Inverse Trigonometric Functions Class 12 Notes: CBSE Maths Chapter 2

Domain and Range of all Inverse Trigonometric Functions

• We must note that inverse trigonometric functions cannot be expressed in terms of trigonometric functions as their reciprocals. For example, ${{\sin }^{-1}}x\ne \dfrac{1}{\sin x}$.

• The principal value of a trigonometric function is that value which lies in the range of principal branch.

• The functions \[{{\sin }^{-1}}x\text{  }\!\!\And\!\!\text{  }{{\tan }^{-1}}x\] are increasing functions in their domain.

• The functions \[{{\cos }^{-1}}x\text{  }\!\!\And\!\!\text{  }{{\cot }^{-1}}x\] are decreasing functions in over domain.

Graphs of Inverse Trigonometric Functions

a) Graph of ${{\sin }^{-1}}x$ is shown below,

b) Graph of ${{\cos }^{-1}}x$ is shown below,

c) Graph of ${{\tan }^{-1}}x$ is shown below,

d) Graph of $\cos e{{c}^{-1}}x$ is shown below,

e) Graph of ${{\sec }^{-1}}x$ is shown below,

f) Graph of ${{\cot }^{-1}}x$ is shown below,

Properties of Inverse Trigonometric Functions

1. Property I

(a). \[{{\sin }^{-1}}\left( \dfrac{1}{x} \right)=\cos e{{c}^{-1}}x\], for all \[x\in \left( -\infty ,1 \right]\cup \left[ 1,\infty  \right)\]

Let us prove this by considering \[\cos e{{c}^{-1}}x=\theta \] ……(i)

Taking \[\cos ec\] on both sides,

\[x=\cos ec\theta \]

Using reciprocal identity,

\[\Rightarrow \dfrac{1}{x}=\sin \theta \]

\[\left\{ \because x\in \left( -\infty ,-1 \right]\cup \left[ 1,\infty  \right) \right\}\Rightarrow \dfrac{1}{x}\in \left[ -1,1 \right]\left\{ 0 \right\}\]

\[\cos e{{c}^{-1}}x=\theta \Rightarrow \theta \in \left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]-\left\{ 0 \right\}\]

\[\Rightarrow \theta ={{\sin }^{-1}}\left( \dfrac{1}{x} \right)\] ……(ii)

From (i) and (ii), we get

\[{{\sin }^{-1}}\left( \dfrac{1}{x} \right)=\cos e{{c}^{-1}}x\]

Hence proved.

b). \[{{\cos }^{-1}}\left( \dfrac{1}{x} \right)={{\sec }^{-1}}x\], for all \[x\in \left( -\infty ,1 \right]\cup \left[ 1,\infty  \right)\]

Let us prove this by taking \[{{\sec }^{-1}}x=\theta \]   ……(i)

Taking \[\sec \] on both sides,

\[\Rightarrow x=\sec \theta \]

Using reciprocal identity,

\[\Rightarrow \dfrac{1}{x}=\cos \theta \]

\[\Rightarrow \theta ={{\cos }^{-1}}\left( \dfrac{1}{x} \right)\] ……(ii)

Then, \[x\in \left( -\infty ,1 \right]\cup \left[ 1,\infty  \right)\] and \[\theta \in \left[ 0,\pi  \right]-\left\{ \dfrac{\pi }{2} \right\}\]

\[\left\{ \begin{align} & \because x=\left( -\infty ,-1 \right]\cup \left[ 1,\infty  \right) \\ & \Rightarrow \dfrac{1}{x}\in \left[ -1,1 \right]-\left\{ 0 \right\}\text{ and }\theta \in \left[ 0,\pi  \right] \\ \end{align} \right.\]

From (i) and (ii), we get

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

Hence proved.

c. \[{{\tan }^{-1}}\left( \dfrac{1}{x} \right)=\left\{ \begin{align} & {{\cot }^{-1}}x,\text{ for }x>0 \\ & -\pi +{{\cot }^{-1}}x,\text{ for }x < 0 \\ \end{align} \right.\] 

Let us prove this by taking \[{{\cot }^{-1}}x=\theta \]. Then \[x\in R,x\ne 0\] and \[\theta \in \left[ 0,\pi  \right]\]   ……(i)

Now there are two cases that arise:

Case I: When \[x>0\]

In this case, we have \[\theta \in \left( 0,\dfrac{\pi }{2} \right)\]

Considering \[{{\cot }^{-1}}x=\theta \] 

Taking \[\cot \] on both sides,

\[\Rightarrow x=\cot \theta \]

Using reciprocal property,                               

\[\Rightarrow \dfrac{1}{x}=\tan \theta \]

\[\theta ={{\tan }^{-1}}\left( \dfrac{1}{x} \right)\]  ……(ii)

