NCERT Solutions for Class 10 Maths Chapter 8: Introduction to Trigonometry - Exercise 8.2

Exercise 8.2

1. Evaluate the following:

(i)  $\sin 60{}^\circ \cos 30{}^\circ +\sin 30{}^\circ \cos 60{}^\circ $ 

Ans: With the help of trigonometric ratio tables we can find the values of standard trigonometric angles. The trigonometric ratio table is as follows:

We have to evaluate $\sin 60{}^\circ \cos 30{}^\circ +\sin 30{}^\circ \cos 60{}^\circ $.

Substitute the values from the above table, we get

$\Rightarrow \dfrac{\sqrt{3}}{2}\times \dfrac{\sqrt{3}}{2}+\dfrac{1}{2}\times \dfrac{1}{2}$

$\Rightarrow \dfrac{3}{4}+\dfrac{1}{4}$ 

$\Rightarrow \dfrac{4}{4}$ 

$\therefore \sin 60{}^\circ \cos 30{}^\circ +\sin 30{}^\circ \cos 60{}^\circ =1$.

(ii)  $2{{\tan }^{2}}45{}^\circ +{{\cos }^{2}}30{}^\circ -{{\sin }^{2}}60{}^\circ $

Ans: With the help of trigonometric ratio tables we can find the values of standard trigonometric angles. The trigonometric ratio table is as follows:

We have to evaluate $2{{\tan }^{2}}45{}^\circ +{{\cos }^{2}}30{}^\circ -{{\sin }^{2}}60{}^\circ $.

Substitute the values from the above table, we get

$\Rightarrow 2{{\left( 1 \right)}^{2}}+{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}-{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}$

$\Rightarrow 2+\dfrac{3}{4}-\dfrac{3}{4}$ 

$\Rightarrow 2$ 

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

(iii)  $\dfrac{\cos 45{}^\circ }{\sec 30{}^\circ +\operatorname{cosec}30{}^\circ }$

Ans: With the help of trigonometric ratio tables we can find the values of standard trigonometric angles. The trigonometric ratio table is as follows:

We have to evaluate $\dfrac{\cos 45{}^\circ }{\sec 30{}^\circ +\operatorname{cosec}30{}^\circ }$.

Substitute the values from the above table, we get

$\Rightarrow \dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{2}{\sqrt{3}}+2}$

$\Rightarrow \dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{2+2\sqrt{3}}{\sqrt{3}}}$ 

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

Multiplying and dividing by \[\sqrt{3}-1\], we get

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

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

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

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

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

$\Rightarrow \dfrac{3-\sqrt{3}}{4\sqrt{2}}$ 

$\therefore \dfrac{\cos 45{}^\circ }{\sec 30{}^\circ +\operatorname{cosec}30{}^\circ }=\dfrac{3-\sqrt{3}}{4\sqrt{2}}$

(iv)  $\dfrac{\sin 30{}^\circ +\tan 45{}^\circ -\operatorname{cosec}60{}^\circ }{\sec 30{}^\circ +\cos 60{}^\circ -\cot 45{}^\circ }$

Ans: With the help of trigonometric ratio tables we can find the values of standard trigonometric angles. The trigonometric ratio table is as follows:

We have to evaluate $\dfrac{\sin 30{}^\circ +\tan 45{}^\circ -\operatorname{cosec}60{}^\circ }{\sec 30{}^\circ +\cos 60{}^\circ -\cot 45{}^\circ }$.

Substitute the values from the above table, we get

$\Rightarrow \dfrac{\dfrac{1}{2}+1-\dfrac{2}{\sqrt{3}}}{\dfrac{2}{\sqrt{3}}+\dfrac{1}{2}+1}$

$\Rightarrow \dfrac{\dfrac{3}{2}-\dfrac{2}{\sqrt{3}}}{\dfrac{2}{\sqrt{3}}+\dfrac{3}{2}}$

$\Rightarrow \dfrac{\dfrac{3\sqrt{3}-4}{2\sqrt{3}}}{\dfrac{3\sqrt{3}+4}{2\sqrt{3}}}$

$\Rightarrow \dfrac{3\sqrt{3}-4}{3\sqrt{3}+4}$

Multiplying and dividing by \[3\sqrt{3}-4\], we get

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

Now, applying the identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$, we get

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

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

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

$\Rightarrow \dfrac{43-24\sqrt{3}}{11}$ 

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

(v) $\dfrac{5{{\cos }^{2}}60{}^\circ +4{{\sec }^{2}}30{}^\circ -{{\tan }^{2}}45{}^\circ }{{{\sec }^{2}}30{}^\circ +{{\cos }^{2}}30{}^\circ }$

Ans: With the help of trigonometric ratio tables we can find the values of standard trigonometric angles. The trigonometric ratio table is as follows:

We have to evaluate $\dfrac{5{{\cos }^{2}}60{}^\circ +4{{\sec }^{2}}30{}^\circ -{{\tan }^{2}}45{}^\circ }{{{\sec }^{2}}30{}^\circ +{{\cos }^{2}}30{}^\circ }$.

