r<3\]

Therefore, \[a=3q\] or \[3q+1\] or \[3q+2\]

Or,

\[\Rightarrow {{a}^{2}}={{(3q)}^{2}}\] or \[{{(3q+1)}^{2}}\] or \[{{(3q+2)}^{2}}\]

\[\Rightarrow {{a}^{2}}={{(9q)}^{2}}\] or \[9{{q}^{2}}+6q+1\] or \[9{{q}^{2}}+12q+4\]

\[\Rightarrow {{a}^{2}}=3\times {{(3q)}^{2}}\] or \[3(3{{q}^{2}}+2q)+1\] or \[3(3{{q}^{2}}+4q+1)+1\]

\[\Rightarrow a=3{{k}_{1}}\] or \[3{{k}_{2}}+1\] or \[3{{k}_{3}}+1\]

Where \[{{k}_{1}},{{k}_{2}},{{k}_{3}}\] are some positive integers 

Hence, it can be said that the square of any positive integer is either of the form \[3m\]or \[3m+1\].

Short Answer Questions (3 Marks)

1. Given \[\sec \theta =\dfrac{13}{12}\], Calculate the Values for All Other Trigonometric Ratios. 

Ans: 

Given:  \[\sec \theta =\dfrac{13}{12}\]

Let us consider a triangle \[ABC\] in which \[\angle A=\theta \] and \[\angle B={{90}^{\circ }}\]

Let \[AB=12k\] and \[AC=13k\]

Then, find the value of $BC$

\[BC=\sqrt{{{(AC)}^{2}}-{{(AB)}^{2}}}\]

\[\begin{align}&\Rightarrow\sqrt{{{(13k)}^{2}}-{{(12k)}^{2}}}\\&\Rightarrow \sqrt{69{{k}^{2}}-144{{k}^{2}}} \\ & \Rightarrow \sqrt{25{{k}^{2}}} \\ & \Rightarrow 5k \\ \end{align}\]

\[\therefore BC=5k\]

Since, \[\sec \theta =\dfrac{13}{12}\]

Similarly, 

\[\sin \theta =\dfrac{BC}{AC}=\dfrac{5k}{13k}=\dfrac{5}{13}\]

\[\cos \theta =\dfrac{AB}{AC}=\dfrac{12k}{13k}=\dfrac{12}{13}\]

\[\tan \theta =\dfrac{BC}{AB}=\dfrac{5k}{12k}=\dfrac{5}{12}\]

\[\cot \theta =\dfrac{AB}{BC}=\dfrac{12k}{5k}=\dfrac{12}{5}\]

\[\cos ec\theta =\dfrac{AC}{BC}=\dfrac{13k}{5k}=\dfrac{13}{5}\]

2. If \[\cot \theta =\dfrac{7}{8}\], then Evaluate the Followings Equations:                                                                                        

i. \[\dfrac{(1+\sin \theta )(1-\sin \theta )}{(1+\cos \theta )(1-\cos \theta )}\]

Ans:

Given: \[\cot \theta =\dfrac{7}{8}\]

Let us consider a triangle \[ABC\] in which \[\angle A=\theta \] and \[\angle B={{90}^{\circ }}\]

Then, \[AB=7k\] and \[BC=8k\]

Using Pythagoras theorem, find $AC$

\[AC=\sqrt{{{(BC)}^{2}}+{{(AB)}^{2}}}\]

\[\Rightarrow \sqrt{{{(8k)}^{2}}+{{(7k)}^{2}}}\]

\[\Rightarrow \sqrt{64{{k}^{2}}+49{{k}^{2}}}\]

\[\Rightarrow \sqrt{113{{k}^{2}}}\]

\[\Rightarrow \sqrt{113}k\]

\[\therefore AC=\sqrt{113}k\]

Now find the value of trigonometric ratios.

\[\sin \theta =\dfrac{BC}{AC}=\dfrac{8k}{\sqrt{113}k}=\dfrac{8}{\sqrt{113}}\] and \[\cos \theta =\dfrac{AB}{AC}=\dfrac{7k}{\sqrt{113}k}=\dfrac{7}{\sqrt{113}}\].

Then, 

\[\dfrac{(1+\sin \theta )(1-\sin \theta )}{(1+\cos \theta )(1-\cos \theta )}\]

We know that, ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$

Then,

\[\Rightarrow \dfrac{1-{{\sin }^{2}}\theta }{1-{{\cos }^{2}}\theta }\Rightarrow \dfrac{{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta }\]

\[\Rightarrow \dfrac{\dfrac{49}{113}}{\dfrac{64}{113}}\Rightarrow \dfrac{49}{113}\cdot \dfrac{113}{64}\Rightarrow \dfrac{49}{64}\]

$\therefore \dfrac{(1+\sin \theta )(1-\sin \theta )}{(1+\cos \theta )(1-\cos \theta )}=\dfrac{49}{64}$

ii. \[{{\cot }^{2}}\theta \]

Ans: 

Given: \[{{\cot }^{2}}\theta \]

We know that, $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$

Then,

\[\Rightarrow \dfrac{{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta }\Rightarrow \dfrac{\dfrac{49}{113}}{\dfrac{64}{113}}\Rightarrow \dfrac{49}{64}\]

$\therefore {{\cot }^{2}}\theta =\dfrac{49}{64}$

Hence, $\dfrac{\left( 1+\sin \theta  \right)\left( 1-\sin \theta  \right)}{\left( 1+\cos \theta  \right)\left( 1-\cos \theta  \right)}$ and ${{\cot }^{2}}\theta $ are same.

3. If \[3\cot A=4\], then show that  \[\dfrac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}={{\cos }^{2}}A-{{\sin }^{2}}A\].              

Ans: 

Given: \[3\cot A=4\]

Let us consider a triangle \[ABC\] in which  \[\angle B={{90}^{\circ }}\]

Then, \[3\cot A=4\] 

\[\Rightarrow \cot A=\dfrac{4}{3}\]

Let \[AB=4k\] and \[BC=3k\]

Using Pythagoras theorem, find $AC$

\[AC=\sqrt{{{(BC)}^{2}}+{{(AB)}^{2}}}\]

 \[\Rightarrow \sqrt{{{(3k)}^{2}}+{{(4k)}^{2}}}=\sqrt{16{{k}^{2}}+9{{k}^{2}}}\]

\[\Rightarrow \sqrt{25{{k}^{2}}}=5k\]

\[\therefore AC=5k\]

Now find the value of trigonometric ratios.

\[\sin A=\dfrac{BC}{AC}=\dfrac{3k}{5k}=\dfrac{3}{5}\]  , \[\cos A=\dfrac{AB}{AC}=\dfrac{4k}{5k}=\dfrac{4}{5}\] and \[\tan A=\dfrac{BC}{AB}=\dfrac{3k}{4k}=\dfrac{3}{4}\].

To prove: \[\dfrac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}={{\cos }^{2}}A-{{\sin }^{2}}A\]

Let us take left-hand side 

L.H.S \[=\dfrac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}\]

Substitute the value of $\tan A$.

\[\Rightarrow \dfrac{1-\dfrac{9}{16}}{1+\dfrac{9}{16}}=\dfrac{16}{25}-\dfrac{9}{25}\]

\[\Rightarrow \dfrac{7}{25}\]

\[\therefore \dfrac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}=\dfrac{7}{25}\]

Then,

R.H.S \[={{\cos }^{2}}A-{{\sin }^{2}}A\]

\[\Rightarrow {{\left( \dfrac{4}{5} \right)}^{2}}-{{\left( \dfrac{3}{5} \right)}^{2}}=\dfrac{16}{25}-\dfrac{9}{25}\]

\[\Rightarrow \dfrac{7}{25}\]

$\therefore {{\cos }^{2}}A-{{\sin }^{2}}A=\dfrac{7}{25}$.

It shows that \[\text{L}\text{.H}\text{.S=R}\text{.H}\text{.S}\]

\[\therefore \dfrac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}={{\cos }^{2}}A-{{\sin }^{2}}A\]

Hence proved.

