Class 12 Important Questions: CBSE Maths Chapter 2 Inverse Trigonometric Functions 2024-25

Very Short Questions and Answers (1 Marks Questions)

1. Write the Principal Value.

(i) Write the Principal Value of ${{\sin }^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right)$.

Ans: Let ${{\sin }^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right)=x$

Taking sine on both sides we get,

$\sin \left( x \right)=-\dfrac{\sqrt{3}}{2}$ 

But, the principal value of $\sin \left( \dfrac{\pi }{3} \right)$ is $\dfrac{\sqrt{3}}{2}$ and we know that and $\sin \left( -x \right)=-\sin \left( x \right)$.

$\Rightarrow \sin \left( -\dfrac{\pi }{3} \right)=-\dfrac{\sqrt{3}}{2}$

$\therefore x=-\dfrac{\pi }{3}$ 

Hence, the principal value of ${{\sin }^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right)$ is $-\dfrac{\pi }{3}$. 

(ii) Write the Principal Value of ${{\cos }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)$.

Ans: Let ${{\cos }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)=x$

Taking cosine on both sides we get,

$\cos \left( x \right)=\dfrac{\sqrt{3}}{2}$ 

But, the principal value of $\cos \left( \dfrac{\pi }{6} \right)$ is $\dfrac{\sqrt{3}}{2}$.

$\therefore x=\dfrac{\pi }{6}$ 

Hence, the principal value of ${{\cos }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)$ is $\dfrac{\pi }{6}$. 

(iii) Write the Principal Value of ${{\tan }^{-1}}\left( -\dfrac{1}{\sqrt{3}} \right)$.

Ans: Let ${{\tan }^{-1}}\left( -\dfrac{1}{\sqrt{3}} \right)=x$

Taking tan on both sides we get,

$\tan \left( x \right)=-\dfrac{1}{\sqrt{3}}$ 

But, the principal value of $\tan \left( \dfrac{\pi }{6} \right)$ is $\dfrac{1}{\sqrt{3}}$ and we know that and $\tan \left( -x \right)=-\tan \left( x \right)$.

$\Rightarrow \tan \left( -\dfrac{\pi }{6} \right)=-\dfrac{1}{\sqrt{3}}$

$\therefore x=-\dfrac{\pi }{6}$ 

Hence, the principal value of ${{\tan }^{-1}}\left( -\dfrac{1}{\sqrt{3}} \right)$ is $-\dfrac{\pi }{6}$. 

(iv) Write the Principal Value of $\text{cose}{{\text{c}}^{-1}}\left( -2 \right)$.

Ans: Let $\text{cose}{{\text{c}}^{-1}}\left( -2 \right)=x$

Taking cosec on both sides we get,

$\text{cosec}\left( x \right)=-2$ 

But, the principal value of $\sin \left( \dfrac{\pi }{6} \right)$ is $\dfrac{1}{2}$ i.e., the principal value of $\text{cosec}\left( \dfrac{\pi }{6} \right)$ is $2$ and we know that and $\text{cosec}\left( -x \right)=-\text{cosec}\left( x \right)$.

$\Rightarrow \text{cosec}\left( -\dfrac{\pi }{6} \right)=-2$

$\therefore x=-\dfrac{\pi }{6}$ 

Hence, the principal value of $\text{cose}{{\text{c}}^{-1}}\left( -2 \right)$ is $-\dfrac{\pi }{6}$. 

(v) Write the Principal Value of $\text{co}{{\text{t}}^{-1}}\left( \dfrac{1}{\sqrt{3}} \right)$.

Ans: Let $\text{co}{{\text{t}}^{-1}}\left( \dfrac{1}{\sqrt{3}} \right)=x$

Taking cot on both sides we get,

$\text{cot}\left( x \right)=\dfrac{1}{\sqrt{3}}$ 

But, the principal value of \[\text{tan}\left( \dfrac{\pi }{3} \right)\] is \[\sqrt{3}\] i.e., the principal value of $\text{cot}\left( \dfrac{\pi }{3} \right)$ is \[\dfrac{1}{\sqrt{3}}\].

$\Rightarrow \text{cot}\left( \dfrac{\pi }{3} \right)=\dfrac{1}{\sqrt{3}}$

$\therefore x=\dfrac{\pi }{3}$ 

Hence, the principal value of $\text{co}{{\text{t}}^{-1}}\left( \dfrac{1}{\sqrt{3}} \right)$ is $\dfrac{\pi }{3}$. 

(vi) Write the Principal Value of $\text{se}{{\text{c}}^{-1}}\left( -2 \right)$.

Ans: Let $\text{se}{{\text{c}}^{-1}}\left( -2 \right)=x$

Taking sec on both sides we get,

$\text{sec}\left( x \right)=-2$ 

But, the principal value of \[\text{cos}\left( \dfrac{2\pi }{3} \right)\] is \[-\dfrac{1}{2}\] i.e., the principal value of $\text{sec}\left( \dfrac{2\pi }{3} \right)$ is \[-2\].

$\Rightarrow \text{sec}\left( \dfrac{2\pi }{3} \right)=-2$

$\therefore x=\dfrac{2\pi }{3}$ 

Hence, the principal value of $\text{se}{{\text{c}}^{-1}}\left( -2 \right)$ is $\dfrac{2\pi }{3}$.

(vii) Write the Principal Value of $\text{si}{{\text{n}}^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right)+\text{co}{{\text{s}}^{-1}}\left( -\dfrac{1}{2} \right)+\text{ta}{{\text{n}}^{-1}}\left( -\dfrac{1}{\sqrt{3}} \right)$.

Ans: Let $\text{si}{{\text{n}}^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right)=x$

Taking sine on both sides we get,

$\sin \left( x \right)=-\dfrac{\sqrt{3}}{2}$ 

But, the principal value of $\sin \left( \dfrac{\pi }{3} \right)$ is $\dfrac{\sqrt{3}}{2}$ and we know that and $\sin \left( -x \right)=-\sin \left( x \right)$

$\Rightarrow \sin \left( -\dfrac{\pi }{3} \right)=-\dfrac{\sqrt{3}}{2}$

$\therefore x=-\dfrac{\pi }{3}$ 

Let $\text{co}{{\text{s}}^{-1}}\left( -\dfrac{1}{2} \right)=y$

Taking cosine on both sides we get,

$\cos \left( y \right)=-\dfrac{1}{2}$ 

But, the principal value of \[\text{cos}\left( \dfrac{2\pi }{3} \right)\] is \[-\dfrac{1}{2}\].

$\Rightarrow \text{cos}\left( \dfrac{2\pi }{3} \right)=-\dfrac{1}{2}$

$\therefore y=\dfrac{2\pi }{3}$

Let ${{\tan }^{-1}}\left( -\dfrac{1}{\sqrt{3}} \right)=z$

Taking tan on both sides we get,

$\tan \left( z \right)=-\dfrac{1}{\sqrt{3}}$ 

But, the principal value of $\tan \left( \dfrac{\pi }{6} \right)$ is $\dfrac{1}{\sqrt{3}}$ and we know that and $\tan \left( -x \right)=-\tan \left( x \right)$.

$\Rightarrow \tan \left( -\dfrac{\pi }{6} \right)=-\dfrac{1}{\sqrt{3}}$

$\therefore z=-\dfrac{\pi }{6}$

Hence, the principal value of $\text{si}{{\text{n}}^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right)+\text{co}{{\text{s}}^{-1}}\left( -\dfrac{1}{2} \right)+\text{ta}{{\text{n}}^{-1}}\left( -\dfrac{1}{\sqrt{3}} \right)$ is \[x+y+z\].

\[\Rightarrow -\dfrac{\pi }{3}+\dfrac{2\pi }{3}-\dfrac{\pi }{6}=\dfrac{\pi }{6}\].

2. Write the Value of the Function.

(i) What is the Value of the Function \[\text{ta}{{\text{n}}^{-1}}\left( \dfrac{1}{\sqrt{3}} \right)-{{\sec }^{-1}}\left( \dfrac{2}{\sqrt{3}} \right)\].

Ans: Let $\text{ta}{{\text{n}}^{-1}}\left( \dfrac{1}{\sqrt{3}} \right)=x$

Taking tan on both sides we get,

$\tan \left( x \right)=\dfrac{1}{\sqrt{3}}$ 

But, the principal value of $\tan \left( \dfrac{\pi }{6} \right)$ is $\dfrac{1}{\sqrt{3}}$.

$\Rightarrow \tan \left( \dfrac{\pi }{6} \right)=\dfrac{1}{\sqrt{3}}$

$\therefore x=\dfrac{\pi }{6}$ 

Let ${{\sec }^{-1}}\left( \dfrac{2}{\sqrt{3}} \right)=y$

Taking sec on both sides we get,

$\sec \left( y \right)=\dfrac{2}{\sqrt{3}}$ 

But, the principal value of \[\text{cos}\left( \dfrac{\pi }{6} \right)\] is \[\dfrac{\sqrt{3}}{2}\] 

$\Rightarrow \sec \left( \dfrac{\pi }{6} \right)=\dfrac{2}{\sqrt{3}}$

$\therefore y=\dfrac{\pi }{6}$

Hence, the principal value of \[\text{ta}{{\text{n}}^{-1}}\left( \dfrac{1}{\sqrt{3}} \right)-{{\sec }^{-1}}\left( \dfrac{2}{\sqrt{3}} \right)\] is $x+y$. 