Substitute the values from the above table, we get

$\Rightarrow \dfrac{5{{\left( \dfrac{1}{2} \right)}^{2}}+4{{\left( \dfrac{2}{\sqrt{3}} \right)}^{2}}-{{1}^{2}}}{{{\left( \dfrac{1}{2} \right)}^{2}}+{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}}$

$\Rightarrow \dfrac{5\left( \dfrac{1}{4} \right)+4\left( \dfrac{4}{3} \right)-1}{\left( \dfrac{1}{4} \right)+\left( \dfrac{3}{4} \right)}$

$\Rightarrow \dfrac{\dfrac{15+64-12}{12}}{\dfrac{1+3}{4}}$

$\Rightarrow \dfrac{\dfrac{15+64-12}{12}}{\dfrac{1+3}{4}}$

$\Rightarrow \dfrac{\dfrac{15+64-12}{12}}{\dfrac{4}{4}}$ 

$\Rightarrow \dfrac{\dfrac{67}{12}}{1}$ 

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

2. Choose the correct option and justify your choice.

(i)  $\dfrac{2\tan 30{}^\circ }{1+{{\tan }^{2}}30{}^\circ }=$ ………

1. $\sin 60{}^\circ $ 

2. $\cos 60{}^\circ $ 

3. $\tan 60{}^\circ $ 

4. $\sin 30{}^\circ $ 

Ans: The given expression is $\dfrac{2\tan 30{}^\circ }{1+{{\tan }^{2}}30{}^\circ }$.

We know that from the trigonometric ratio table we have $\tan 30{}^\circ =\dfrac{1}{\sqrt{3}}$.

Substitute the value in the given expression we get

$\dfrac{2\tan 30{}^\circ }{1+{{\tan }^{2}}30{}^\circ }=\dfrac{2\left( \dfrac{1}{\sqrt{3}} \right)}{1+{{\left( \dfrac{1}{\sqrt{3}} \right)}^{2}}}$

$\Rightarrow \dfrac{2\tan 30{}^\circ }{1+{{\tan }^{2}}30{}^\circ }=\dfrac{\dfrac{2}{\sqrt{3}}}{1+\dfrac{1}{3}}$

$\Rightarrow \dfrac{2\tan 30{}^\circ }{1+{{\tan }^{2}}30{}^\circ }=\dfrac{\dfrac{2}{\sqrt{3}}}{\dfrac{4}{3}}$

$\Rightarrow \dfrac{2\tan 30{}^\circ }{1+{{\tan }^{2}}30{}^\circ }=\dfrac{\sqrt{3}}{2}$

From the trigonometric table we know that 

$\sin 60{}^\circ =\dfrac{\sqrt{3}}{2}$ 

$\cos 60{}^\circ =\dfrac{1}{2}$ 

$\tan 60{}^\circ =\sqrt{3}$ 

$\sin 30{}^\circ =\dfrac{1}{2}$

Hence, $\dfrac{2\tan 30{}^\circ }{1+{{\tan }^{2}}30{}^\circ }=\sin 60{}^\circ $.

Therefore, option (A) is the correct answer.

(ii)  $\dfrac{1-{{\tan }^{2}}45{}^\circ }{1+{{\tan }^{2}}45{}^\circ }=$ ………

1. $\tan 90{}^\circ $ 

2. $1$ 

3. $\sin 45{}^\circ $ 

4. $0$

Ans: The given expression is $\dfrac{1-{{\tan }^{2}}45{}^\circ }{1+{{\tan }^{2}}45{}^\circ }$.

We know that from the trigonometric ratio table we have $\tan 45{}^\circ =1$.