4. In \[\Delta ABC\] Right Angles at \[B\], if \[A=\dfrac{1}{\sqrt{3}}\], then Find Value of the Following Equations:                                  

(i) \[\sin A\cos C+\cos A\sin C\]

Ans: 

Let us consider a triangle \[ABC\] in which  \[\angle B={{90}^{\circ }}\]

Let \[BC=k\]and \[AB=\sqrt{3}k\]

Then, using Pythagoras theorem find $AC$

\[AC=\sqrt{{{(BC)}^{2}}+{{(AB)}^{2}}}\]

\[\Rightarrow \sqrt{{{(k)}^{2}}+{{(\sqrt{3}k)}^{2}}}\Rightarrow \sqrt{{{k}^{2}}+3{{k}^{2}}}\Rightarrow \sqrt{4{{k}^{2}}}\Rightarrow 2k\]

\[\therefore AC=2k\]

Now find the value of trigonometric ratios.

\[\sin A=\dfrac{BC}{AC}=\dfrac{k}{2k}=\dfrac{1}{2}\] and \[\cos A=\dfrac{AB}{AC}=\dfrac{\sqrt{3}k}{2k}=\dfrac{\sqrt{3}}{2}\]

For \[\angle C\], adjacent \[=BC\], opposite \[=AB\], and hypotenuse \[=AC\]

\[\sin C=\dfrac{AB}{AC}=\dfrac{\sqrt{3}k}{2k}=\dfrac{\sqrt{3}}{2}\] and \[\cos A=\dfrac{BC}{AC}=\dfrac{k}{2k}=\dfrac{1}{2}\]

Now find the values of the following equations,

\[\sin A\cos C+\cos A\sin C\]

\[\Rightarrow \dfrac{1}{2}\times \dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\times \dfrac{\sqrt{3}}{2}\Rightarrow \dfrac{1}{4}+\dfrac{3}{4}\Rightarrow \dfrac{4}{4}\Rightarrow 1\]

\[\therefore \sin A\cos C+\cos A\sin C=1\]

(ii) \[\cos A\cos C-\sin A\sin C\]

Ans:

\[\Rightarrow \dfrac{\sqrt{3}}{2}\times \dfrac{1}{2}-\dfrac{1}{2}\times \dfrac{\sqrt{3}}{2}\Rightarrow \dfrac{\sqrt{3}}{4}-\dfrac{\sqrt{3}}{4}\Rightarrow 0\]

\[\therefore \cos A\cos C-\sin A\sin C=0\]

5. In \[\Delta PQR\], right angled at \[Q\],\[PR+QR=25cm\] and \[PQ=5cm\]. Determine the Values of  \[\sin P,\cos P\]  and \[\tan P\].                                                                      

Ans: 

Given: In \[\Delta PQR\], right angled at \[Q\]

And \[PR+QR=25cm\], \[PQ=5cm\]

Let us take \[QR=xcm\] and \[PR=(25-x)cm\]

By using Pythagoras theorem, find the value of $x$.

\[R{{P}^{2}}=R{{Q}^{2}}+Q{{P}^{2}}\]

\[\Rightarrow {{(25-x)}^{2}}={{(x)}^{2}}+{{(5)}^{2}}\Rightarrow 625-50x+{{x}^{2}}={{x}^{2}}+25\]

\[\Rightarrow -50x=-600\Rightarrow x=12\]

Hence, \[RQ=12cm\]and \[RP=25-12=13cm\]

Now, find the values of \[\sin P,\cos P\]  and \[\tan P\].    

\[\therefore \sin P=\dfrac{RQ}{RP}=\dfrac{12}{13}\], \[\cos P=\dfrac{PQ}{RP}=\dfrac{5}{13}\] and \[\tan P=\dfrac{RQ}{PQ}=\dfrac{12}{5}\].

6. If \[\tan (A+B)=\sqrt{3}\] and \[\tan (A-B)=\dfrac{1}{\sqrt{3}}\]; \[{{0}^{\circ }}<A+B\le {{90}^{\circ }}\]; \[A>B\]. Find \[A\]and \[B\].  

Ans: 

Given: $\tan \left( A+B \right)=\sqrt{3}$ and $\tan \left( A-B \right)=\dfrac{1}{\sqrt{3}}$.

We know that,$\tan 60{}^\circ =\sqrt{3}$ and $\tan 30{}^\circ =\dfrac{1}{\sqrt{3}}$.

Then,

$\tan \left( A+B \right)=\tan 60{}^\circ $

$\Rightarrow A+B=60{}^\circ $          …… (1)

$\tan \left( A-B \right)=\tan 30{}^\circ $

$\Rightarrow A-B=30{}^\circ $          …… (2)

Adding equation (1) and (2). We get,

$\begin{align} & A+B+A-B=60{}^\circ +30{}^\circ \\ & \Rightarrow 2A=90{}^\circ \Rightarrow A=45{}^\circ \\ \end{align}$

$\therefore A=45{}^\circ $

Put $A=45{}^\circ $ in equation (1).

$A+B=60{}^\circ $

$\Rightarrow 45{}^\circ +B=60{}^\circ \Rightarrow B=60{}^\circ -45{}^\circ \Rightarrow B=15{}^\circ $

$\therefore B=15{}^\circ $

Hence, $A=45{}^\circ $ and $B=15{}^\circ $.

7. Choose the Correct Option. Justify Your Choice:   

(i) \[9{{\sec }^{2}}A-9{{\tan }^{2}}A\]\[=\]

1. \[1\] 

2. \[9\] 

3. \[8\] 

4. \[0\]

Ans: (B) $9$

\[9{{\sec }^{2}}A-9{{\tan }^{2}}A\]

\[\Rightarrow 9({{\sec }^{2}}A-{{\tan }^{2}}A)\Rightarrow 9\times 1\Rightarrow 9\]

(ii) \[(1+\tan \theta +\sec \theta )(1+\cot \theta -\cos ec\theta )\]\[=\]

1. \[0\] 

2. \[1\] 

3. \[2\] 

4. none of these

Ans: (C) \[2\]

\[(1+\tan \theta +\sec \theta )(1+\cot \theta -\cos ec\theta )\]

\[\Rightarrow \left( 1+\dfrac{\sin \theta }{\cos \theta }+\dfrac{1}{\cos \theta } \right)\left( 1+\dfrac{\cos \theta }{\sin \theta }-\dfrac{1}{\sin \theta } \right)\Rightarrow \left( \dfrac{\cos \theta +\sin \theta +1}{\cos \theta } \right)\left( \dfrac{\sin \theta +\cos \theta -1}{\sin \theta } \right)\]

We know that, \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]

\[\Rightarrow \dfrac{{{(\cos \theta +\sin \theta )}^{2}}-{{(1)}^{2}}}{\cos \theta .\sin \theta }\Rightarrow \dfrac{{{\cos }^{2}}\theta +{{\sin }^{2}}\theta +2\cos \theta \sin \theta -1}{\cos \theta .\sin \theta }\]        

\[\Rightarrow \dfrac{1+2\cos \theta \sin \theta -1}{\cos \theta .\sin \theta }\Rightarrow \dfrac{2\cos \theta \sin \theta }{\cos \theta .\sin \theta }\Rightarrow 2\]

(iii) \[(\sec A+\tan A)(1-\sin A)\]\[=\]

1. \[\sec A\] 

2. \[\sin A\] 

3. \[\cos ecA\] 

4. \[\cos A\]

Ans: (D) \[\cos A\]

\[(\sec A+\tan A)(1-\sin A)\]

\[\Rightarrow \left( \dfrac{1}{\cos A}+\dfrac{\sin A}{\cos A} \right)(1-\sin A)\Rightarrow \left( \dfrac{1+\sin A}{\cos A} \right)(1-\sin A)\]

We know that, \[1-{{\sin }^{2}}A={{\cos }^{2}}A\]

\[\Rightarrow \dfrac{1-{{\sin }^{2}}A}{\cos A}\Rightarrow \dfrac{{{\cos }^{2}}A}{\cos A}\Rightarrow \cos A\]  

(iv) \[\dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}=\]

1. \[{{\sec }^{2}}A\] 

2. \[-1\] 

3. \[{{\cot }^{2}}A\] 

4. none of these

Ans: (D) ${{\tan }^{2}}A$

\[\dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}=\dfrac{{{\sec }^{2}}A-{{\tan }^{2}}A+{{\tan }^{2}}A}{\cos e{{c}^{2}}A-{{\cot }^{2}}A+{{\cot }^{2}}A}\]

\[\Rightarrow \dfrac{{{\sec }^{2}}A}{\cos e{{c}^{2}}A}\Rightarrow \dfrac{\dfac{rac{1}{{{\cos }^{2}}A}}{\dfrac{1}{{{\sin }^{2}}A}}\]

\[\Rightarrow \dfr{{\sin }^{2}}A}{{{\cos }^{2}}A}\Rightarrow {{\tan }^{2}}A\].