$\Rightarrow \dfrac{\pi }{6}-\dfrac{\pi }{6}=0$.

(ii) What is the Value of the Function \[{{\sin }^{-1}}\left( -\dfrac{1}{2} \right)-{{\cos }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)\].

Ans: Let ${{\sin }^{-1}}\left( -\dfrac{1}{2} \right)=x$

Taking sine on both sides we get,

$\sin \left( x \right)=-\dfrac{1}{2}$ 

But, the principal value of $\sin \left( \dfrac{\pi }{6} \right)$ is $\dfrac{1}{2}$ and we know that $\sin \left( -x \right)=-\sin \left( x \right)$.

$\Rightarrow \sin \left( -\dfrac{\pi }{6} \right)=-\dfrac{1}{2}$

$\therefore x=-\dfrac{\pi }{6}$ 

Let ${{\cos }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)=y$

Taking cos on both sides we get,

$\cos \left( y \right)=\dfrac{\sqrt{3}}{2}$ 

But, the principal value of \[\text{cos}\left( \dfrac{\pi }{6} \right)\] is \[\dfrac{\sqrt{3}}{2}\] 

\[\Rightarrow \text{cos}\left( \dfrac{\pi }{6} \right)=\dfrac{\sqrt{3}}{2}\]

$\therefore y=\dfrac{\pi }{6}$

Hence, the principal value of \[{{\sin }^{-1}}\left( -\dfrac{1}{2} \right)-{{\cos }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)\] is $x-y$. 

$\Rightarrow -\dfrac{\pi }{6}-\dfrac{\pi }{6}=-\dfrac{\pi }{3}$.

(iii) What is the Value of the Function \[{{\tan }^{-1}}\left( 1 \right)-{{\cot }^{-1}}\left( -1 \right)\].

Ans: Let ${{\tan }^{-1}}\left( 1 \right)=x$

Taking tan on both sides we get, $\tan \left( x \right)=1$ 

But, the principal value of $\tan \left( \dfrac{\pi }{4} \right)$ is $1$.

$\Rightarrow \tan \left( \dfrac{\pi }{4} \right)=1$

$\therefore x=\dfrac{\pi }{4}$ 

Let ${{\cot }^{-1}}\left( -1 \right)=y$

Taking cot on both sides we get, \[\cot \left( y \right)=-1\] 

But we know that the principal value of $\cot \left( \dfrac{3\pi }{4} \right)=-1$. 

\[\Rightarrow \cot \left( \dfrac{3\pi }{4} \right)=-1\] 

$\therefore y=\dfrac{3\pi }{4}$

Hence, the principal value of \[{{\tan }^{-1}}\left( 1 \right)-{{\cot }^{-1}}\left( -1 \right)\] is $x-y$ 

$\Rightarrow \dfrac{\pi }{4}-\dfrac{3\pi }{4}=-\dfrac{\pi }{2}$.

(iv) What is the Value of the Function \[\text{cose}{{\text{c}}^{-1}}\left( \sqrt{2} \right)+{{\sec }^{-1}}\left( \sqrt{2} \right)\].

Ans: We know that the principal value of $\sin \left( \dfrac{\pi }{4} \right),\cos \left( \dfrac{\pi }{4} \right)$ is $\dfrac{1}{\sqrt{2}}$. 

Let $\text{cose}{{\text{c}}^{-1}}\left( \sqrt{2} \right)=x$ and $\text{se}{{\text{c}}^{-1}}\left( \sqrt{2} \right)=y$ 

Taking cosec on both sides of first equation and sec on both sides of second equation we get, $\text{cosec}\left( x \right)=\sqrt{2}$ and $\sec \left( y \right)=\sqrt{2}$. 

But, the principal value of $\sin \left( \dfrac{\pi }{4} \right),\cos \left( \dfrac{\pi }{4} \right)$ is $\dfrac{1}{\sqrt{2}}$ and $\text{cosec}\left( \text{x} \right)\text{=}\dfrac{\text{1}}{\text{sinx}}\text{,sec}\left( \text{x} \right)\text{=}\dfrac{\text{1}}{\text{cosx}}$. 

$\text{cosec}\left( \dfrac{\pi }{4} \right)=\sqrt{2}$ and $\sec \left( \dfrac{\pi }{4} \right)=\sqrt{2}$.

$\therefore x=y=\dfrac{\pi }{4}$ 

Hence, the principal value of \[\text{cose}{{\text{c}}^{-1}}\left( \sqrt{2} \right)+{{\sec }^{-1}}\left( \sqrt{2} \right)\] is $x+y$. 

 $\Rightarrow \dfrac{\pi }{4}+\dfrac{\pi }{4}=\dfrac{\pi }{2}$.

(v) What is the Value of the Function \[\text{ta}{{\text{n}}^{-1}}\left( 1 \right)+{{\cot }^{-1}}\left( 1 \right)+{{\sin }^{-1}}\left( 1 \right)\].

Ans: Let ${{\tan }^{-1}}\left( 1 \right)=x$

Taking tan on both sides we get, $\tan \left( x \right)=1$ 

But, the principal value of $\tan \left( \dfrac{\pi }{4} \right)$ is $1$.

$\Rightarrow \tan \left( \dfrac{\pi }{4} \right)=1$

$\therefore x=\dfrac{\pi }{4}$ 

Let ${{\cot }^{-1}}\left( 1 \right)=y$

Using the trigonometric identity ${{\cot }^{-1}}\left( x \right)=\dfrac{\pi }{2}-{{\tan }^{-1}}\left( x \right)$ we get

$y=\dfrac{\pi }{2}-{{\tan }^{-1}}\left( 1 \right)$ 

But, ${{\tan }^{-1}}\left( 1 \right)=\dfrac{\pi }{4}$ (proved above)

$\therefore y=\dfrac{\pi }{4}$

Let ${{\sin }^{-1}}\left( 1 \right)=z$

Taking sine on both sides we get, $\sin \left( z \right)=1$ 

But, the principal value of $\sin \left( \dfrac{\pi }{2} \right)$ is $1$.

$\Rightarrow \sin \left( \dfrac{\pi }{2} \right)=1$

$\therefore z=\dfrac{\pi }{2}$ 

Hence, the principal value of \[\text{ta}{{\text{n}}^{-1}}\left( 1 \right)+{{\cot }^{-1}}\left( 1 \right)+{{\sin }^{-1}}\left( 1 \right)\] is $x+y+z$.

$\Rightarrow \dfrac{\pi }{4}+\dfrac{\pi }{4}+\dfrac{\pi }{2}=\pi $.

(vi) What is the Value of the Function \[{{\sin }^{-1}}\left( \sin \dfrac{4\pi }{5} \right)\].

Ans: Let \[{{\sin }^{-1}}\left( \sin \dfrac{4\pi }{5} \right)=x\]

Taking sine on both sides we get, \[\sin \dfrac{4\pi }{5}=\sin x\]. 

Writing $\sin \left( \dfrac{4\pi }{5} \right)$ as $\sin \left( \pi -\dfrac{\pi }{5} \right)=\sin \left( \dfrac{\pi }{5} \right)$. 

$\therefore x=\dfrac{\pi }{5}$

Hence, the principal value of \[{{\sin }^{-1}}\left( \sin \dfrac{4\pi }{5} \right)\] is \[\dfrac{\pi }{5}\].

(vii) What is the Value of the Function \[{{\tan }^{-1}}\left( \tan \dfrac{5\pi }{6} \right)\].

Ans: Let \[{{\tan }^{-1}}\left( \tan \dfrac{5\pi }{6} \right)=x\]

Taking tan on both sides we get, \[\tan \dfrac{5\pi }{6}=\tan x\] 

Writing $\tan \left( \dfrac{5\pi }{6} \right)$ as $\tan \left( \pi -\dfrac{\pi }{6} \right)=-\tan \left( \dfrac{\pi }{6} \right)$. Also, we know that $\tan \left( -x \right)=-\tan \left( x \right)$.

$\therefore x=-\dfrac{\pi }{6}$

Hence, the principal value of \[{{\tan }^{-1}}\left( \tan \dfrac{5\pi }{6} \right)\] is \[-\dfrac{\pi }{6}\].

(viii) What is the Value of the Function \[\text{cose}{{\text{c}}^{-1}}\left( \text{cosec}\dfrac{3\pi }{4} \right)\].

Ans: Let \[\text{cose}{{\text{c}}^{-1}}\left( \text{cosec}\dfrac{3\pi }{4} \right)\]

Taking cosec on both sides we get, \[\text{cosec}\dfrac{3\pi }{4}=\cos \text{ec}\left( x \right)\] 

Writing \[\text{cosec}\left( \dfrac{3\pi }{4} \right)\] as $\text{cosec}\left( \pi -\dfrac{\pi }{4} \right)=\text{cosec}\left( \dfrac{\pi }{4} \right)$. 

$\therefore x=\dfrac{\pi }{4}$

Hence, the principal value of \[\text{cose}{{\text{c}}^{-1}}\left( \text{cosec}\dfrac{3\pi }{4} \right)\] is \[\dfrac{\pi }{4}\].

Long Questions and Answers (4 Marks Questions)

3. Show that \[{{\tan }^{-1}}\left( \dfrac{\sqrt{1+\cos x}+\sqrt{1-\cos x}}{\sqrt{1+\cos x}-\sqrt{1-\cos x}} \right)=\dfrac{\pi }{4}+\dfrac{x}{2}\], $x\in \left[ 0,\pi  \right]$. 