Substitute the value in the given expression we get

$\dfrac{1-{{\tan }^{2}}45{}^\circ }{1+{{\tan }^{2}}45{}^\circ }=\dfrac{1-{{1}^{2}}}{1+{{1}^{2}}}$

$\Rightarrow \dfrac{1-{{\tan }^{2}}45{}^\circ }{1+{{\tan }^{2}}45{}^\circ }=\dfrac{1-1}{1+1}$

$\Rightarrow \dfrac{1-{{\tan }^{2}}45{}^\circ }{1+{{\tan }^{2}}45{}^\circ }=\dfrac{0}{2}$

$\therefore \dfrac{1-{{\tan }^{2}}45{}^\circ }{1+{{\tan }^{2}}45{}^\circ }=0$

Therefore, option (D) is the correct answer.

(iii)  $\sin 2A=2\sin A$ is true when $A=$ ……..

1. $0{}^\circ $ 

2. $30{}^\circ $ 

3. $45{}^\circ $ 

4. $60{}^\circ $ 

Ans: The given expression is $\sin 2A=2\sin A$.

We know that from the trigonometric ratio table we have 

$\sin 0{}^\circ =0$

$\sin 30{}^\circ =\dfrac{1}{2}$

$\sin 45{}^\circ =\dfrac{1}{\sqrt{2}}$

$\sin 60{}^\circ =\dfrac{\sqrt{3}}{2}$  

$\sin 90{}^\circ =1$ 

The given statement is true when $A=0{}^\circ $.

Substitute the value in the given expression we get

$\Rightarrow \sin 2A=2\sin A$

$\Rightarrow \sin 2\times 0{}^\circ =2\sin 0{}^\circ $

$0=0$ 

Therefore, option (A) is the correct answer.

(iv)  $\dfrac{2\tan 30{}^\circ }{1-{{\tan }^{2}}30{}^\circ }=$………

1. $\sin 60{}^\circ $ 

2. $\cos 60{}^\circ $ 

3. $\tan 60{}^\circ $ 

4. $\sin 30{}^\circ $ 

Ans: The given expression is $\dfrac{2\tan 30{}^\circ }{1-{{\tan }^{2}}30{}^\circ }$.

We know that from the trigonometric ratio table we have $\tan 30{}^\circ =\dfrac{1}{\sqrt{3}}$.

Substitute the value in the given expression we get

$\dfrac{2\tan 30{}^\circ }{1-{{\tan }^{2}}30{}^\circ }=\dfrac{2\left( \dfrac{1}{\sqrt{3}} \right)}{1-{{\left( \dfrac{1}{\sqrt{3}} \right)}^{2}}}$

$\Rightarrow \dfrac{2\tan 30{}^\circ }{1-{{\tan }^{2}}30{}^\circ }=\dfrac{\dfrac{2}{\sqrt{3}}}{1-\dfrac{1}{3}}$

$\Rightarrow \dfrac{2\tan 30{}^\circ }{1-{{\tan }^{2}}30{}^\circ }=\dfrac{\dfrac{2}{\sqrt{3}}}{\dfrac{2}{3}}$

$\Rightarrow \dfrac{2\tan 30{}^\circ }{1-{{\tan }^{2}}30{}^\circ }=\sqrt{3}$

From the trigonometric table we know that 

$\sin 60{}^\circ =\dfrac{\sqrt{3}}{2}$ 

$\cos 60{}^\circ =\dfrac{1}{2}$ 

$\tan 60{}^\circ =\sqrt{3}$ 

$\sin 30{}^\circ =\dfrac{1}{2}$

Hence, $\dfrac{2\tan 30{}^\circ }{1-{{\tan }^{2}}30{}^\circ }=\tan 60{}^\circ $.

Therefore, option (C) is the correct answer.

3. If $\tan \left( A+B \right)=\sqrt{3}$ and $\tan \left( A-B \right)=\dfrac{1}{\sqrt{3}}$, $0{}^\circ <A+B\le 90{}^\circ $. Find $A$ and $B$.

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

From the trigonometric ratio table we know that $\tan 60{}^\circ =\sqrt{3}$ and $\tan 30{}^\circ =\dfrac{1}{\sqrt{3}}$.

Then we get

$\tan \left( A+B \right)=\sqrt{3}$

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

$\Rightarrow A+B=60{}^\circ $ ……….(1)

Also, $\tan \left( A-B \right)=\dfrac{1}{\sqrt{3}}$

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

$\Rightarrow A-B=30{}^\circ $ ……….(2)

Adding eq. (1) and (2), we get

$2A=90{}^\circ $

$\therefore A=45{}^\circ $ 

Substitute the obtained value in eq. (1), we get

$45{}^\circ +B=60{}^\circ $ 

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

$\therefore B=15{}^\circ $ 

Therefore, the values of $A$ and $B$ is $45{}^\circ $ and $15{}^\circ $ respectively.