From (i) and (ii), we get   \[\left\{ \because \theta \in \left( 0,\dfrac{\pi }{2} \right) \right\}\]

\[{{\tan }^{-1}}\left( \dfrac{1}{x} \right)={{\cot }^{-1}}x\], for all \[x>0\]

Case II: When \[x < 0\]

In this case, we have \[\theta \in \left( \dfrac{\pi }{2},\pi \right)\text{ }\left\{ \because x=\cot \theta < 0 \right\}\]

Now, \[\dfrac{\pi }{2} < \theta  < \pi \]

\[\Rightarrow -\dfrac{\pi }{2} < \theta -\pi  < 0\]

\[\Rightarrow \theta -\pi \in \left( -\dfrac{\pi }{2},0 \right)\]

\[\therefore {{\cot }^{-1}}x=\theta \]

Taking \[\cot \] on both sides,

\[\Rightarrow x=\cot \theta \]

Using reciprocal property,

\[\Rightarrow \dfrac{1}{x}=\tan \theta \]

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

\[\Rightarrow \dfrac{1}{x}=\tan \left( \theta -\pi \right)\text{ }\left\{ \because \tan \left( \pi -\theta \right)=-\tan \theta \right\}\]

\[\Rightarrow \theta -\pi ={{\tan }^{-1}}\left( \dfrac{1}{x} \right)\text{ }\left\{ \because \theta -\pi \in \left( -\dfrac{\pi }{2},0 \right) \right\}\]

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

From (i) and (iii), we get

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

Hence it is proved that \[{{\tan }^{-1}}\left( \dfrac{1}{x} \right)=\left\{ \begin{align} & {{\cot }^{-1}}x,\text{ for }x>0 \\ & -\pi +{{\cot }^{-1}}x,\text{ for }x < 0 \\ \end{align} \right.\].

2. Property II

1. \[{{\sin }^{-1}}\left( -x \right)=-{{\sin }^{-1}}\left( x \right)\], for all \[x\in \left[ -1,1 \right]\]

2. \[{{\tan }^{-1}}\left( -x \right)=-{{\tan }^{-1}}x\], for all \[x\in R\]

3. \[\cos e{{c}^{-1}}\left( -x \right)=-\cos e{{c}^{-1}}x\], for all \[x\in \left( -\infty ,-1 \right]\cup \left[ 1,\infty  \right)\]

Clearly, \[-x\in \left[ -1,1 \right]\] for all \[x\in \left[ -1,1 \right]\]

Let us prove a) by taking \[{{\sin }^{-1}}\left( -x \right)=\theta \]

Then, taking $\sin $ on both sides, we get

\[-x=\sin \theta \]   ……(i)

\[\Rightarrow x=-\sin \theta \]

\[\Rightarrow x=\sin \left( -\theta  \right)\]

\[\Rightarrow -\theta ={{\sin }^{-1}}x\]

\[\left\{ \because x\in \left[ -1,1 \right]\text{ and }-\theta \in \left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]\text{ for all }\theta \in \left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right] \right\}\]

\[\Rightarrow \theta =-{{\sin }^{-1}}x\]  ……(ii)

From (i) and (ii), we get

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

Hence proved.

The b) and c) properties can also be proved in the similar manner.

3. Property III

1. \[{{\cos }^{-1}}\left( -x \right)=\pi -{{\cos }^{-1}}\left( x \right)\], for all \[x\in \left[ -1,1 \right]\]

2. \[{{\sec }^{-1}}\left( -x \right)=\pi -{{\sec }^{-1}}x\], for all \[x\in \left( -\infty ,-1 \right]\cup \left[ 1,\infty  \right)\]

3. \[{{\cot }^{-1}}\left( -x \right)=\pi -{{\cot }^{-1}}x\], for all \[x\in R\]

Clearly, \[-x\in \left[ -1,1 \right]\] for all \[x\in \left[ -1,1 \right]\]

Let us prove it by taking \[{{\cos }^{-1}}\left( -x \right)=\theta \]  ……(i)

Then, taking $\cos $ on both sides, we get

\[-x=\cos \theta \]

\[\Rightarrow x=-\cos \theta \]

\[\Rightarrow x=\cos \left( \pi -\theta  \right)\]

\[\left\{ \because x\in \left[ -1,1 \right]\text{ and }\pi -\theta \in \left[ 0,\pi  \right]\text{ for all }\theta \in \left[ 0,\pi  \right] \right\}\]

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

\[\Rightarrow \theta =\pi -{{\cos }^{-1}}x\] ……(ii)

From (i) and (ii), we get

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

Hence Proved.