Ans: Using the trigonometric identities

$\sqrt{1+\cos x}=2\cos \left( \dfrac{x}{2} \right)$ and $\sqrt{1-\cos x}=2\sin \left( \dfrac{x}{2} \right)$ we get,

\[{{\tan }^{-1}}\left( \dfrac{\sqrt{1+\cos x}+\sqrt{1-\cos x}}{\sqrt{1+\cos x}-\sqrt{1-\cos x}} \right)={{\tan }^{-1}}\left( \dfrac{2\cos \left( \dfrac{x}{2} \right)+2\sin \left( \dfrac{x}{2} \right)}{2\cos \left( \dfrac{x}{2} \right)-2\sin \left( \dfrac{x}{2} \right)} \right)\]      …..(1)

Dividing the numerator and denominator of (1) by $2\cos \left( \dfrac{x}{2} \right)$ we get,

\[{{\tan }^{-1}}\left( \dfrac{\sqrt{1+\cos x}+\sqrt{1-\cos x}}{\sqrt{1+\cos x}-\sqrt{1-\cos x}} \right)={{\tan }^{-1}}\left( \dfrac{1+\tan \left( \dfrac{x}{2} \right)}{1-\tan \left( \dfrac{x}{2} \right)} \right)\]                      …..(2)

Now using the identity, ${{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)={{\tan }^{-1}}\left( x \right)+{{\tan }^{-1}}\left( y \right)$ on equation (2) we get, 

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

\[\Rightarrow \dfrac{\pi }{4}+\dfrac{x}{2}\] 

4. Show that \[{{\tan }^{-1}}\left( \dfrac{\cos x}{1-\sin x} \right)-{{\cot }^{-1}}\left( \sqrt{\dfrac{1+\cos x}{1-\cos x}} \right)=\dfrac{\pi }{4}\], $x\in \left( 0,\dfrac{\pi }{2} \right)$. 

Ans: Using the trigonometric identities

\[\sqrt{1+\cos x}=2\cos \left( \dfrac{x}{2} \right)\] and \[\sqrt{1-\cos x}=2\sin \left( \dfrac{x}{2} \right)\] on \[{{\cot }^{-1}}\left( \sqrt{\dfrac{1+\cos x}{1-\cos x}} \right)\] we get, 

\[{{\cot }^{-1}}\left( \sqrt{\dfrac{1+\cos x}{1-\cos x}} \right)={{\cot }^{-1}}\left( \dfrac{2\cos \left( \dfrac{x}{2} \right)}{2\sin \left( \dfrac{x}{2} \right)} \right)\]

\[\Rightarrow {{\cot }^{-1}}\left( \cot \left( \dfrac{x}{2} \right) \right)\] 

But we know that ${{\cot }^{-1}}\left( \cot x \right)=x$, therefore, 

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

\[\Rightarrow \dfrac{x}{2}\] 

Let it be known as equation (1)

Now let us solve \[{{\tan }^{-1}}\left( \dfrac{\cos x}{1-\sin x} \right)\]                                                                               ….. (2)

Using the identities \[1={{\sin }^{2}}\left( \dfrac{x}{2} \right)+{{\cos }^{2}}\left( \dfrac{x}{2} \right)\] and \[\sin x=2\sin \left( \dfrac{x}{2} \right)\cos \left( \dfrac{x}{2} \right)\] in the denominator of (2) we get,

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

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

Let it be known as equation (3).

Now using the identity \[\cos x={{\cos }^{2}}\left( \dfrac{x}{2} \right)-{{\sin }^{2}}\left( \dfrac{x}{2} \right)\] in the numerator of (3) we get,

\[{{\tan }^{-1}}\left( \dfrac{\cos x}{1-\sin x} \right)={{\tan }^{-1}}\left( \dfrac{\left( \cos \left( \dfrac{x}{2} \right)-\sin \left( \dfrac{x}{2} \right) \right)\cdot \left( \cos \left( \dfrac{x}{2} \right)+\sin \left( \dfrac{x}{2} \right) \right)}{{{\left( \cos \left( \dfrac{x}{2} \right)-\sin \left( \dfrac{x}{2} \right) \right)}^{2}}} \right)\]  …..(4)

Cancelling the \[\left( \cos \left( \dfrac{x}{2} \right)-\sin \left( \dfrac{x}{2} \right) \right)\] term from numerator and denominator of (4) we get 

\[{{\tan }^{-1}}\left( \dfrac{\cos x}{1-\sin x} \right)={{\tan }^{-1}}\left( \dfrac{\cos \left( \dfrac{x}{2} \right)+\sin \left( \dfrac{x}{2} \right)}{\cos \left( \dfrac{x}{2} \right)-\sin \left( \dfrac{x}{2} \right)} \right)\]                                          …..(5)

Simplifying it further by taking $\cos \left( \dfrac{x}{2} \right)$ common from both numerator and denominator of (5) we get, 

\[{{\tan }^{-1}}\left( \dfrac{\cos x}{1-\sin x} \right)={{\tan }^{-1}}\left( \dfrac{1+\tan \left( \dfrac{x}{2} \right)}{1-\tan \left( \dfrac{x}{2} \right)} \right)\]                                                       …..(6)

Now using the identity, ${{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)={{\tan }^{-1}}\left( x \right)+{{\tan }^{-1}}\left( y \right)$ on (6) we get, 

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

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

Let it be known as equation (7).

Using equations (1) and (7) we get 

\[{{\tan }^{-1}}\left( \dfrac{\cos x}{1-\sin x} \right)-{{\cot }^{-1}}\left( \sqrt{\dfrac{1+\cos x}{1-\cos x}} \right)=\dfrac{\pi }{4}+\dfrac{x}{2}-\dfrac{x}{2}\]

\[\therefore {{\tan }^{-1}}\left( \dfrac{\cos x}{1-\sin x} \right)-{{\cot }^{-1}}\left( \sqrt{\dfrac{1+\cos x}{1-\cos x}} \right)=\dfrac{\pi }{4}\] 

5. Show that 

\[{{\tan }^{-1}}\left( \dfrac{x}{\sqrt{{{a}^{2}}-{{x}^{2}}}} \right)={{\sin }^{-1}}\left( \dfrac{x}{a} \right)={{\cos }^{-1}}\left( \dfrac{\sqrt{{{a}^{2}}-{{x}^{2}}}}{a} \right)\]

Ans: Using the trigonometric substitution $x=a\sin \theta $ in \[{{\tan }^{-1}}\left( \dfrac{x}{\sqrt{{{a}^{2}}-{{x}^{2}}}} \right)\] we get, 

\[{{\tan }^{-1}}\left( \dfrac{x}{\sqrt{{{a}^{2}}-{{x}^{2}}}} \right)={{\tan }^{-1}}\left( \dfrac{a\sin \theta }{\sqrt{{{a}^{2}}-{{a}^{2}}{{\sin }^{2}}\theta }} \right)\]                                      …..(1) 

Taking out $a$ common from the denominator of (1) and cancelling out with numerator we get

\[{{\tan }^{-1}}\left( \dfrac{x}{\sqrt{{{a}^{2}}-{{x}^{2}}}} \right)={{\tan }^{-1}}\left( \dfrac{\sin \theta }{\sqrt{1-{{\sin }^{2}}\theta }} \right)\]                                                   …..(2)

Using the trigonometric identity $\sqrt{1-{{\sin }^{2}}x}=\cos x$ in the denominator of (2) we get,

\[{{\tan }^{-1}}\left( \dfrac{x}{\sqrt{{{a}^{2}}-{{x}^{2}}}} \right)={{\tan }^{-1}}\left( \dfrac{\sin \theta }{\cos \theta } \right)\] 

\[\Rightarrow {{\tan }^{-1}}\left( \dfrac{x}{\sqrt{{{a}^{2}}-{{x}^{2}}}} \right)={{\tan }^{-1}}\left( \tan (\theta ) \right)\] 

Let it be known as equation (3).

Using the result ${{\tan }^{-1}}\left( \tan x \right)=x$ in (3) we get

\[{{\tan }^{-1}}\left( \dfrac{x}{\sqrt{{{a}^{2}}-{{x}^{2}}}} \right)=\theta \]                                                                        …..(4)

Let us now re-substitute

$\begin{align}& x=a\sin \theta  \\ & \Rightarrow \theta ={{\sin }^{-1}}\left(\dfrac{x}{a} \right) \\ \end{align}$ 

In equation (4) we get,

\[{{\tan }^{-1}}\left( \dfrac{x}{\sqrt{{{a}^{2}}-{{x}^{2}}}} \right)={{\sin }^{-1}}\left( \dfrac{x}{a} \right)\]                                                           …..(5)

Now, from $x=a\sin \theta $ we get 

$\sin \theta =\dfrac{x}{a}$. Hence, using the trigonometric identity 

${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ we get,

${{\left( \dfrac{x}{a} \right)}^{2}}+{{\cos }^{2}}\theta =1$

$\Rightarrow {{\cos }^{2}}\theta =1-\dfrac{{{x}^{2}}}{{{a}^{2}}}$ 

$\Rightarrow \cos \theta =\sqrt{\dfrac{{{a}^{2}}-{{x}^{2}}}{{{a}^{2}}}}$ 

Let this be known as equation (6).