4. State whether the following are true or false. Justify your answer.

(i)  $\sin \left( A+B \right)=\sin A+\sin B$.

Ans: Let us assume $A=30{}^\circ $ and $B=60{}^\circ $.

Now, let us consider LHS of the given expression, we get

$\sin \left( A+B \right)$

Substitute the assumed values in the LHS, we get

$\sin \left( A+B \right)=\sin \left( 30{}^\circ +60{}^\circ  \right)$

$\Rightarrow \sin \left( A+B \right)=\sin \left( 90{}^\circ  \right)$ 

From the trigonometric ratio table we know that $\sin 90{}^\circ =1$, we get

$\Rightarrow \sin \left( A+B \right)=1$

Now, let us consider the RHS of the given expression and substitute the values, we get

$\sin A+\sin B=\sin 30{}^\circ +\sin 60{}^\circ $

From the trigonometric ratio table we know that $\sin 30{}^\circ =\dfrac{1}{2}$ and $\sin 60{}^\circ =\dfrac{\sqrt{3}}{2}$, we get

$\Rightarrow \sin A+\sin B=\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}$

$\Rightarrow \sin A+\sin B=\dfrac{1+\sqrt{3}}{2}$

Thus, $LHS\ne RHS$.

Therefore, the given statement is false.

(ii)  The value of $\sin \theta $ increases as $\theta $ increases. 

Ans: The value of sine from the trigonometric ratio table is as follows:

$\sin 0{}^\circ =0$

$\sin 30{}^\circ =\dfrac{1}{2}=0.5$

$\sin 45{}^\circ =\dfrac{1}{\sqrt{2}}=0.707$

$\sin 60{}^\circ =\dfrac{\sqrt{3}}{2}=0.866$  

$\sin 90{}^\circ =1$ 

Therefore, we can conclude that the value of $\sin \theta $ increases as $\theta $ increases. 

Therefore, the given statement is true.

(iii)  The value of $\cos \theta $ increases as $\theta $ increases. 

Ans: The value of cosine from the trigonometric ratio table is as follows:

$\cos 0{}^\circ =1$

$\cos 30{}^\circ =\dfrac{\sqrt{3}}{2}=0.866$

$\cos 45{}^\circ =\dfrac{1}{\sqrt{2}}=0.707$

$\cos 60{}^\circ =\dfrac{1}{2}=0.5$  

$\cos 90{}^\circ =0$ 

Therefore, we can conclude that the value of $\cos \theta $ decreases as $\theta $ increases. 

Therefore, the given statement is false.

(iv)  \[\sin \theta =\cos \theta \] for all values of \[\theta \].

Ans: The trigonometric ratio table is given as follows:

From the above table we can conclude that \[\sin \theta =\cos \theta \] is true only for $\theta =45{}^\circ $.

\[\sin \theta =\cos \theta \] is not true for all values of $\theta $.

Therefore, the given statement is false.

(v)  $\cot A$ is not defined for $A=0{}^\circ $.

Ans: We know that $\cot A=\dfrac{\cos A}{\sin A}$ .

If $A=0{}^\circ $, then $\cot 0{}^\circ =\dfrac{\cos 0{}^\circ }{\sin 0{}^\circ }$

From trigonometric ratio table we get

$\sin 0{}^\circ =0$ and $\cos 0{}^\circ =1$

We get

$\cot 0{}^\circ =\dfrac{1}{0}$, which is undefined.

Therefore, the given statement is true.

Conclusion

For a thorough understanding of trigonometry fundamentals, NCERT Solutions for Ex 8.2 Class 10 Maths Chapter 8, "Introduction to Trigonometry," is an excellent resource. This chapter is essential because it establishes a number of ideas that are fundamental to further research in engineering, physics, and mathematics. In particular, students should concentrate on becoming skilled in trigonometric identities and ratios, as well as how to use them to solve problems that include heights and distances. It is necessary to fully understand these ideas in order to answer a variety of exam questions.

NCERT Solutions for Class 10 Maths Chapter 8 Exercises

CBSE Class 10 Maths Chapter 8 Other Study Materials

Chapter-Specific NCERT Solutions for Class 10 Maths

Given below are the chapter-wise NCERT Solutions for Class 10 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

Study Resources for Class 10 Maths

For complete preparation of Maths for CBSE Class 10 board exams, check out the following links for different study materials available at Vedantu.