Long Answer Questions (4 Marks)

1. Express the Trigonometric Ratios \[\sin A,\sec A\] and \[\tan A\] in Terms of \[\cot A\].  

Ans: 

Find the value for \[\sin A\] in terms of \[\cot A\]

By using identity \[\cos e{{c}^{2}}A-{{\cot }^{2}}A=1\]

Then,

\[\Rightarrow {{\operatorname{cosec}}^{2}}A=1+{{\cot }^{2}}A\]

\[\Rightarrow \dfrac{1}{{{\sin }^{2}}A}=1+{{\cot }^{2}}A\]

\[\Rightarrow {{\sin }^{2}}A=\dfrac{1}{1+{{\cot }^{2}}A}\]

\[\Rightarrow \sin A=\dfrac{1}{\sqrt{1+{{\cot }^{2}}A}}\]

$\therefore \sin A=\dfrac{1}{\sqrt{1+{{\cot }^{2}}A}}$

Express the value of \[\sec A\] in terms of \[\cot A\]

By using identity \[{{\sec }^{2}}A-{{\tan }^{2}}A=1\]

Then,

\[\Rightarrow {{\sec }^{2}}A=1+{{\tan }^{2}}A\]

\[\Rightarrow {{\sec }^{2}}A=1+\dfrac{1}{{{\cot }^{2}}A}\]

\[\Rightarrow {{\sec }^{2}}A=\dfrac{1+{{\cot }^{2}}A}{{{\cot }^{2}}A}\]

\[\Rightarrow \sec A=\dfrac{\sqrt{1+{{\cot }^{2}}A}}{\cot A}\]

\[\therefore \sec A=\dfrac{\sqrt{1+{{\cot }^{2}}A}}{\cot A}\]

Express the value of \[\tan A\] in terms of \[\cot A\]

We know that, $\tan A=\dfrac{1}{\cot A}$

\[\therefore \tan A=\dfrac{1}{\cot A}\]

2. Write the Other Trigonometric Ratios of \[A\] in Terms of \[\sec A\].                 

Ans: 

Express the value of \[\sin A\] in terms of \[\sec A\]

By using identity, \[{{\sin }^{2}}A+{{\cos }^{2}}A=1\]

\[\Rightarrow {{\sin }^{2}}A=1-{{\cos }^{2}}A\Rightarrow 1-\dfrac{1}{{{\sec }^{2}}A}\Rightarrow \dfrac{{{\sec }^{2}}A-1}{{{\sec }^{2}}A}\]

\[\Rightarrow \sin A=\dfrac{\sqrt{{{\sec }^{2}}A-1}}{\sec A}\]

\[\therefore \sin A=\dfrac{\sqrt{{{\sec }^{2}}A-1}}{\sec A}\]

Express the value of \[\cos A\] in terms of \[\sec A\]

We know that, \[\cos A=\dfrac{1}{\sec A}\]

\[\therefore \cos A=\dfrac{1}{\sec A}\]

Express the value of \[\tan A\] in terms of \[\sec A\]

By using identity \[{{\sec }^{2}}A-{{\tan }^{2}}A=1\]

Then,

\[\Rightarrow {{\tan }^{2}}A={{\sec }^{2}}A-1\]

\[\Rightarrow \tan A=\sqrt{{{\sec }^{2}}A-1}\]

$\therefore \tan A=\sqrt{{{\sec }^{2}}A-1}$

Express the value of \[\cos ecA\] in terms of \[\sec A\]

We know that, $\operatorname{cosec}A=\dfrac{1}{\sin A}$

Then,

Substitute the value of \[\sin A\]

\[\Rightarrow \cos ecA=\dfrac{1}{\dfrac{\sqrt{{{\sec }^{2}}A-1}}{\sec A}}\Rightarrow \dfrac{\sec A}{\sqrt{{{\sec }^{2}}A-1}}\]

$\therefore \operatorname{cosec}A=\dfrac{\sec A}{\sqrt{{{\sec }^{2}}A-1}}$

Express the value of \[\cot A\] in terms of \[\sec A\]

We know that, \[\cot A=\dfrac{1}{\tan A}\]

Substitute the value of \[\tan A\]

\[\Rightarrow \cot A=\dfrac{1}{\sqrt{{{\sec }^{2}}A-1}}\]

$\therefore \cot A=\dfrac{1}{\sqrt{{{\sec }^{2}}A-1}}$

3. Evaluate the following equations:                                                                                                       

(i). \[\dfrac{{{\sin }^{2}}{{63}^{\circ }}+{{\sin }^{2}}{{27}^{\circ }}}{{{\cos }^{2}}{{17}^{\circ }}+{{\cos }^{2}}{{73}^{\circ }}}\]

Ans:

Given: \[\dfrac{{{\sin }^{2}}{{63}^{\circ }}+{{\sin }^{2}}{{27}^{\circ }}}{{{\cos }^{2}}{{17}^{\circ }}+{{\cos }^{2}}{{73}^{\circ }}}\]

We know that, \[\sin ({{90}^{\circ }}-\theta )=\cos \theta ,\,\cos ({{90}^{\circ }}-\theta )=\sin \theta \] and \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]

Then,

\[\Rightarrow \dfrac{{{\sin }^{2}}{{63}^{\circ }}+{{\sin }^{2}}({{90}^{\circ }}-{{63}^{\circ }})}{{{\cos }^{2}}({{90}^{\circ }}-{{73}^{\circ }})+{{\cos }^{2}}{{73}^{\circ }}}\]         

\[\Rightarrow \dfrac{{{\sin }^{2}}{{63}^{\circ }}+{{\cos }^{2}}{{63}^{\circ }}}{{{\sin }^{2}}{{73}^{\circ }}+{{\cos }^{2}}{{73}^{\circ }}}\]                   

\[\Rightarrow \dfrac{1}{1}\Rightarrow 1\]

$\therefore \dfrac{{{\sin }^{2}}{{63}^{\circ }}+{{\sin }^{2}}{{27}^{\circ }}}{{{\cos }^{2}}{{17}^{\circ }}+{{\cos }^{2}}{{73}^{\circ }}}=1$

(ii). \[\sin {{25}^{\circ }}\cos {{65}^{\circ }}+\cos {{25}^{\circ }}\sin {{65}^{\circ }}\]

Ans: 

Given: \[\sin {{25}^{\circ }}\cos {{65}^{\circ }}+\cos {{25}^{\circ }}\sin {{65}^{\circ }}\]

We know that, \[\sin ({{90}^{\circ }}-\theta )=\cos \theta ,\,\cos ({{90}^{\circ }}-\theta )=\sin \theta \] and \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]

\[\Rightarrow \sin {{25}^{\circ }}\cos ({{90}^{\circ }}-{{25}^{\circ }})+\cos {{25}^{\circ }}\sin ({{90}^{\circ }}-{{25}^{\circ }})\]

\[\Rightarrow \sin {{25}^{\circ }}.\sin {{25}^{\circ }}+\cos {{25}^{\circ }}.\cos {{25}^{\circ }}\]                   

\[\Rightarrow {{\sin }^{2}}{{25}^{\circ }}+{{\cos }^{2}}{{25}^{\circ }}=1\]      

\[\therefore \sin {{25}^{\circ }}\cos {{65}^{\circ }}+\cos {{25}^{\circ }}\sin {{65}^{\circ }}=1\]

4. Choose the Correct Option. Justify Your Choice:  

(i) \[9{{\sec }^{2}}A-9{{\tan }^{2}}A\]\[=\]

1. \[1\] 

2. \[9\] 

3. \[8\] 

4. \[0\]

Ans: (B) $9$

\[9{{\sec }^{2}}A-9{{\tan }^{2}}A\]

\[\Rightarrow 9({{\sec }^{2}}A-{{\tan }^{2}}A)\Rightarrow 9\times 1\Rightarrow 9\]

(ii) \[(1+\tan \theta +\sec \theta )(1+\cot \theta -\cos ec\theta )\]\[=\]

1. \[0\] 

2. \[1\] 

3. \[2\] 

4. none of these

Ans: (C) \[2\]

\[(1+\tan \theta +\sec \theta )(1+\cot \theta -\cos ec\theta )\]

\[\Rightarrow \left( 1+\dfrac{\sin \theta }{\cos \theta }+\dfrac{1}{\cos \theta } \right)\left( 1+\dfrac{\cos \theta }{\sin \theta }-\dfrac{1}{\sin \theta } \right)\Rightarrow \left( \dfrac{\cos \theta +\sin \theta +1}{\cos \theta } \right)\left( \dfrac{\sin \theta +\cos \theta -1}{\sin \theta } \right)\]

We know that, \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]

\[\Rightarrow \dfrac{{{(\cos \theta +\sin \theta )}^{2}}-{{(1)}^{2}}}{\cos \theta .\sin \theta }\Rightarrow \dfrac{{{\cos }^{2}}\theta +{{\sin }^{2}}\theta +2\cos \theta \sin \theta -1}{\cos \theta .\sin \theta }\]        

\[\Rightarrow \dfrac{1+2\cos \theta \sin \theta -1}{\cos \theta .\sin \theta }\Rightarrow \dfrac{2\cos \theta \sin \theta }{\cos \theta .\sin \theta }\Rightarrow 2\]