Taking ${{\cos }^{-1}}$ on both sides of (6) we get,

 ${{\cos }^{-1}}\left( \cos \theta  \right)={{\cos }^{-1}}\left( \dfrac{\sqrt{{{a}^{2}}-{{x}^{2}}}}{a} \right)$

$\Rightarrow \theta ={{\cos }^{-1}}\left( \dfrac{\sqrt{{{a}^{2}}-{{x}^{2}}}}{a} \right)$ 

From equation (4) we get,

\[{{\tan }^{-1}}\left( \dfrac{x}{\sqrt{{{a}^{2}}-{{x}^{2}}}} \right)={{\cos }^{-1}}\left( \dfrac{\sqrt{{{a}^{2}}-{{x}^{2}}}}{a} \right)\].

6. Show that \[{{\cot }^{-1}}\left[ 2\tan \left( {{\cos }^{-1}}\dfrac{8}{17} \right) \right]+{{\tan }^{-1}}\left[ 2\tan \left( {{\sin }^{-1}}\dfrac{8}{17} \right) \right]={{\tan }^{-1}}\left( \dfrac{300}{161} \right)\]

Ans: Let us first change \[{{\cos }^{-1}}\dfrac{8}{17}\] into the form of tan inverse. 

Let \[{{\cos }^{-1}}\dfrac{8}{17}=x\]

Taking cosine on both sides we get 

\[\cos \left( {{\cos }^{-1}}\dfrac{8}{17} \right)=\cos x\] 

\[\Rightarrow \cos x=\dfrac{8}{17}\] 

Let this be known as equation (1).

Now using the trigonometric identity ${{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta $ we get,

 ${{\sin }^{2}}x=1-{{\left( \dfrac{8}{17} \right)}^{2}}$

$\Rightarrow \sin x=\dfrac{15}{17}$ 

Let this be known as equation (2).

From (1) and (2), 

$\tan x=\dfrac{\sin x}{\cos x}$ 

$\Rightarrow \tan x=\dfrac{15}{8}$ 

Let this be known as equation (3).

Taking ${{\tan }^{-1}}$ on both sides of equation (3) we get, 

${{\tan }^{-1}}\left( \tan x \right)={{\tan }^{-1}}\left( \dfrac{15}{8} \right)$ 

$\Rightarrow x={{\tan }^{-1}}\left( \dfrac{15}{8} \right)$ 

Let this be known as equation (4).

Similarly let us now change \[{{\sin }^{-1}}\dfrac{8}{17}\] into the form of tan inverse. 

Let \[{{\sin }^{-1}}\dfrac{8}{17}=y\]

Taking sine on both sides we get 

\[\sin \left( {{\sin }^{-1}}\dfrac{8}{17} \right)=\sin y\] 

\[\Rightarrow \sin y=\dfrac{8}{17}\] 

Let this be known as equation (5).

Now using the trigonometric identity ${{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta $ on (5) we get,

${{\cos }^{2}}y=1-{{\left( \dfrac{8}{17} \right)}^{2}}$

$\Rightarrow \cos y=\dfrac{15}{17}$ 

Let this be known as equation (6).

From (5) and (6), 

$\tan y=\dfrac{\sin y}{\cos y}$ 

$\Rightarrow \tan y=\dfrac{8}{15}$ 

Let this be known as equation (7).

Taking ${{\tan }^{-1}}$ on both sides of equation (7) we get, 

${{\tan }^{-1}}\left( \tan y \right)={{\tan }^{-1}}\left( \dfrac{8}{15} \right)$ 

$\Rightarrow y={{\tan }^{-1}}\left( \dfrac{8}{15} \right)$ 

Let this be known as equation (8).

Substituting these values from (4) and (8) in the original equation we get, 

\[\begin{align} & {{\cot }^{-1}}\left[ 2\tan \left( {{\cos }^{-1}}\dfrac{8}{17} \right) \right]+{{\tan }^{-1}}\left[ 2\tan \left( {{\sin }^{-1}}\dfrac{8}{17} \right) \right] \\ & \text{  }={{\cot }^{-1}}\left[ 2\tan \left( {{\tan }^{-1}}\dfrac{15}{8} \right) \right]+{{\tan }^{-1}}\left[ 2\tan \left( {{\tan }^{-1}}\dfrac{8}{15} \right) \right] \\ \end{align}\]

\[\Rightarrow {{\cot }^{-1}}\left[ 2\tan \left( {{\cos }^{-1}}\dfrac{8}{17} \right) \right]+{{\tan }^{-1}}\left[ 2\tan \left( {{\sin }^{-1}}\dfrac{8}{17} \right) \right]={{\cot }^{-1}}\left( \dfrac{15}{4} \right)+{{\tan }^{-1}}\left(\dfrac{16}{15} \right)\] 

Let this be known as equation (9).

Now using the trigonometric identity ${{\tan }^{-1}}x+{{\cot }^{-1}}x=\dfrac{\pi }{2}$ on equation (9) we get, 

\[{{\cot }^{-1}}\left[ 2\tan \left( {{\cos }^{-1}}\dfrac{8}{17} \right) \right]+{{\tan }^{-1}}\left[ 2\tan \left( {{\sin }^{-1}}\dfrac{8}{17} \right) \right]={{\cot }^{-1}}\left( \dfrac{15}{4} \right)+\dfrac{\pi }{2}-{{\cot }^{-1}}\left( \dfrac{16}{15} \right)\]

Let this be known as equation (10).

Again, using the trigonometric identity ${{\cot }^{-1}}\left( \dfrac{xy+1}{y-x} \right)={{\cot }^{-1}}\left( x \right)-{{\cot }^{-1}}\left( y \right)$ on (10) we get, 

\[{{\cot }^{-1}}\left[ 2\tan \left( {{\cos }^{-1}}\dfrac{8}{17} \right) \right]+{{\tan }^{-1}}\left[ 2\tan \left( {{\sin }^{-1}}\dfrac{8}{17} \right) \right]=\dfrac{\pi }{2}+{{\cot }^{-1}}\left( \dfrac{5}{\dfrac{16}{15}-\dfrac{15}{4}} \right)\]

\[\Rightarrow {{\cot }^{-1}}\left[ 2\tan \left( {{\cos }^{-1}}\dfrac{8}{17} \right) \right]+{{\tan }^{-1}}\left[ 2\tan \left( {{\sin }^{-1}}\dfrac{8}{17} \right) \right]=\dfrac{\pi }{2}+{{\cot }^{-1}}\left( \dfrac{300}{-161} \right)\] 

Let this be known as equation (11).

Using ${{\cot }^{-1}}\left( x \right)=-{{\cot }^{-1}}\left( x \right)$ on (11) we get,

\[{{\cot }^{-1}}\left[ 2\tan \left( {{\cos }^{-1}}\dfrac{8}{17} \right) \right]+{{\tan }^{-1}}\left[ 2\tan \left( {{\sin }^{-1}}\dfrac{8}{17} \right) \right]=\dfrac{\pi }{2}-{{\cot }^{-1}}\left( \dfrac{300}{161} \right)\]         …..(12)

Again, using the trigonometric identity ${{\tan }^{-1}}x+{{\cot }^{-1}}x=\dfrac{\pi }{2}$ on (12) we get, 

\[{{\cot }^{-1}}\left[ 2\tan \left( {{\cos }^{-1}}\dfrac{8}{17} \right) \right]+{{\tan }^{-1}}\left[ 2\tan \left( {{\sin }^{-1}}\dfrac{8}{17} \right) \right]={{\tan }^{-1}}\left( \dfrac{300}{161} \right)\]

Hence proved.

7. Show that \[{{\tan }^{-1}}\left( \dfrac{\sqrt{1+{{x}^{2}}}+\sqrt{1-{{x}^{2}}}}{\sqrt{1+{{x}^{2}}}-\sqrt{1-{{x}^{2}}}} \right)=\dfrac{\pi }{4}+\dfrac{1}{2}{{\cos }^{-1}}{{x}^{2}}\]

Ans: To solve this question, substitute ${{x}^{2}}=\cos \theta $, therefore, 

 \[{{\tan }^{-1}}\left( \dfrac{\sqrt{1+{{x}^{2}}}+\sqrt{1-{{x}^{2}}}}{\sqrt{1+{{x}^{2}}}-\sqrt{1-{{x}^{2}}}} \right)={{\tan }^{-1}}\left( \dfrac{\sqrt{1+\cos \theta }+\sqrt{1-\cos \theta }}{\sqrt{1+\cos \theta }-\sqrt{1-\cos \theta }} \right)\]           …..(1)

Using the trigonometric identities $\sqrt{1+\cos x}=2\cos \left( \dfrac{x}{2} \right)$ and $\sqrt{1-\cos x}=2\sin \left( \dfrac{x}{2} \right)$ on equation (1) we get,

\[{{\tan }^{-1}}\left( \dfrac{\sqrt{1+{{x}^{2}}}+\sqrt{1-{{x}^{2}}}}{\sqrt{1+{{x}^{2}}}-\sqrt{1-{{x}^{2}}}} \right)={{\tan }^{-1}}\left( \dfrac{2\cos \left( \dfrac{\theta }{2} \right)+2\sin \left( \dfrac{\theta }{2} \right)}{2\cos \left( \dfrac{\theta }{2} \right)-2\sin \left( \dfrac{\theta }{2} \right)} \right)\]                      …..(2)

Taking \[2\cos \left( \dfrac{\theta }{2} \right)\] common from numerator and denominator of (2) we get, 

\[{{\tan }^{-1}}\left( \dfrac{\sqrt{1+{{x}^{2}}}+\sqrt{1-{{x}^{2}}}}{\sqrt{1+{{x}^{2}}}-\sqrt{1-{{x}^{2}}}} \right)={{\tan }^{-1}}\left( \dfrac{1+\tan \left( \dfrac{\theta }{2} \right)}{1-\tan \left( \dfrac{\theta }{2} \right)} \right)\]                                      …..(3)

Now using the identity, ${{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)={{\tan }^{-1}}\left( x \right)+{{\tan }^{-1}}\left( y \right)$ on (3) we get, 

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

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

Let this be known as equation (4).