The b) and c) properties can also be proved in the similar manner.

4.  Property IV

a) \[{{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2}\], for all \[x\in \left[ -1,1 \right]\]

Let us prove it by taking \[{{\sin }^{-1}}x=\theta \]  ……(i)

Then, \[\theta \in \left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]\text{ }\left[ \because x\in \left[ -1,1 \right] \right]\]

\[\Rightarrow -\dfrac{\pi }{2}\le \theta \le \dfrac{\pi }{2}\]

\[\Rightarrow -\dfrac{\pi }{2}\le -\theta \le \dfrac{\pi }{2}\]

\[\Rightarrow 0\le \dfrac{\pi }{2}-\theta \le \pi \]

\[\Rightarrow \dfrac{\pi }{2}-\theta \in \left[ 0,\pi  \right]\]

Now we consider \[{{\sin }^{-1}}x=\theta \]

Taking $\sin $ on both sides, we get

\[\Rightarrow x=\sin \theta \]

Changing functions, we get

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

\[\Rightarrow {{\cos }^{-1}}x=\dfrac{\pi }{2}-\theta \]

\[\left\{ \because x\in \left[ -1,1 \right]\text{ and }\left( \dfrac{\pi }{2}-\theta  \right)\in \left[ 0,\pi  \right] \right\}\]                                                

\[\Rightarrow \theta +{{\cos }^{-1}}x=\dfrac{\pi }{2}\] ……(ii)

From (i) and (ii), we get

\[{{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2}\]

Hence proved.

b) \[{{\tan }^{-1}}x+{{\cot }^{-1}}x=\dfrac{\pi }{2}\], for all \[x\in R\]

Let us prove it by taking \[{{\tan }^{-1}}x=\theta \] ……(i)

Then, \[\theta \in \left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)\text{ }\left\{ \because x\in R \right\}\]

\[\Rightarrow -\dfrac{\pi }{2} < \theta  < \dfrac{\pi }{2}\]

\[\Rightarrow -\dfrac{\pi }{2} < -\theta  < \dfrac{\pi }{2}\]

\[\Rightarrow 0 < \dfrac{\pi }{2}-\theta  < \pi \]

\[\Rightarrow \left( \dfrac{\pi }{2}-\theta  \right)\in \left( 0,\pi  \right)\]

Now consider \[{{\tan }^{-1}}x=\theta \]

Taking $\tan $ on both sides, we get

\[\Rightarrow x=\tan \theta \]

\[\Rightarrow x=\cot \left( \dfrac{\pi }{2}-\theta  \right)\]

\[\Rightarrow {{\cot }^{-1}}x=\dfrac{\pi }{2}-\theta \text{ }\left\{ \because \dfrac{\pi }{2}-\theta \in \left( 0,\pi \right) \right\}\]

\[\Rightarrow \theta +{{\cot }^{-1}}x=\dfrac{\pi }{2}\]   ……(ii)

From (i) and (ii), we get

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

c) \[{{\sec }^{-1}}x+\cos e{{c}^{-1}}x=\dfrac{\pi }{2}\], for all \[x\in \left( -\infty ,-1 \right]\cup \left[ 1,\infty  \right)\]

Let us prove it by taking \[{{\sec }^{-1}}x=\theta \]   ……(i)

Then, \[\theta \in \left[ 0,\pi \right]-\left\{ \dfrac{\pi }{2} \right\}\text{ }\left\{ \because x\in \left( -\infty ,-1 \right]\cup \left[ 1,\infty \right) \right\}\]

\[\Rightarrow 0\le \theta \le \pi ,\theta \ne \dfrac{\pi }{2}\]

\[\Rightarrow -\pi \le -\theta \le 0,\theta \ne \dfrac{\pi }{2}\]

\[\Rightarrow -\dfrac{\pi }{2}\le \dfrac{\pi }{2}-\theta \le \dfrac{\pi }{2},\dfrac{\pi }{2}-\theta \ne 0\]

\[\Rightarrow \left( \dfrac{\pi }{2}-\theta  \right)\in \left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right],\dfrac{\pi }{2}-\theta \ne 0\]

Now considering \[{{\sec }^{-1}}x=\theta \]

Taking $\sec $ on both sides, we get

\[\Rightarrow x=\sec \theta \]

\[\Rightarrow x=\cos ec\left( \dfrac{\pi }{2}-\theta  \right)\]

\[\Rightarrow \cos e{{c}^{-1}}x=\dfrac{\pi }{2}-\theta \]

\[\left\{ \because \left( \dfrac{\pi }{2}-\theta  \right)\in \left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right],\dfrac{\pi }{2}-\theta \ne 0 \right\}\]

\[\Rightarrow \theta +\cos e{{c}^{-1}}x=\dfrac{\pi }{2}\]   ….…(ii)