(iii) \[(\sec A+\tan A)(1-\sin A)\]\[=\]

1. \[\sec A\] 

2. \[\sin A\] 

3. \[\cos ecA\] 

4. \[\cos A\]

Ans: (D) \[\cos A\]

\[(\sec A+\tan A)(1-\sin A)\]

\[\Rightarrow \left( \dfrac{1}{\cos A}+\dfrac{\sin A}{\cos A} \right)(1-\sin A)\Rightarrow \left( \dfrac{1+\sin A}{\cos A} \right)(1-\sin A)\]

We know that, \[1-{{\sin }^{2}}A={{\cos }^{2}}A\]

\[\Rightarrow \dfrac{1-{{\sin }^{2}}A}{\cos A}\Rightarrow \dfrac{{{\cos }^{2}}A}{\cos A}\Rightarrow \cos A\] 

(iv) \[\dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}=\]

1. \[{{\sec }^{2}}A\] 

2. \[-1\] 

3. \[{{\cot }^{2}}A\] 

4. none of these

Ans: (D) ${{\tan }^{2}}A$

\[\dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}=\dfrac{{{\sec }^{2}}A-{{\tan }^{2}}A+{{\tan }^{2}}A}{\cos e{{c}^{2}}A-{{\cot }^{2}}A+{{\cot }^{2}}A}\]

\[\Rightarrow \dfrac{{{\sec }^{2}}A}{\cos e{{c}^{2}}A}\Rightarrow \dfrac{\dfrac{1}{{{\cos }^{2}}A}}{\dfrac{1}{{{\sin }^{2}}A}}\]

\[\Rightarrow \dfrac{{{\sin }^{2}}A}{{{\cos }^{2}}A}\Rightarrow {{\tan }^{2}}A\].

5. Prove the Following Identities, Where the Angles Involved are Acute Angles for Which the Expressions are Defined:                      

(i). ${{(\operatorname{cosec}\theta -\cot \theta )}^{2}}=\dfrac{1-\cos \theta }{1+\cos \theta }$

Ans:

Given: ${{(\operatorname{cosec}\theta -\cot \theta )}^{2}}=\dfrac{1-\cos \theta }{1+\cos \theta }$

We know that, \[{{(a-b)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\], $\operatorname{cosec}\theta =\dfrac{1}{\sin \theta }$ and $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$

Then, let us take left-hand side

\[\Rightarrow {{(\operatorname{cosec}\theta -\cot \theta )}^{2}}={{\operatorname{cosec}}^{2}}\theta +{{\cot }^{2}}\theta -2\operatorname{cosec}\theta \cot \theta \] 

\[\Rightarrow \dfrac{1}{{{\sin }^{2}}\theta }+\dfrac{{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta }-2\times \dfrac{1}{\sin \theta }.\dfrac{\cos \theta }{\sin \theta }\Rightarrow \dfrac{1+{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta }-\dfrac{2\cos \theta }{{{\sin }^{2}}\theta }\]

\[\Rightarrow \dfrac{1+{{\cos }^{2}}\theta -2\cos \theta }{{{\sin }^{2}}\theta }\Rightarrow \dfrac{{{(1-\cos \theta )}^{2}}}{{{\sin }^{2}}\theta }\]

\[\Rightarrow \dfrac{(1-\cos \theta )(1-\cos \theta )}{1-{{\cos }^{2}}\theta }\Rightarrow \dfrac{(1-\cos \theta )(1-\cos \theta )}{(1+\cos \theta )(1-\cos \theta )}\]

\[\Rightarrow \dfrac{1-\cos \theta }{1+\cos \theta }=\text{R}\text{.H}\text{.S}\]

$\therefore {{(\operatorname{cosec}\theta -\cot \theta )}^{2}}=\dfrac{1-\cos \theta }{1+\cos \theta }$

Hence proved.

(ii) \[\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=2\sec A\]

Ans:

Given: \[\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=2\sec A\]

We know that, \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]

Then, let us take left-hand side

\[\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}\Rightarrow \dfrac{{{\cos }^{2}}A+1+{{\sin }^{2}}A+2\sin A}{(1+\sin A)\cos A}\]

\[\Rightarrow \dfrac{{{\cos }^{2}}A+{{\sin }^{2}}A+1+2\sin A}{(1+\sin A)\cos A}\]      

\[\Rightarrow \dfrac{1+1+2\sin A}{(1+\sin A)\cos A}\]

\[\Rightarrow \dfrac{2+2\sin A}{(1+\sin A)\cos A}\]

\[\Rightarrow \dfrac{2(1+\sin A)}{(1+\sin A)\cos A}\]

\[\Rightarrow \dfrac{2}{\cos A}\Rightarrow 2\sec A=\text{R}\text{.H}\text{.S}\]

$\therefore \dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=2\sec A$

Hence proved.

(iii). \[\dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=1+\sec \theta \operatorname{cosec}\theta \]

Ans:

Given: \[\dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=1+\sec \theta \operatorname{cosec}\theta \]

We know that, \[{{a}^{3}}-{{b}^{3}}=(a-b)({{a}^{2}}+{{b}^{2}}+ab)\] and \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]

Then, let us take L.H.S

\[\dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=\dfrac{\dfrac{\sin \theta }{\cos \theta }}{1-\dfrac{\cos \theta }{\sin \theta }}+\dfrac{\dfrac{\cos \theta }{\sin \theta }}{1-\dfrac{\sin \theta }{\cos \theta }}\]

\[\Rightarrow \dfrac{\sin \theta }{\cos \theta }\times \dfrac{\sin \theta }{\sin \theta -\cos \theta }+\dfrac{\cos \theta }{\sin \theta }\times \dfrac{\cos \theta }{\cos \theta -\sin \theta }\]

\[\Rightarrow \dfrac{{{\sin }^{2}}\theta }{\cos \theta (\sin \theta -\cos \theta )}-\dfrac{{{\cos }^{2}}\theta }{\sin \theta (\sin \theta -\cos \theta )}\]

\[\Rightarrow \dfrac{{{\sin }^{3}}\theta -{{\cos }^{3}}\theta }{\sin \theta \cos \theta (\sin \theta -\cos \theta )}\]                     

\[\Rightarrow \dfrac{1+\sin \theta \cos \theta }{\sin \theta \cos \theta }\]

\[\Rightarrow \dfrac{1}{\sin \theta \cos \theta }+1\]

\[\Rightarrow 1+\dfrac{1}{\sin \theta \cos \theta }\Rightarrow 1+\sec \theta \cos ec\theta \] = R.H.S

$\therefore \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=1+\sec \theta \operatorname{cosec}\theta $

Hence proved.

(iv) \[\dfrac{1+\sec A}{\sec A}=\dfrac{{{\sin }^{2}}A}{1-\cos A}\]

Ans:

Given: \[\dfrac{1+\sec A}{\sec A}=\dfrac{{{\sin }^{2}}A}{1-\cos A}\]

Then, let us take L.H.S 

\[\dfrac{1+\sec A}{\sec A}=\dfrac{1+\dfrac{1}{\cos A}}{\dfrac{1}{\cos A}}\]

\[\Rightarrow \dfrac{\cos A+1}{\cos A}\times \dfrac{\cos A}{1}\Rightarrow 1+\cos A\]

\[\Rightarrow 1+\cos A\times \dfrac{1-\cos A}{1-\cos A}\Rightarrow \dfrac{1-{{\cos }^{2}}A}{1-\cos A}\]

\[\Rightarrow \dfrac{{{\sin }^{2}}A}{1-\cos A}=\text{R}\text{.H}\text{.S}\]

$\therefore \dfrac{1+\sec A}{\sec A}=\dfrac{{{\sin }^{2}}A}{1-\cos A}$

Hence proved.

(v) \[\dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\cos ecA+\cot A\], using the identity \[\cos e{{c}^{2}}A=1+{{\cot }^{2}}A\]

Ans:

Given: \[\dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\cos ecA+\cot A\]

We know that, \[\cos e{{c}^{2}}A=1+{{\cot }^{2}}A\]

Then, let us take L.H.S 

\[\dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}\]

Dividing all terms by \[\sin A\]

\[\Rightarrow \dfrac{\cot A-1+\cos ecA}{\cot A+1-\cos ecA}\]

\[\Rightarrow \dfrac{\cot A+\cos ecA-1}{\cot A-\cos ecA+1}\]

\[\Rightarrow \dfrac{(\cot A+\cos ecA)-(\cos e{{c}^{2}}A-{{\cot }^{2}}A)}{(1+\cot A-\cos ecA)}\]

\[\Rightarrow \dfrac{(\cot A+\cos ecA)+({{\cot }^{2}}A-\cos e{{c}^{2}}A)}{(1+\cot A-\cos ecA)}\]

\[\Rightarrow \dfrac{(\cot A+\cos ecA)(1+\cot A-\cos ecA)}{(1+\cot A-\cos ecA)}\]

\[\Rightarrow \cot A+\cos ecA=\text{R}\text{.H}\text{.S}\]

$\therefore \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\cos ecA+\cot A$

Hence proved.