Re-substituting ${{x}^{2}}=\cos \theta $ in (4) we get, 

 \[{{\tan }^{-1}}\left( \dfrac{\sqrt{1+{{x}^{2}}}+\sqrt{1-{{x}^{2}}}}{\sqrt{1+{{x}^{2}}}-\sqrt{1-{{x}^{2}}}} \right)=\dfrac{\pi }{4}+\dfrac{1}{2}{{\cos }^{-1}}{{x}^{2}}\]

Hence Proved.

8. Solve the following for $x$: \[{{\cot }^{-1}}2x+{{\cot }^{-1}}3x=\dfrac{\pi }{4}\]

Ans: To solve this question, use the identity ${{\cot }^{-1}}\left( \dfrac{xy-1}{x+y} \right)={{\cot }^{-1}}\left( x \right)+{{\cot }^{-1}}\left( y \right)$

\[{{\cot }^{-1}}2x+{{\cot }^{-1}}3x={{\cot }^{-1}}\left( \dfrac{6{{x}^{2}}-1}{5x} \right)\]

Hence, 

\[{{\cot }^{-1}}\left( \dfrac{6{{x}^{2}}-1}{5x} \right)=\dfrac{\pi }{4}\]                                                                                …..(1)

Taking cot on both sides we get, 

\[\begin{align}& \cot \left( {{\cot }^{-1}}\left( \dfrac{6{{x}^{2}}-1}{5x} \right) \right)=\cot \left( \dfrac{\pi }{4} \right) \\ & \Rightarrow \dfrac{6{{x}^{2}}-1}{5x}=\cot \left( \dfrac{\pi }{4} \right) \\ \end{align}\] 

Let it be known as equation (2).

Solving the equation (2) by substituting the principal value of $\cot \left( \dfrac{\pi }{4} \right)$ we get, 

\[\begin{align}& 6{{x}^{2}}-1=5x \\ & \Rightarrow 6{{x}^{2}}-5x-1=0 \\ \end{align}\] 

Simplifying it further we get, 

\[\begin{align} & 6{{x}^{2}}+\left( -6x+x \right)-1=0 \\  & \left( 6x+1 \right)\left( x-1 \right)=0 \\ & \Rightarrow x=1,-\dfrac{1}{6} \\ \end{align}\] 

But $x=-\dfrac{1}{6}$ is not possible. Therefore, 

$x=1$. 

9. Show that \[{{\tan }^{-1}}\left( \dfrac{m}{n} \right)-{{\tan }^{-1}}\left( \dfrac{m-n}{m+n} \right)=\dfrac{\pi }{4},\text{  }m,n>0\]

Ans: To solve this problem use the trigonometric identity, ${{\tan }^{-1}}\left( \dfrac{x-y}{1+xy} \right)={{\tan }^{-1}}\left( x \right)-{{\tan }^{-1}}\left( y \right)$

Hence, 

\[{{\tan }^{-1}}\left( \dfrac{m}{n} \right)-{{\tan }^{-1}}\left( \dfrac{m-n}{m+n} \right)={{\tan }^{-1}}\left( \dfrac{\dfrac{m}{n}-\dfrac{m-n}{m+n}}{1+\left( \dfrac{m-n}{m+n} \right)\left( \dfrac{m}{n} \right)} \right)\]                                   …..(1)

Simplifying the fractions of equation (1) we get

\[{{\tan }^{-1}}\left( \dfrac{m}{n} \right)-{{\tan }^{-1}}\left( \dfrac{m-n}{m+n} \right)={{\tan }^{-1}}\left( \dfrac{\dfrac{{{m}^{2}}+{{n}^{2}}}{n\left( m+n \right)}}{1+\left( \dfrac{{{m}^{2}}-mn}{n\left( m+n \right)} \right)} \right)\]

\[\Rightarrow {{\tan }^{-1}}\left( \dfrac{m}{n} \right)-{{\tan }^{-1}}\left( \dfrac{m-n}{m+n} \right)={{\tan }^{-1}}\left( \dfrac{{{m}^{2}}+{{n}^{2}}}{{{m}^{2}}+{{n}^{2}}} \right)\] 

\[\Rightarrow {{\tan }^{-1}}\left( \dfrac{m}{n} \right)-{{\tan }^{-1}}\left( \dfrac{m-n}{m+n} \right)={{\tan }^{-1}}\left( 1 \right)\] 

Let this be known as equation (2).

But we know that,

$\tan \left( \dfrac{\pi }{4} \right)=1$ 

$\Rightarrow {{\tan }^{-1}}\left( 1 \right)=\dfrac{\pi }{4}$ 

Hence from equation (2) it is proved that \[{{\tan }^{-1}}\left( \dfrac{m}{n} \right)-{{\tan }^{-1}}\left( \dfrac{m-n}{m+n} \right)=\dfrac{\pi }{4},\text{  }m,n>0\]

10. Prove that \[\tan \left[ \dfrac{1}{2}{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)+\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-{{y}^{2}}}{1+{{y}^{2}}} \right) \right]=\dfrac{x+y}{1-xy}\]

Ans: To solve this problem use the substitution, 

x = tanθ

y = tanα

In the LHS of the given expression.

\[\tan \left[ \dfrac{1}{2}{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)+\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-{{y}^{2}}}{1+{{y}^{2}}} \right) \right]=\tan \left[ \dfrac{1}{2}{{\sin }^{-1}}\left( \dfrac{2\tan \theta }{1+{{\tan }^{2}}\theta } \right)+\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-{{\tan }^{2}}\alpha }{1+{{\tan }^{2}}\alpha } \right) \right]\]

Let this be known as equation (1).

Using the trigonometric identity 

\[\begin{align} & \dfrac{2\tan \theta }{1+{{\tan }^{2}}\theta }=\sin 2\theta  \\  & \dfrac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta }=\cos 2\theta  \\ \end{align}\] 

From (1) we get, 

\[\tan \left[ \dfrac{1}{2}{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)+\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-{{y}^{2}}}{1+{{y}^{2}}} \right) \right]=\tan \left[ \dfrac{1}{2}{{\sin }^{-1}}\left( \sin 2\theta  \right)+\dfrac{1}{2}{{\cos }^{-1}}\left( \cos 2\alpha  \right) \right]\] 

\[\Rightarrow \tan \left[ \dfrac{1}{2}{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)+\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-{{y}^{2}}}{1+{{y}^{2}}} \right) \right]=\tan \left[ \theta +\alpha  \right]\] 

Let this be known as equation (2).

Now using the trigonometric identity, \[\tan \left( x+y \right)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}\] on equation (2) we get, 

\[\tan \left[ \dfrac{1}{2}{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)+\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-{{y}^{2}}}{1+{{y}^{2}}} \right) \right]=\dfrac{\tan \theta +\tan \alpha }{1-\tan \theta \tan \alpha }\]                    …..(3)

Re-substituting $\theta ={{\tan }^{-1}}\left( x \right)$ and $\alpha ={{\tan }^{-1}}\left( y \right)$ in equation (3) we get, 

\[\tan \left[ \dfrac{1}{2}{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)+\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-{{y}^{2}}}{1+{{y}^{2}}} \right) \right]=\dfrac{\tan \left( {{\tan }^{-1}}x \right)+\tan \left( {{\tan }^{-1}}y \right)}{1-\tan \left( {{\tan }^{-1}}x \right)\cdot \tan \left( {{\tan }^{-1}}y \right)}\] 

\[\Rightarrow \tan \left[ \dfrac{1}{2}{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)+\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-{{y}^{2}}}{1+{{y}^{2}}} \right) \right]=\dfrac{x+y}{1-xy}\] 

11. Solve the following for $x$: \[{{\cos }^{-1}}\left( \dfrac{{{x}^{2}}-1}{{{x}^{2}}+1} \right)+\dfrac{1}{2}{{\tan }^{-1}}\left( \dfrac{-2x}{1-{{x}^{2}}} \right)=\dfrac{2\pi }{3}\]

Ans: To solve this question, the substitution $x=\tan \theta $ in the LHS of the given equation 

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

Let this be known as equation (1).

Using the trigonometric identities \[\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta }=\tan 2\theta \] and \[\dfrac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta }=\cos 2\theta \] 

From (1) we get, 

\[{{\cos }^{-1}}\left( \dfrac{{{x}^{2}}-1}{{{x}^{2}}+1} \right)+\dfrac{1}{2}{{\tan }^{-1}}\left( \dfrac{-2x}{1-{{x}^{2}}} \right)={{\cos }^{-1}}\left( -\cos 2\theta  \right)+\dfrac{1}{2}{{\tan }^{-1}}\left( -\tan 2\theta  \right)\]            …..(2)

But, 

${{\cos }^{-1}}\left( -x \right)=\pi -{{\cos }^{-1}}\left( x \right)$ 

${{\tan }^{-1}}\left( -x \right)=-{{\tan }^{-1}}\left( x \right)$ 

Therefore from (2) we get,

\[{{\cos }^{-1}}\left( \dfrac{{{x}^{2}}-1}{{{x}^{2}}+1} \right)+\dfrac{1}{2}{{\tan }^{-1}}\left( \dfrac{-2x}{1-{{x}^{2}}} \right)=\pi -{{\cos }^{-1}}\left( \cos 2\theta  \right)-\dfrac{1}{2}{{\tan }^{-1}}\left( \tan 2\theta  \right)\]

\[\Rightarrow {{\cos }^{-1}}\left( \dfrac{{{x}^{2}}-1}{{{x}^{2}}+1} \right)+\dfrac{1}{2}{{\tan }^{-1}}\left( \dfrac{-2x}{1-{{x}^{2}}} \right)=\pi -2\theta -\theta \] 

\[\Rightarrow {{\cos }^{-1}}\left( \dfrac{{{x}^{2}}-1}{{{x}^{2}}+1} \right)+\dfrac{1}{2}{{\tan }^{-1}}\left( \dfrac{-2x}{1-{{x}^{2}}} \right)=\pi -3\theta \] 

Let this be known as equation (3).

Re substituting $\theta ={{\tan }^{-1}}\left( x \right)$ in equation (3) we get,

\[{{\cos }^{-1}}\left( \dfrac{{{x}^{2}}-1}{{{x}^{2}}+1} \right)+\dfrac{1}{2}{{\tan }^{-1}}\left( \dfrac{-2x}{1-{{x}^{2}}} \right)=\pi -3{{\tan }^{-1}}x\]                                         …..(4)

But it is given that \[{{\cos }^{-1}}\left( \dfrac{{{x}^{2}}-1}{{{x}^{2}}+1} \right)+\dfrac{1}{2}{{\tan }^{-1}}\left( \dfrac{-2x}{1-{{x}^{2}}} \right)=\dfrac{2\pi }{3}\]. Therefore from (4) we get, 

\[\pi -3{{\tan }^{-1}}x=\dfrac{2\pi }{3}\] 

\[\Rightarrow 3{{\tan }^{-1}}x=\dfrac{\pi }{3}\] 

\[\Rightarrow x=\tan \left( \dfrac{\pi }{9} \right)\] 

12. Prove that \[{{\tan }^{-1}}\left( \dfrac{1}{3} \right)+{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\tan }^{-1}}\left( \dfrac{1}{7} \right)+{{\tan }^{-1}}\left( \dfrac{1}{8} \right)=\dfrac{\pi }{4}\]

Ans: To solve this problem use the trigonometric identity, ${{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)={{\tan }^{-1}}\left( x \right)+{{\tan }^{-1}}\left( y \right)$ in the LHS of the given expression

Hence, 

\[{{\tan }^{-1}}\left( \dfrac{1}{3} \right)+{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\tan }^{-1}}\left( \dfrac{1}{7} \right)+{{\tan }^{-1}}\left( \dfrac{1}{8} \right)={{\tan }^{-1}}\left( \dfrac{\dfrac{1}{3}+\dfrac{1}{5}}{1-\dfrac{1}{15}} \right)+{{\tan }^{-1}}\left( \dfrac{\dfrac{1}{7}+\dfrac{1}{8}}{1-\dfrac{1}{56}} \right)\]

\[\Rightarrow {{\tan }^{-1}}\left( \dfrac{1}{3} \right)+{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\tan }^{-1}}\left( \dfrac{1}{7} \right)+{{\tan }^{-1}}\left( \dfrac{1}{8} \right)={{\tan }^{-1}}\left( \dfrac{4}{7} \right)+{{\tan }^{-1}}\left( \dfrac{3}{11} \right)\] 

Let this be known as equation (1).

Again, using the trigonometric identity ${{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)={{\tan }^{-1}}\left( x \right)+{{\tan }^{-1}}\left( y \right)$ on (1) we get,

\[{{\tan }^{-1}}\left( \dfrac{1}{3} \right)+{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\tan }^{-1}}\left( \dfrac{1}{7} \right)+{{\tan }^{-1}}\left( \dfrac{1}{8} \right)={{\tan }^{-1}}\left( \dfrac{\dfrac{4}{7}+\dfrac{3}{11}}{1-\dfrac{12}{77}} \right)\]

\[\Rightarrow {{\tan }^{-1}}\left( \dfrac{1}{3} \right)+{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\tan }^{-1}}\left( \dfrac{1}{7} \right)+{{\tan }^{-1}}\left( \dfrac{1}{8} \right)={{\tan }^{-1}}\left( \dfrac{65}{65} \right)\] 

\[\Rightarrow {{\tan }^{-1}}\left( \dfrac{1}{3} \right)+{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\tan }^{-1}}\left( \dfrac{1}{7} \right)+{{\tan }^{-1}}\left( \dfrac{1}{8} \right)={{\tan }^{-1}}\left( 1 \right)\] 

Let this be known as equation (2).

But we know that,

$\tan \left( \dfrac{\pi }{4} \right)=1$

$\Rightarrow {{\tan }^{-1}}\left( 1 \right)=\dfrac{\pi }{4}$ 

Hence from equation (2) it is proved that \[{{\tan }^{-1}}\left( \dfrac{1}{3} \right)+{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\tan }^{-1}}\left( \dfrac{1}{7} \right)+{{\tan }^{-1}}\left( \dfrac{1}{8} \right)=\dfrac{\pi }{4}\]

13. Solve the following for $x$: \[\tan \left( {{\cos }^{-1}}x \right)=\sin \left( {{\tan }^{-1}}2 \right),\text{ }x>0\]

Ans: To solve this question, write ${{\cos }^{-1}}x$ in LHS of the given expression in terms of tan inverse. Substitute, 

\[{{\cos }^{-1}}x=\theta \]

\[\Rightarrow \cos \theta =x\] 

Let this be known as equation (1).

Now using the trigonometric identity $\sin \theta =\sqrt{1-{{\cos }^{2}}\theta }$ we get from (1),

 $\sin \theta =\sqrt{1-{{x}^{2}}}$                                                                                            …..(2)

Hence from (1) and (2) we get,

$\tan \theta =\dfrac{\sin \theta }{\cos \theta }$

$\Rightarrow \tan \theta =\dfrac{\sqrt{1-{{x}^{2}}}}{x}$ 

Let this be known as equation (3).

Taking tan inverse on both sides of equation (3) we get

${{\tan }^{-1}}\left( \tan \theta  \right)={{\tan }^{-1}}\left( \dfrac{\sqrt{1-{{x}^{2}}}}{x} \right)$ 

$\Rightarrow \theta ={{\tan }^{-1}}\left( \dfrac{\sqrt{1-{{x}^{2}}}}{x} \right)$ 

Let this be known as equation (4).

Hence from (4), we get the given equation as

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

\[\Rightarrow \dfrac{\sqrt{1-{{x}^{2}}}}{x}=\sin \left( {{\tan }^{-1}}2 \right)\] 

Let this be known as equation (5).

Now again write ${{\tan }^{-1}}2$ from RHS of equation (5) in terms of sine inverse. Substitute, 

\[{{\tan }^{-1}}2=y\]

\[\Rightarrow \tan y=2\] 

Let this be known as equation (6).

Now using the trigonometric identity $\sqrt{1+{{\tan }^{2}}\theta }=\sec \theta $ from (6) we get,

 $\sec y=\sqrt{1+{{\left( 2 \right)}^{2}}}$

$\Rightarrow \sec y=\sqrt{5}$

$\Rightarrow \cos y=\dfrac{1}{\sqrt{5}}$ 

Let this be known as equation (7).

Using the trigonometric identity $\sin \theta =\sqrt{1-{{\cos }^{2}}\theta }$ we get from (7),

 $\sin y=\sqrt{1-{{\left( \dfrac{1}{\sqrt{5}} \right)}^{2}}}$

$\Rightarrow \sin y=\dfrac{2}{\sqrt{5}}$

$\Rightarrow y={{\sin }^{-1}}\left( \dfrac{2}{\sqrt{5}} \right)$ 

Let this be known as equation (8).

 Hence from equations (5), (6) and (8) we get,

\[\dfrac{\sqrt{1-{{x}^{2}}}}{x}=\dfrac{2}{\sqrt{5}}\]                                                                                   …..(9)

Squaring both sides and cross multiplying we get,

\[5\left( 1-{{x}^{2}} \right)=4{{x}^{2}}\]

\[\Rightarrow 5=9{{x}^{2}}\]

\[\Rightarrow x=\dfrac{\sqrt{5}}{3}\] 

(Since $x>0$)

14. Prove that \[2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\tan }^{-1}}\left( \dfrac{1}{4} \right)={{\tan }^{-1}}\left( \dfrac{32}{43} \right)\]

Ans: To solve this question, use the trigonometric identity, ${{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)=2{{\tan }^{-1}}\left( x \right)$ in the LHS of the given expression.

\[2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\tan }^{-1}}\left( \dfrac{1}{4} \right)={{\tan }^{-1}}\left( \dfrac{2\left( \dfrac{1}{5} \right)}{1-{{\left( \dfrac{1}{5} \right)}^{2}}} \right)+{{\tan }^{-1}}\left( \dfrac{1}{4} \right)\]

\[\Rightarrow 2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\tan }^{-1}}\left( \dfrac{1}{4} \right)={{\tan }^{-1}}\left( \dfrac{5}{12} \right)+{{\tan }^{-1}}\left( \dfrac{1}{4} \right)\] 

Let this be known as equation (1).