(vi) \[\sqrt{\dfrac{1+\sin A}{1-\sin A}}=\sec A+\tan A\]

Ans:

Given: \[\sqrt{\dfrac{1+\sin A}{1-\sin A}}=\sec A+\tan A\]

We know that, \[1-{{\sin }^{2}}\theta ={{\cos }^{2}}\theta \] and \[(a+b)(a-b)={{a}^{2}}-{{b}^{2}}\]

Then, let us take L.H.S 

\[\sqrt{\dfrac{1+\sin A}{1-\sin A}}\]

Let us take conjugate of the term. Then,

\[\Rightarrow \sqrt{\dfrac{1+\sin A}{1-\sin A}}\times \sqrt{\dfrac{1+\sin A}{1+\sin A}}\]

\[\Rightarrow \sqrt{\dfrac{{{(1+\sin A)}^{2}}}{1-{{\sin }^{2}}A}}\]

\[\Rightarrow \sqrt{\dfrac{{{\left( 1+\sin A \right)}^{2}}}{{{\operatorname{Cos}}^{2}}A}}\]

\[\Rightarrow \dfrac{1+\sin A}{\cos A}\Rightarrow \dfrac{1}{\cos A}+\dfrac{\sin A}{\cos A}\]                           

\[\Rightarrow \sec A+\tan A=\text{R}\text{.H}\text{.S}\]           

$\therefore \sqrt{\dfrac{1+\sin A}{1-\sin A}}=\sec A+\tan A$       

Hence proved.        

(vii) \[\dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2{{\cos }^{3}}\theta -\cos \theta }=\tan \theta \]

Ans:

Given: \[\dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2{{\cos }^{3}}\theta -\cos \theta }=\tan \theta \]

We know that, \[1-{{\sin }^{2}}\theta ={{\cos }^{2}}\theta \]

Then, let us take L.H.S

\[\dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2{{\cos }^{3}}\theta -\cos \theta }=\dfrac{\sin \theta (1-2{{\sin }^{2}}\theta )}{\cos \theta (2{{\cos }^{2}}\theta -1)}\]

\[\Rightarrow \dfrac{\sin \theta (1-2{{\sin }^{2}}\theta )}{\cos \theta \left[ 2(1-{{\sin }^{2}}\theta )-1 \right]}\]

\[\Rightarrow \dfrac{\sin \theta (1-2{{\sin }^{2}}\theta )}{\cos \theta (2-2{{\sin }^{2}}\theta -1)}\]

\[\Rightarrow \dfrac{\sin \theta (1-2{{\operatorname{Sin}}^{2}}\theta )}{\cos \theta (1-2{{\operatorname{Sin}}^{2}}\theta )}\]

\[\Rightarrow \dfrac{\sin \theta }{\cos \theta }=\tan \theta =\text{R}\text{.H}\text{.S}\]

$\therefore \dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2{{\cos }^{3}}\theta -\cos \theta }=\tan \theta $

Hence proved.

(viii) \[{{(\sin A+\cos ecA)}^{2}}+{{(\cos A+\sec A)}^{2}}=7+{{\tan }^{2}}A+{{\cot }^{2}}A\]

Ans:

Given: \[{{(\sin A+\cos ecA)}^{2}}+{{(\cos A+\sec A)}^{2}}=7+{{\tan }^{2}}A+{{\cot }^{2}}A\]

We know that, \[\cos e{{c}^{2}}\theta =1+{{\cot }^{2}}\theta \] and \[{{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta \]

Then, let us take L.H.S 

\[{{(\sin A+\cos ecA)}^{2}}+{{(\cos A+\sec A)}^{2}}={{\left( \sin A+\dfrac{1}{\sin A} \right)}^{2}}+{{\left( \cos A+\dfrac{1}{\cos A} \right)}^{2}}\]

\[={{\sin }^{2}}A+\dfrac{1}{{{\sin }^{2}}A}+2\sin A.\dfrac{1}{\sin A}+{{\cos }^{2}}A+\dfrac{1}{{{\cos }^{2}}A}+2\cos A.\dfrac{1}{\cos A}\]

\[=2+2+{{\sin }^{2}}A+{{\cos }^{2}}A+\dfrac{1}{{{\sin }^{2}}A}+\dfrac{1}{{{\cos }^{2}}A}\]

\[=4+1+\dfrac{1}{{{\sin }^{2}}A}+\dfrac{1}{{{\cos }^{2}}A}\]

\[=5+\cos e{{c}^{2}}A+{{\sec }^{2}}A\]

\[=5+1+{{\cot }^{2}}A+1+{{\tan }^{2}}A\]

\[=7+{{\tan }^{2}}A+{{\cot }^{2}}A=\text{R}\text{.H}\text{.S}\]

$\therefore {{(\sin A+\cos ecA)}^{2}}+{{(\cos A+\sec A)}^{2}}=7+{{\tan }^{2}}A+{{\cot }^{2}}A$

Hence proved.

(ix) \[(\cos ecA-\sin A)(\sec A-\cos A)=\dfrac{1}{\tan A+\cot A}\]

Ans:

Given: \[(\cos ecA-\sin A)(\sec A-\cos A)=\dfrac{1}{\tan A+\cot A}\]

We know that, \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]

Then, let us take L.H.S 

\[(\cos ecA-\sin A)(\sec A-\cos A)=\left( \dfrac{1}{\sin A}-\sin A \right)\left( \dfrac{1}{\cos A}-\cos A \right)\]

\[=\left( \dfrac{1-{{\sin }^{2}}A}{\sin A} \right)\left( \dfrac{1-{{\cos }^{2}}A}{\cos A} \right)\]

\[=\dfrac{{{\cos }^{2}}A}{\sin A}\times \dfrac{{{\sin }^{2}}A}{\cos A}\]

\[=\sin A.\cos A\]

\[=\dfrac{\sin A.\cos A}{{{\sin }^{2}}A+{{\cos }^{2}}A}\]

Dividing all the terms by \[\sin A.\cos A\]

\[=\dfrac{\dfrac{\sin A.\cos A}{\sin A.\cos A}}{\dfrac{{{\sin }^{2}}A}{\sin A.\cos A}+\dfrac{{{\cos }^{2}}A}{\sin A.\cos A}}\]

\[=\dfrac{1}{\dfrac{\sin A}{\cos A}+\dfrac{\cos A}{\sin A}}\]

\[=\dfrac{1}{\tan A+\cot A}=\text{R}\text{.H}\text{.S}\]

$\therefore (\cos ecA-\sin A)(\sec A-\cos A)=\dfrac{1}{\tan A+\cot A}$

Hence proved.

(x) \[\left( \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A} \right)={{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}={{\tan }^{2}}A\]

Ans:

Given:\[\left( \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A} \right)={{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}={{\tan }^{2}}A\]

We know that, \[1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta \] and \[1+{{\cot }^{2}}\theta =\cos e{{c}^{2}}A\]

Then, let us take L.H.S

\[\left( \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A} \right)=\dfrac{{{\sec }^{2}}A}{\cos e{{c}^{2}}A}\]

\[=\dfrac{1}{{{\cos }^{2}}A}\times \dfrac{{{\sin }^{2}}A}{1}\]

\[={{\tan }^{2}}A=\text{R}\text{.H}\text{.S}\]

Now, prove the Middle side 

\[{{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}={{\left( \dfrac{1-\tan A}{1-\dfrac{1}{\tan A}} \right)}^{2}}\]

\[={{\left( \dfrac{1-\tan A}{\dfrac{\tan A-1}{\tan A}} \right)}^{2}}\]

\[=\left( \dfrac{1-\tan A}{\dfrac{-(1-\tan A)}{\tan A}} \right)\]

\[={{(-\tan A)}^{2}}\]

\[={{\tan }^{2}}A=\text{R}\text{.H}\text{.S}\]

$\therefore \left( \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A} \right)={{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}={{\tan }^{2}}A$

Hence proved.

6. Use Euclid’s Division Algorithm to Find the \[\text{HCF}\] of:                                     

(i) \[\mathbf{135}\] and \[\mathbf{225}\]

Ans:  

Given: \[135\] and \[225\]

We have \[225>135\],

So, we have to apply the division lemma to \[225\] and \[135\] to obtain

\[225=135\times 1+90\]

Here remainder\[90\ne 0\], again we are applying the division lemma to \[135\] and \[90\] to obtain 

\[135=90\times 1+45\]

Again, the remainder\[45\ne 0\], then apply the division lemma to obtain 

\[90=2\times 45+0\]

Since now we got remainder as zero. Here, the process get stops. 