Now use the trigonometric identity, ${{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)={{\tan }^{-1}}\left( x \right)+{{\tan }^{-1}}\left( y \right)$ on equation (1) we get,

\[2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\tan }^{-1}}\left( \dfrac{1}{4} \right)={{\tan }^{-1}}\left( \dfrac{\dfrac{5}{12}+\dfrac{1}{4}}{1-\dfrac{5}{48}} \right)\]

\[\Rightarrow 2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\tan }^{-1}}\left( \dfrac{1}{4} \right)={{\tan }^{-1}}\left( \dfrac{32}{43} \right)\] 

15. Evaluate \[\tan \left[ \dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{3}{\sqrt{11}} \right) \right]\]

Ans: To solve this question, use the substitution \[\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{3}{\sqrt{11}} \right)=x\] in the LHS of the given expression. Therefore, 

\[\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{3}{\sqrt{11}} \right)=x\]

\[\Rightarrow \cos 2x=\dfrac{3}{\sqrt{11}}\] 

Let this be known as equation (1).

But we know that \[\dfrac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x}=\cos 2x\]. Hence from (1) we get,

\[\dfrac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x}=\dfrac{3}{\sqrt{11}}\]                                                                    …..(2)

Applying the rule of component and divided on equation (2) we get,

\[\dfrac{1-{{\tan }^{2}}x+1+{{\tan }^{2}}x}{1-{{\tan }^{2}}x-1-{{\tan }^{2}}x}=\dfrac{3+\sqrt{11}}{3-\sqrt{11}}\]

\[\Rightarrow \dfrac{2}{-2{{\tan }^{2}}x}=\dfrac{3+\sqrt{11}}{3-\sqrt{11}}\] 

Let this be known as equation (3).

Taking reciprocal of equation (3) we get,

\[-{{\tan }^{2}}x=\dfrac{3-\sqrt{11}}{3+\sqrt{11}}\]

\[\Rightarrow {{\tan }^{2}}x=\dfrac{\sqrt{11}-3}{3+\sqrt{11}}\]

\[\Rightarrow \tan x=\sqrt{\dfrac{\sqrt{11}-3}{3+\sqrt{11}}}\] 

Let this be known as equation (4).

Now re-substituting \[\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{3}{\sqrt{11}} \right)=x\] in (4) we get,

\[\tan \left[ \dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{3}{\sqrt{11}} \right) \right]=\sqrt{\dfrac{\sqrt{11}-3}{3+\sqrt{11}}}\]

16. Prove that \[{{\tan }^{-1}}\left( \dfrac{a\cos x-b\sin x}{b\cos x+a\sin x} \right)={{\tan }^{-1}}\left( \dfrac{a}{b} \right)-x\]

Ans: To solve this question, take $b\cos x$ common from both numerator and denominator from the LHS of the given equation. 

\[{{\tan }^{-1}}\left( \dfrac{a\cos x-b\sin x}{b\cos x+a\sin x} \right)={{\tan }^{-1}}\left( \dfrac{\dfrac{a}{b}-\tan x}{1+\dfrac{a}{b}\tan x} \right)\]                          …..(1)

Now use the trigonometric identity, ${{\tan }^{-1}}\left( \dfrac{x-y}{1+xy} \right)={{\tan }^{-1}}\left( x \right)-{{\tan }^{-1}}\left( y \right)$ on equation (1) we get,

\[{{\tan }^{-1}}\left( \dfrac{a\cos x-b\sin x}{b\cos x+a\sin x} \right)={{\tan }^{-1}}\left( \dfrac{a}{b} \right)-{{\tan }^{-1}}\left( \tan x \right)\]

\[\Rightarrow {{\tan }^{-1}}\left( \dfrac{a\cos x-b\sin x}{b\cos x+a\sin x} \right)={{\tan }^{-1}}\left( \dfrac{a}{b} \right)-x\] 

17. Prove that \[\cot \left\{ {{\tan }^{-1}}x+{{\tan }^{-1}}\left( \dfrac{1}{x} \right) \right\}+{{\cos }^{-1}}\left( 1-2{{x}^{2}} \right)+{{\cos }^{-1}}\left( 2{{x}^{2}}-1 \right)=\pi \], $x>0$ 

Ans: To solve this question, use the trigonometric identity \[{{\tan }^{-1}}x+{{\cot }^{-1}}x=\dfrac{\pi }{2}\] on \[\cot \left\{ {{\tan }^{-1}}x+{{\tan }^{-1}}\left( \dfrac{1}{x} \right) \right\}\] . Therefore, 

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

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

Let this be known as equation (1).

Now using the trigonometric identity \[{{\cot }^{-1}}\left( \dfrac{xy-1}{x+y} \right)={{\cot }^{-1}}\left( x \right)+{{\cot }^{-1}}\left( y \right)\] on equation (1) we get, 

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

\[\Rightarrow \left\{ {{\tan }^{-1}}x+{{\tan }^{-1}}\left( \dfrac{1}{x} \right) \right\}=\cot \left\{ \pi -\left[ {{\cot }^{-1}}\left( 0 \right) \right] \right\}\] 

Let this be known as equation (2).

But we know that the principal value of

 $\cot \left( \dfrac{\pi }{2} \right)=0$

$\Rightarrow {{\cot }^{-1}}\left( 0 \right)=\dfrac{\pi }{2}$ 

Hence, from equation (2) we get,

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

\[\Rightarrow \cot \left\{ {{\tan }^{-1}}x+{{\tan }^{-1}}\left( \dfrac{1}{x} \right) \right\}=0\] 

Let this be known as equation (3).

Also, we know that ${{\cos }^{-1}}\left( -x \right)=\pi -{{\cos }^{-1}}\left( x \right)$. Therefore, 

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

\[\Rightarrow {{\cos }^{-1}}\left( 1-2{{x}^{2}} \right)+{{\cos }^{-1}}\left( 2{{x}^{2}}-1 \right)={{\cos }^{-1}}\left( 1-2{{x}^{2}} \right)+\pi -{{\cos }^{-1}}\left( 2{{x}^{2}}-1 \right)\]

\[\Rightarrow {{\cos }^{-1}}\left( 1-2{{x}^{2}} \right)+{{\cos }^{-1}}\left( 2{{x}^{2}}-1 \right)=\pi \] 

Let this be known as equation (4).

Hence from equations (3) and (4) we get,

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

18. Prove that \[{{\tan }^{-1}}\left( \dfrac{a-b}{1+ab} \right)+{{\tan }^{-1}}\left( \dfrac{b-c}{1+bc} \right)+{{\tan }^{-1}}\left( \dfrac{c-a}{1+ac} \right)=0\] where $a,b,c>0$ 

Ans: To solve this question, use the trigonometric identity, ${{\tan }^{-1}}\left( \dfrac{x-y}{1+xy} \right)={{\tan }^{-1}}\left( x \right)-{{\tan }^{-1}}\left( y \right)$ on the LHS of the given expression. Therefore,

\[{{\tan }^{-1}}\left( \dfrac{a-b}{1+ab} \right)+{{\tan }^{-1}}\left( \dfrac{b-c}{1+bc} \right)+{{\tan }^{-1}}\left( \dfrac{c-a}{1+ac} \right)\]

\[\text{ }=\left( {{\tan }^{-1}}\left( a \right)-{{\tan }^{-1}}\left( b \right) \right)+\left( {{\tan }^{-1}}\left( b \right)-{{\tan }^{-1}}\left( c \right) \right)+\left( {{\tan }^{-1}}\left( c \right)-{{\tan }^{-1}}\left( a \right) \right)\]

\[\text{ }=0\] 

19. Solve the following for $x$: \[2{{\tan }^{-1}}\left( \cos x \right)={{\tan }^{-1}}\left( 2\text{cosec }x \right)\]

Ans: To solve this question, use the trigonometric identity, $2{{\tan }^{-1}}\left( x \right)={{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)$ on the LHS of the given expression.

 \[2{{\tan }^{-1}}\left( \cos x \right)={{\tan }^{-1}}\left( \dfrac{2\cos x}{1-{{\cos }^{2}}x} \right)\]                                                         …..(1)

Using the trigonometric identity ${{\sin }^{2}}x=1-{{\cos }^{2}}x$ in the denominator of (1) we get, 

\[2{{\tan }^{-1}}\left( \cos x \right)={{\tan }^{-1}}\left( \dfrac{2\cos x}{{{\sin }^{2}}x} \right)\]                                                            …..(2)

Substituting the value from equation (2) into the given expression we get,

\[{{\tan }^{-1}}\left( \dfrac{2\cos x}{{{\sin }^{2}}x} \right)={{\tan }^{-1}}\left( 2\text{cosec }x \right)\]                                                        …..(3)

Taking tan on both the sides of equation (3) we get, 

\[\tan \left( {{\tan }^{-1}}\left( \dfrac{2\cos x}{{{\sin }^{2}}x} \right) \right)=\tan \left( {{\tan }^{-1}}\left( 2\text{cosec }x \right) \right)\]

\[\Rightarrow \dfrac{2\cos x}{{{\sin }^{2}}x}=2\text{cosec }x\]

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

\[\Rightarrow \cos x=\sin x\] 

It is possible only when $x=\dfrac{\pi }{4}$. 