The divisor at this stage is \[45\]

Therefore, the HCF of \[135\] and \[225\] is \[45\].

(ii) \[\mathbf{196}\] and\[~\mathbf{38220}\] 

Ans:

Given: \[196\] and \[38220\]

We have \[38220>196\], 

So, we have to apply the division lemma to \[38220\] and \[196\] to obtain 

\[38220=196\times 195+0\]

Since we get the remainder as zero, the process stops here. 

The divisor at this stage is \[196\], 

Therefore, HCF of \[196\] and \[38220\] is \[196\].  

(iii) \[~\mathbf{867}\] and \[\mathbf{255}\]

Ans:

Given: \[867\] and \[255\]

We have 867 > 255, 

So, we have to apply the division lemma to \[867\]and \[255\] to obtain 

\[867=255\times 3+102\]

Here remainder \[102\ne 0\], again apply the division lemma to\[255\] and \[102\] to obtain 

\[255=102\times 2\text{ }51\]

Again, remainder \[51\ne 0\], again apply the division lemma to \[102\] and \[51\] to obtain 

\[102=51\times 2+0\]

Since we get the remainder as zero, the process stops here. 

The divisor at this stage is \[51\], 

Therefore, HCF of \[867\] and \[255\] is \[51\].

7. Evaluate the Following Equations:                                                                                                              

i. \[\sin {{60}^{\circ }}\cos {{30}^{\circ }}+\sin {{30}^{\circ }}\cos {{60}^{\circ }}\]

Ans: 

Given: \[\sin {{60}^{\circ }}\cos {{30}^{\circ }}+\sin {{30}^{\circ }}\cos {{60}^{\circ }}\]

We know that, $\sin 60{}^\circ =\dfrac{\sqrt{3}}{2},\,\sin 30{}^\circ =\dfrac{1}{2}$ and $\cos 60{}^\circ =\dfrac{1}{2},\,\cos 30{}^\circ =\dfrac{\sqrt{3}}{2}$

Then,

\[\sin {{60}^{\circ }}\cos {{30}^{\circ }}+\sin {{30}^{\circ }}\cos {{60}^{\circ }}\]

\[=\dfrac{\sqrt{3}}{2}\times \dfrac{\sqrt{3}}{2}+\dfrac{1}{2}\times \dfrac{1}{2}\]

\[=\dfrac{3}{4}+\dfrac{1}{4}\]

\[=\dfrac{4}{4}=1\]

\[\therefore \sin {{60}^{\circ }}\cos {{30}^{\circ }}+\sin {{30}^{\circ }}\cos {{60}^{\circ }}=1\]

ii. \[2{{\tan }^{2}}{{45}^{\circ }}+{{\cos }^{2}}{{30}^{\circ }}-{{\sin }^{2}}{{60}^{\circ }}\]

Ans:

Given: \[2{{\tan }^{2}}{{45}^{\circ }}+{{\cos }^{2}}{{30}^{\circ }}-{{\sin }^{2}}{{60}^{\circ }}\]

We know that, $\tan 45{}^\circ =1$, $\sin 60{}^\circ =\dfrac{\sqrt{3}}{2}$ and $\cos 30{}^\circ =\dfrac{\sqrt{3}}{2}$

Then, \[2{{\tan }^{2}}{{45}^{\circ }}+{{\cos }^{2}}{{30}^{\circ }}-{{\sin }^{2}}{{60}^{\circ }}\]

\[=2{{(1)}^{2}}+{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}-{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}\]

\[=2+\dfrac{3}{4}-\dfrac{3}{4}=2\]

$\therefore 2{{\tan }^{2}}{{45}^{\circ }}+{{\cos }^{2}}{{30}^{\circ }}-{{\sin }^{2}}{{60}^{\circ }}=2$

iii. \[\dfrac{\cos {{45}^{\circ }}}{\sec {{30}^{\circ }}+\cos ec{{30}^{\circ }}}\]

Ans:

Given: \[\dfrac{\cos {{45}^{\circ }}}{\sec {{30}^{\circ }}+\cos ec{{30}^{\circ }}}\]

We know that, $\cos 45{}^\circ =\dfrac{1}{\sqrt{2}},\,\sec 30{}^\circ =\dfrac{2}{\sqrt{3}}$ and $\operatorname{cosec}30{}^\circ =2$

\[=\dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{2}{\sqrt{3}}+2}\]

\[=\dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{2+2\sqrt{3}}{\sqrt{3}}}\]

\[=\dfrac{1}{\sqrt{2}}\times \dfrac{\sqrt{3}}{2+2\sqrt{3}}\]

\[=\dfrac{\sqrt{3}}{\sqrt{2}\times 2(\sqrt{3}+1)}\]

\[=\dfrac{\sqrt{3}}{\sqrt{2}\times 2(\sqrt{3}+1)}\times \dfrac{\sqrt{3}-1}{\sqrt{3}-1}\]

\[=\dfrac{\sqrt{3}\left( \sqrt{3}-1 \right)}{\sqrt{2}\times 2(3-1)}\]

\[=\dfrac{\sqrt{3}(\sqrt{3}-1)}{4\sqrt{2}}\times \dfrac{\sqrt{2}}{\sqrt{2}}\]\[\]

\[=\dfrac{3\sqrt{2}-\sqrt{6}}{8}\]

$\therefore \dfrac{\cos {{45}^{\circ }}}{\sec {{30}^{\circ }}+\cos ec{{30}^{\circ }}}=\dfrac{3\sqrt{2}-\sqrt{6}}{8}$

(iv) \[\dfrac{\sin {{30}^{\circ }}+\tan {{45}^{\circ }}-\cos ec{{60}^{\circ }}}{\sec {{30}^{\circ }}+\cos {{60}^{\circ }}+\cot {{45}^{\circ }}}\]

Ans:

Given: \[\dfrac{\sin {{30}^{\circ }}+\tan {{45}^{\circ }}-\cos ec{{60}^{\circ }}}{\sec {{30}^{\circ }}+\cos {{60}^{\circ }}+\cot {{45}^{\circ }}}\]

We know that,  $\sin 30{}^\circ =\dfrac{1}{2},\,\tan 45{}^\circ =1,\,\operatorname{cosec}60{}^\circ =\dfrac{2}{\sqrt{3}},\,\sec 30{}^\circ =\dfrac{2}{\sqrt{3}},\,\cos 60{}^\circ =\dfrac{1}{2}$ and $\cot 45{}^\circ =1$.

Then, \[\dfrac{\sin {{30}^{\circ }}+\tan {{45}^{\circ }}-\cos ec{{60}^{\circ }}}{\sec {{30}^{\circ }}+\cos {{60}^{\circ }}+\cot {{45}^{\circ }}}\]

$=\dfrac{{1}/{2}\;+1-{2}/{\sqrt{3}}\;}{{2}/{\sqrt{3}}\;+{1}/{2}\;+1}$

\[=\dfrac{\dfrac{\sqrt{3}+2\sqrt{3}-4}{2\sqrt{3}}}{\dfrac{4+\sqrt{3}+2\sqrt{3}}{2\sqrt{3}}}\]

\[=\dfrac{3\sqrt{3}-4}{3\sqrt{3}+4}\]

\[=\dfrac{3\sqrt{3}-4}{3\sqrt{3}+4}\times \dfrac{3\sqrt{3}-4}{3\sqrt{3}-4}\]

\[=\dfrac{27+16-24\sqrt{3}}{27-16}\]

\[=\dfrac{43-24\sqrt{3}}{11}\]

$\therefore \dfrac{\sin {{30}^{\circ }}+\tan {{45}^{\circ }}-\cos ec{{60}^{\circ }}}{\sec {{30}^{\circ }}+\cos {{60}^{\circ }}+\cot {{45}^{\circ }}}=\dfrac{43-24\sqrt{3}}{11}$

(v) \[\dfrac{5{{\cos }^{2}}{{60}^{\circ }}+4{{\sin }^{2}}{{30}^{\circ }}-{{\tan }^{2}}{{45}^{\circ }}}{{{\sin }^{2}}{{30}^{\circ }}+{{\cos }^{2}}{{30}^{\circ }}}\]

Ans:

Given: \[\dfrac{5{{\cos }^{2}}{{60}^{\circ }}+4{{\sin }^{2}}{{30}^{\circ }}-{{\tan }^{2}}{{45}^{\circ }}}{{{\sin }^{2}}{{30}^{\circ }}+{{\cos }^{2}}{{30}^{\circ }}}\]. Then,

\[=\dfrac{5{{\left( \dfrac{1}{2} \right)}^{2}}+{{\left( \dfrac{3}{\sqrt{3}} \right)}^{2}}-{{(1)}^{2}}}{{{\left( \dfrac{1}{2} \right)}^{2}}+{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}}\]

\[=\dfrac{5\times \dfrac{1}{4}+4\times \dfrac{4}{3}-1}{\dfrac{1}{4}+\dfrac{3}{4}}\]

\[=\dfrac{\dfrac{1}{12}\times 67}{\dfrac{4}{4}}\]

\[=\dfrac{67}{12}\]

$\therefore \dfrac{5{{\cos }^{2}}{{60}^{\circ }}+4{{\sin }^{2}}{{30}^{\circ }}-{{\tan }^{2}}{{45}^{\circ }}}{{{\sin }^{2}}{{30}^{\circ }}+{{\cos }^{2}}{{30}^{\circ }}}=\dfrac{67}{12}$

8. Prove the Following Identities, Where the Angles Involved are Acute Angles for Which the Expressions are Defined:                                      

(i) ${{(\operatorname{cosec}\theta -\cot \theta )}^{2}}=\dfrac{1-\cos \theta  }{1+\cos \theta }$

Ans:

Given: ${{(\operatorname{cosec}\theta -\cot \theta )}^{2}}=\dfrac{1-\cos \theta }{1+\cos \theta }$

We know that, \[{{(a-b)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\], $\operatorname{cosec}\theta =\dfrac{1}{\sin \theta }$ and $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$

Then, let us take left-hand side

\[\Rightarrow {{(\operatorname{cosec}\theta -\cot \theta )}^{2}}={{\operatorname{cosec}}^{2}}\theta +{{\cot }^{2}}\theta -2\operatorname{cosec}\theta \cot \theta \] 

\[\Rightarrow \dfrac{1}{{{\sin }^{2}}\theta }+\dfrac{{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta }-2\times \dfrac{1}{\sin \theta }.\dfrac{\cos \theta }{\sin \theta }\Rightarrow \dfrac{1+{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta }-\dfrac{2\cos \theta }{{{\sin }^{2}}\theta }\]

\[\Rightarrow \dfrac{1+{{\cos }^{2}}\theta -2\cos \theta }{{{\sin }^{2}}\theta }\Rightarrow \dfrac{{{(1-\cos \theta )}^{2}}}{{{\sin }^{2}}\theta }\]

\[\Rightarrow \dfrac{(1-\cos \theta )(1-\cos \theta )}{1-{{\cos }^{2}}\theta }\Rightarrow \dfrac{(1-\cos \theta )(1-\cos \theta )}{(1+\cos \theta )(1-\cos \theta )}\]

\[\Rightarrow \dfrac{1-\cos \theta }{1+\cos \theta }=\text{R}\text{.H}\text{.S}\]

$\therefore {{(\operatorname{cosec}\theta -\cot \theta )}^{2}}=\dfrac{1-\cos \theta }{1+\cos \theta }$

Hence proved.

(ii) \[\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=2\sec A\]

Ans:

Given: \[\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=2\sec A\]

We know that, \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]

Then, let us take left-hand side

\[\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}\Rightarrow \dfrac{{{\cos }^{2}}A+1+{{\sin }^{2}}A+2\sin A}{(1+\sin A)\cos A}\]

\[\Rightarrow \dfrac{{{\cos }^{2}}A+{{\sin }^{2}}A+1+2\sin A}{(1+\sin A)\cos A}\]      

\[\Rightarrow \dfrac{1+1+2\sin A}{(1+\sin A)\cos A}\]

\[\Rightarrow \dfrac{2+2\sin A}{(1+\sin A)\cos A}\]

\[\Rightarrow \dfrac{2(1+\sin A)}{(1+\sin A)\cos A}\]

\[\Rightarrow \dfrac{2}{\cos A}\Rightarrow 2\sec A=\text{R}\text{.H}\text{.S}\]

$\therefore \dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=2\sec A$

Hence proved.

(iii) \[\dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=1+\sec \theta \operatorname{cosec}\theta \]

Ans:

Given: \[\dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=1+\sec \theta \operatorname{cosec}\theta \]

We know that, \[{{a}^{3}}-{{b}^{3}}=(a-b)({{a}^{2}}+{{b}^{2}}+ab)\] and \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]

Then, let us take L.H.S

\[\dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=\dfrac{\dfrac{\sin \theta }{\cos \theta }}{1-\dfrac{\cos \theta }{\sin \theta }}+\dfrac{\dfrac{\cos \theta }{\sin \theta }}{1-\dfrac{\sin \theta }{\cos \theta }}\]

\[\Rightarrow \dfrac{\sin \theta }{\cos \theta }\times \dfrac{\sin \theta }{\sin \theta -\cos \theta }+\dfrac{\cos \theta }{\sin \theta }\times \dfrac{\cos \theta }{\cos \theta -\sin \theta }\]

\[\Rightarrow \dfrac{{{\sin }^{2}}\theta }{\cos \theta (\sin \theta -\cos \theta )}-\dfrac{{{\cos }^{2}}\theta }{\sin \theta (\sin \theta -\cos \theta )}\]

\[\Rightarrow \dfrac{{{\sin }^{3}}\theta -{{\cos }^{3}}\theta }{\sin \theta \cos \theta (\sin \theta -\cos \theta )}\]                     

\[\Rightarrow \dfrac{1+\sin \theta \cos \theta }{\sin \theta \cos \theta }\]

\[\Rightarrow \dfrac{1}{\sin \theta \cos \theta }+1\]

\[\Rightarrow 1+\dfrac{1}{\sin \theta \cos \theta }\Rightarrow 1+\sec \theta \cos ec\theta \] = R.H.S

$\therefore \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=1+\sec \theta \operatorname{cosec}\theta $

Hence proved.

(iv) \[\dfrac{1+\sec A}{\sec A}=\dfrac{{{\sin }^{2}}A}{1-\cos A}\]

Ans:

Given: \[\dfrac{1+\sec A}{\sec A}=\dfrac{{{\sin }^{2}}A}{1-\cos A}\]

Then, let us take L.H.S 

\[\dfrac{1+\sec A}{\sec A}=\dfrac{1+\dfrac{1}{\cos A}}{\dfrac{1}{\cos A}}\]

\[\Rightarrow \dfrac{\cos A+1}{\cos A}\times \dfrac{\cos A}{1}\Rightarrow 1+\cos A\]

\[\Rightarrow 1+\cos A\times \dfrac{1-\cos A}{1-\cos A}\Rightarrow \dfrac{1-{{\cos }^{2}}A}{1-\cos A}\]

\[\Rightarrow \dfrac{{{\sin }^{2}}A}{1-\cos A}=\text{R}\text{.H}\text{.S}\]

$\therefore \dfrac{1+\sec A}{\sec A}=\dfrac{{{\sin }^{2}}A}{1-\cos A}$

Hence proved.

(v) \[\dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\cos ecA+\cot A\], using the identity \[\cos e{{c}^{2}}A=1+{{\cot }^{2}}A\]

Ans:

Given: \[\dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\cos ecA+\cot A\]

We know that, \[\cos e{{c}^{2}}A=1+{{\cot }^{2}}A\]

Then, let us take L.H.S 

\[\dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}\]

Dividing all terms by \[\sin A\]

\[\Rightarrow \dfrac{\cot A-1+\cos ecA}{\cot A+1-\cos ecA}\]

\[\Rightarrow \dfrac{\cot A+\cos ecA-1}{\cot A-\cos ecA+1}\]

\[\Rightarrow \dfrac{(\cot A+\cos ecA)-(\cos e{{c}^{2}}A-{{\cot }^{2}}A)}{(1+\cot A-\cos ecA)}\]

\[\Rightarrow \dfrac{(\cot A+\cos ecA)+({{\cot }^{2}}A-\cos e{{c}^{2}}A)}{(1+\cot A-\cos ecA)}\]

\[\Rightarrow \dfrac{(\cot A+\cos ecA)(1+\cot A-\cos ecA)}{(1+\cot A-\cos ecA)}\]

\[\Rightarrow \cot A+\cos ecA=\text{R}\text{.H}\text{.S}\]

$\therefore \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\cos ecA+\cot A$

Hence proved.

(vi) \[\sqrt{\dfrac{1+\sin A}{1-\sin A}}=\sec A+\tan A\]

Ans:

Given: \[\sqrt{\dfrac{1+\sin A}{1-\sin A}}=\sec A+\tan A\]

We know that, \[1-{{\sin }^{2}}\theta ={{\cos }^{2}}\theta \] and \[(a+b)(a-b)={{a}^{2}}-{{b}^{2}}\]

Then, let us take L.H.S 

\[\sqrt{\dfrac{1+\sin A}{1-\sin A}}\]

Let us take conjugate of the term. Then,

\[\Rightarrow \sqrt{\dfrac{1+\sin A}{1-\sin A}}\times \sqrt{\dfrac{1+\sin A}{1+\sin A}}\]

\[\Rightarrow \sqrt{\dfrac{{{(1+\sin A)}^{2}}}{1-{{\sin }^{2}}A}}\]

\[\Rightarrow \sqrt{\dfrac{{{\left( 1+\sin A \right)}^{2}}}{{{\operatorname{Cos}}^{2}}A}}\]

\[\Rightarrow \dfrac{1+\sin A}{\cos A}\Rightarrow \dfrac{1}{\cos A}+\dfrac{\sin A}{\cos A}\]                           

\[\Rightarrow \sec A+\tan A=\text{R}\text{.H}\text{.S}\]           

$\therefore \sqrt{\dfrac{1+\sin A}{1-\sin A}}=\sec A+\tan A$       

Hence proved.        

(vii) \[\dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2{{\cos }^{3}}\theta -\cos \theta }=\tan \theta \]

Ans:

Given: \[\dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2{{\cos }^{3}}\theta -\cos \theta }=\tan \theta \]

We know that, \[1-{{\sin }^{2}}\theta ={{\cos }^{2}}\theta \]

Then, let us take L.H.S

\[\dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2{{\cos }^{3}}\theta -\cos \theta }=\dfrac{\sin \theta (1-2{{\sin }^{2}}\theta )}{\cos \theta (2{{\cos }^{2}}\theta -1)}\]

\[\Rightarrow \dfrac{\sin \theta (1-2{{\sin }^{2}}\theta )}{\cos \theta \left[ 2(1-{{\sin }^{2}}\theta )-1 \right]}\]

\[\Rightarrow \dfrac{\sin \theta (1-2{{\sin }^{2}}\theta )}{\cos \theta (2-2{{\sin }^{2}}\theta -1)}\]

\[\Rightarrow \dfrac{\sin \theta (1-2{{\operatorname{Sin}}^{2}}\theta )}{\cos \theta (1-2{{\operatorname{Sin}}^{2}}\theta )}\]

\[\Rightarrow \dfrac{\sin \theta }{\cos \theta }=\tan \theta =\text{R}\text{.H}\text{.S}\]

$\therefore \dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2{{\cos }^{3}}\theta -\cos \theta }=\tan \theta $

Hence proved.

(viii) \[{{(\sin A+\cos ecA)}^{2}}+{{(\cos A+\sec A)}^{2}}=7+{{\tan }^{2}}A+{{\cot }^{2}}A\]

Ans:

Given: \[{{(\sin A+\cos ecA)}^{2}}+{{(\cos A+\sec A)}^{2}}=7+{{\tan }^{2}}A+{{\cot }^{2}}A\]

We know that, \[\cos e{{c}^{2}}\theta =1+{{\cot }^{2}}\theta \] and \[{{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta \]

Then, let us take L.H.S 

\[{{(\sin A+\cos ecA)}^{2}}+{{(\cos A+\sec A)}^{2}}={{\left( \sin A+\dfrac{1}{\sin A} \right)}^{2}}+{{\left( \cos A+\dfrac{1}{\cos A} \right)}^{2}}\]

\[={{\sin }^{2}}A+\dfrac{1}{{{\sin }^{2}}A}+2\sin A.\dfrac{1}{\sin A}+{{\cos }^{2}}A+\dfrac{1}{{{\cos }^{2}}A}+2\cos A.\dfrac{1}{\cos A}\]

\[=2+2+{{\sin }^{2}}A+{{\cos }^{2}}A+\dfrac{1}{{{\sin }^{2}}A}+\dfrac{1}{{{\cos }^{2}}A}\]

\[=4+1+\dfrac{1}{{{\sin }^{2}}A}+\dfrac{1}{{{\cos }^{2}}A}\]

\[=5+\cos e{{c}^{2}}A+{{\sec }^{2}}A\]

\[=5+1+{{\cot }^{2}}A+1+{{\tan }^{2}}A\]

\[=7+{{\tan }^{2}}A+{{\cot }^{2}}A=\text{R}\text{.H}\text{.S}\]

$\therefore {{(\sin A+\cos ecA)}^{2}}+{{(\cos A+\sec A)}^{2}}=7+{{\tan }^{2}}A+{{\cot }^{2}}A$

Hence proved.

(ix) \[(\cos ecA-\sin A)(\sec A-\cos A)=\dfrac{1}{\tan A+\cot A}\]

Ans:

Given: \[(\cos ecA-\sin A)(\sec A-\cos A)=\dfrac{1}{\tan A+\cot A}\]

We know that, \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]

Then, let us take L.H.S 

\[(\cos ecA-\sin A)(\sec A-\cos A)=\left( \dfrac{1}{\sin A}-\sin A \right)\left( \dfrac{1}{\cos A}-\cos A \right)\]

\[=\left( \dfrac{1-{{\sin }^{2}}A}{\sin A} \right)\left( \dfrac{1-{{\cos }^{2}}A}{\cos A} \right)\]

\[=\dfrac{{{\cos }^{2}}A}{\sin A}\times \dfrac{{{\sin }^{2}}A}{\cos A}\]

\[=\sin A.\cos A\]

\[=\dfrac{\sin A.\cos A}{{{\sin }^{2}}A+{{\cos }^{2}}A}\]

Dividing all the terms by \[\sin A.\cos A\]

\[=\dfrac{\dfrac{\sin A.\cos A}{\sin A.\cos A}}{\dfrac{{{\sin }^{2}}A}{\sin A.\cos A}+\dfrac{{{\cos }^{2}}A}{\sin A.\cos A}}\]

\[=\dfrac{1}{\dfrac{\sin A}{\cos A}+\dfrac{\cos A}{\sin A}}\]

\[=\dfrac{1}{\tan A+\cot A}=\text{R}\text{.H}\text{.S}\]

$\therefore (\cos ecA-\sin A)(\sec A-\cos A)=\dfrac{1}{\tan A+\cot A}$

Hence proved.

(x) \[\left( \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A} \right)={{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}={{\tan }^{2}}A\]

Ans:

Given: \[\left( \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A} \right)={{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}={{\tan }^{2}}A\]

We know that, \[1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta \] and \[1+{{\cot }^{2}}\theta =\cos e{{c}^{2}}A\]

Then, let us take L.H.S

\[\left( \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A} \right)=\dfrac{{{\sec }^{2}}A}{\cos e{{c}^{2}}A}\]

\[=\dfrac{1}{{{\cos }^{2}}A}\times \dfrac{{{\sin }^{2}}A}{1}\]

\[={{\tan }^{2}}A=\text{R}\text{.H}\text{.S}\]

Now, prove the Middle side 

\[{{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}={{\left( \dfrac{1-\tan A}{1-\dfrac{1}{\tan A}} \right)}^{2}}\]

\[={{\left( \dfrac{1-\tan A}{\dfrac{\tan A-1}{\tan A}} \right)}^{2}}\]

\[=\left( \dfrac{1-\tan A}{\dfrac{-(1-\tan A)}{\tan A}} \right)\]

\[={{(-\tan A)}^{2}}\]

\[={{\tan }^{2}}A=\text{R}\text{.H}\text{.S}\]

$\therefore \left( \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A} \right)={{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}={{\tan }^{2}}A$

Hence proved.

Key Topics Covered in Chapter 8

Chapter 8 of Class 10 Mathematics consists of the discussion of the following key concepts.

• Basic Trigonometry

• Opposite & Adjacent Sides In A Right-Angled Triangle

• Basic Trigonometric Ratios

• Standard Values Of Trigonometric Ratios 

• Complementary Trigonometric Ratios 

Significance of Important Questions of Trigonometry Class 10

Trigonometry is one of the most significant parts of Important Questions for Class 10 Maths Chapter 8. This chapter revolves around the memorization and conceptual understanding of the user’s problem-solving ability of a given space, preferably a triangle. These Class 10 Maths Trigonometry Important Questions help students to have a better understanding and application of trigonometry in the real world and clear the base around its introductory stage, to help students retain its benefits in the long run.

Did You Know?

Trigonometry has different applications and a few of them are mentioned below:

• It is used in cartography for the creation of maps.

• It has applications in the aviation industry and satellite systems.

• It is also used to describe light and sound waves. 

This was the complete discussion on CBSE Class 10 Maths Chapter 8 Introduction to Trigonometry important questions. If you are a Class 10 student we highly recommend downloading and practising the important questions on trigonometry. Practising these important questions will not only help you understand the concepts but will also help you in analysing the important topics and exam patterns. 

We wish you all the very best! Be exam ready with Vedantu!

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