20. Express \[{{\sin }^{-1}}\left( x\sqrt{1-x}-\sqrt{x}\sqrt{1-{{x}^{2}}} \right)\] in the simplest form. 

Ans: To solve this question, use the substitution $x=\sin \theta $ and $\sqrt{x}=\sin \phi $ . Therefore,

\[{{\sin }^{-1}}\left( x\sqrt{1-x}-\sqrt{x}\sqrt{1-{{x}^{2}}} \right)={{\sin }^{-1}}\left( \sin \theta \sqrt{1-{{\sin }^{2}}\phi }-\sin \phi \sqrt{1-{{\sin }^{2}}\theta } \right)\]

\[\Rightarrow {{\sin }^{-1}}\left( x\sqrt{1-x}-\sqrt{x}\sqrt{1-{{x}^{2}}} \right)={{\sin }^{-1}}\left( \sin \theta \cos \phi -\sin \phi \cos \theta  \right)\] 

Let this be known as equation (1).

But we know that $\sin a-\sin b=\sin a\cos b-\cos a\sin b$. Hence from (1) we get, 

\[{{\sin }^{-1}}\left( x\sqrt{1-x}-\sqrt{x}\sqrt{1-{{x}^{2}}} \right)={{\sin }^{-1}}\left( \sin \left( \theta -\phi  \right) \right)\]

\[\Rightarrow {{\sin }^{-1}}\left( x\sqrt{1-x}-\sqrt{x}\sqrt{1-{{x}^{2}}} \right)=\theta -\phi \] 

Let this be known as equation (2).

Re-substituting $x=\sin \theta $ and $\sqrt{x}=\sin \phi $ in equation (2) we get,

\[{{\sin }^{-1}}\left( x\sqrt{1-x}-\sqrt{x}\sqrt{1-{{x}^{2}}} \right)={{\sin }^{-1}}\left( x \right)-{{\sin }^{-1}}\left( \sqrt{x} \right)\]

21. If \[{{\tan }^{-1}}\left( a \right)+{{\tan }^{-1}}\left( b \right)+{{\tan }^{-1}}\left( c \right)=\pi \] then prove that $a+b+c=abc$. 

Ans: To solve this question, use the trigonometric identity, ${{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)={{\tan }^{-1}}\left( x \right)+{{\tan }^{-1}}\left( y \right)$ on the LHS of the given expression. Therefore,

 \[{{\tan }^{-1}}\left( a \right)+{{\tan }^{-1}}\left( b \right)={{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right)\]                                               …..(1)

Again, using the same trigonometric identity on the LHS of the given expression and using (1), we get, 

\[{{\tan }^{-1}}\left( a \right)+{{\tan }^{-1}}\left( b \right)+{{\tan }^{-1}}\left( c \right)={{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right)+{{\tan }^{-1}}\left( c \right)\]

\[\Rightarrow {{\tan }^{-1}}\left( a \right)+{{\tan }^{-1}}\left( b \right)+{{\tan }^{-1}}\left( c \right)={{\tan }^{-1}}\left( \dfrac{\dfrac{a+b}{1-ab}+c}{1-\dfrac{c\left( a+b \right)}{1-ab}} \right)\]

\[\Rightarrow {{\tan }^{-1}}\left( a \right)+{{\tan }^{-1}}\left( b \right)+{{\tan }^{-1}}\left( c \right)={{\tan }^{-1}}\left( \dfrac{a+b+c-abc}{1-ab-ca-bc} \right)\] 

Let this be known as equation (2).

But it is given that \[{{\tan }^{-1}}\left( a \right)+{{\tan }^{-1}}\left( b \right)+{{\tan }^{-1}}\left( c \right)=\pi \]. Therefore, from equation (2) we get, 

\[{{\tan }^{-1}}\left( \dfrac{a+b+c-abc}{1-ab-ca-bc} \right)=\pi \]                                                                      …..(3)

Taking tan on both sides of the equation (3) we get, 

\[\tan \left( {{\tan }^{-1}}\left( \dfrac{a+b+c-abc}{1-ab-ca-bc} \right) \right)=\tan \left( \pi  \right)\]

\[\Rightarrow \dfrac{a+b+c-abc}{1-ab-ca-bc}=0\]

\[\Rightarrow a+b+c-abc=0\] 

Hence proved that if \[{{\tan }^{-1}}\left( a \right)+{{\tan }^{-1}}\left( b \right)+{{\tan }^{-1}}\left( c \right)=\pi \] then $a+b+c=abc$.

22. If \[{{\sin }^{-1}}x>{{\cos }^{-1}}x\] then \[x\] belong to which category? 

Ans: To solve this question, use the trigonometric identity, ${{\sin }^{-1}}\left( x \right)+{{\cos }^{-1}}\left( x \right)=\dfrac{\pi }{2}$.

 \[{{\sin }^{-1}}x>{{\cos }^{-1}}x\]

\[\Rightarrow {{\sin }^{-1}}x>\dfrac{\pi }{2}-{{\sin }^{-1}}x\]

\[\Rightarrow 2{{\sin }^{-1}}x>\dfrac{\pi }{2}\]

\[\Rightarrow {{\sin }^{-1}}x>\dfrac{\pi }{4}\] 

Let this be known as expression (1).

Taking sine on both the sides of expression (1) we get, 

\[\sin \left( {{\sin }^{-1}}x \right)>\sin \left( \dfrac{\pi }{4} \right)\]

\[\Rightarrow x>\sin \left( \dfrac{\pi }{4} \right)\]

\[\Rightarrow x>\dfrac{1}{\sqrt{2}}\] 

Let this be known as expression (2).

Also, we know that $-1\le x\le 1$ (since range of $\sin $ and $\cos $ is $\left[ -1,1 \right]$.              …..(3)

Hence from (2) and (3), the interval in which $x$ belongs is $\left( \dfrac{1}{\sqrt{2}},1 \right]$.  

Inverse Trigonometric Functions Class 12 Important Questions - Free PDF Download

Chapter 2 - Inverse Trigonometry

Introduction

The opposite operations that the sine, cosine, tangent, secant, cosecant, and cotangent perform are provided by the inverse trigonometric functions. They are used in a right triangle to find the measure of an angle when two of the three side lengths are identified.

Principal Inverse Trigonometric Functions with Domain and Range

Properties of  Inverse Trigonometric Functions 

In order to not only solve problems but also to provide a deeper understanding of this idea, the properties of the 6 inverse trigonometric functions are important. The properties of inverse trigonometric relations are nothing but the relationship between the 6 fundamental trigonometric functions.

Here is the representation of important properties involving inverse trigonometric functions for quick reference:

The solution to Class 12 Maths Chapter 2 Important Questions provided by Vedantu has steps to derive the properties of the above-mentioned inverse trigonometric functions.

Benefits of Class 12 Science Chapter 2 Inverse Trigonometric Functions

• Foundation for Advanced Mathematics: Inverse trigonometric functions are a cornerstone for higher-level math concepts like calculus, integration, and differential equations, which are critical in engineering, physics, and other technical fields.

• Real-Life Applications: Understanding this chapter helps in solving real-world problems involving angles, distances, and periodic phenomena, such as navigation, signal processing, and wave mechanics.

• Improved Problem-Solving Skills: By mastering the properties and formulas, students can tackle complex trigonometric and algebraic problems more effectively, enhancing their analytical abilities.

• Competitive Exam Preparation: Topics in this chapter frequently appear in competitive exams like JEE, NEET, and other entrance tests, making it essential for scoring well.

• Conceptual Clarity: Learning about the relationships between different trigonometric functions and their inverses sharpens conceptual understanding, aiding in better comprehension of subsequent chapters.

Tips to Study Class 12 Maths Chapter 2 Inverse Trigonometric Functions

• Understand Basic Trigonometric Functions: Before diving into inverse trigonometric functions, ensure a strong grasp of basic trigonometric concepts and their properties, as they form the foundation for this chapter.

• Memorise Key Formulas: Create a formula sheet for inverse trigonometric properties, such as addition/subtraction formulas and relationships between functions, and revise it regularly for quick recall.

• Learn Domain and Range: Pay close attention to the domain and range of each inverse trigonometric function. This understanding is critical when solving equations and graphing functions.

• Practice Graphs: Practice sketching the graphs of inverse trigonometric functions. This will help in visualizing their behavior and solving graphical questions efficiently.

• Focus on Properties: Familiarise yourself with important properties like $\sin^{-1}(-x) = -\sin^{-1}(x)$ and $\tan^{-1}(x) + \cot^{-1}(x) = \frac{\pi}{2}$​. These are frequently used in proofs and simplifications.

• Work on Conceptual Clarity: Avoid rote memorisation. Instead, focus on understanding why properties and formulas work through examples and derivations.

Conclusion

Vedantu's online learning platform provides important questions for Class 12 Maths Chapter 2 - Inverse Trigonometric Functions. These questions have been curated by subject matter experts to help students revise the chapter thoroughly. These questions cover all the important topics of the chapter, such as the concept of inverse trigonometric functions and their properties, and help students to prepare well for their exams. The questions are designed to test the students' understanding of the concepts and their problem-solving skills. Students can practice these questions to improve their performance in the exams and gain a better understanding of the chapter.

Related Study Materials for Class 12 Maths Chapter 2 Inverse Trigonometric Functions

CBSE Class 12 Maths Chapter-wise Important Questions

CBSE Class 12 Maths Chapter-wise Important Questions and Answers cover topics from all 13 chapters, helping students prepare thoroughly by focusing on key topics for easier revision.

Additional Study Materials for Class 12 Maths