From (i) and (ii), we get

\[{{\sec }^{-1}}x+\cos e{{c}^{-1}}x=\dfrac{\pi }{2}\]

5. Property V

1. \[{{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\dfrac{x+y}{1-xy},xy < 1\]

2. \[{{\tan }^{-1}}x-{{\tan }^{-1}}y={{\tan }^{-1}}\dfrac{x-y}{1+xy},xy>-1\]

3. \[{{\tan }^{-1}}x+{{\tan }^{-1}}y=\pi +{{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right),xy>1;x,y=0\]

Let us prove a) by taking ${{\tan }^{-1}}x=\theta $ and ${{\tan }^{-1}}y=\phi $.

Taking $\tan $ on both sides for both terms, we get $x=\tan \theta $ and $y=\tan \phi $.

Using formula for $\tan \left( A+B \right)=\dfrac{\tan A+tanB}{1-\tan A\tan B}$, we can write

$\tan \left( \theta +\phi  \right)=\dfrac{\tan \theta +tan\phi }{1-\tan \theta \tan \phi }$

Writing in terms of $x\text{ }and\text{ }y$,

$\tan \left( \theta +\phi  \right)=\dfrac{x+y}{1-xy}$

$\Rightarrow \theta +\phi ={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)$

Therefore \[{{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\dfrac{x+y}{1-xy},xy < 1\].

Hence proved.

The properties b) and c) can be proved in similar manner by considering $y$ as $-y$ and $y$ as $x$ respectively in the above proof.

6.  Property VI

1. \[2{{\tan }^{-1}}x={{\sin }^{-1}}\dfrac{2x}{1+{{x}^{2}}},\left| x \right|\le 1\]

2. \[2{{\tan }^{-1}}x={{\cos }^{-1}}\dfrac{1-{{x}^{2}}}{1+{{x}^{2}}},x\ge 0\]

3. \[2{{\tan }^{-1}}x={{\tan }^{-1}}\dfrac{2x}{1-{{x}^{2}}},-1 < x < 1\]

Let us prove a) by taking ${{\tan }^{-1}}x=y$.

Taking $\tan $ on both sides, we get

$x=\tan y$

We can write ${{\sin }^{-1}}\dfrac{2x}{1+{{x}^{2}}}$ as ${{\sin }^{-1}}\dfrac{2\tan y}{1+{{\tan }^{2}}y}$.

Using formula $\sin 2x=\dfrac{2\tan x}{1+{{\tan }^{2}}x}$, we get

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

Using ${{\sin }^{-1}}\left( \sin x \right)=x$, this can be written as

${{\sin }^{-1}}\dfrac{2x}{1+{{x}^{2}}}=2y$

$\Rightarrow {{\sin }^{-1}}\dfrac{2x}{1+{{x}^{2}}}=2{{\tan }^{-1}}x$

Hence proved.

The same process can be followed to prove properties b) and c) as well.

7.  Property VII

1. \[\sin \left( {{\sin }^{-1}}x \right)=x\], for all \[x\in \left[ -1,1 \right]\]

2. \[\cos \left( {{\cos }^{-1}}x \right)=x\], for all \[x\in \left[ -1,1 \right]\]

3. \[\tan \left( {{\tan }^{-1}}x \right)=x\], for all \[x\in R\]

4. \[\cos ec\left( \cos e{{c}^{-1}}x \right)=x\], for all \[x\in \left( -\infty ,-1 \right]\cup \left[ 1,\infty  \right)\]

5. \[\sec \left( {{\sec }^{-1}}x \right)=x\], for all \[x\in \left( -\infty ,-1 \right]\cup \left[ 1,\infty  \right)\]

6. \[\cot \left( {{\cot }^{-1}}x \right)=x\], for all \[x\in R\]

Let us prove a). We know that, if \[f:A\to B\] is a bijection, then \[{{f}^{-1}}:B\to A\] exists such that \[fo{{f}^{-1}}\left( y \right)=f\left( {{f}^{-1}}\left( y \right) \right)=y\] for all \[y\in B\].

Clearly, all these results are direct consequences of this property.

Aliter: Let \[\theta \in \left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]\] and \[x\in \left[ -1,1 \right]\] such that \[\sin \theta =x\].

Taking $\sin $ on both sides, \[\theta ={{\sin }^{-1}}x\]

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

Hence, \[\sin \left( {{\sin }^{-1}}x \right)=x\] for all \[x\in \left[ -1,1 \right]\] and we proved it.

We can prove properties from b) to f) in a similar manner.

It should be noted that, \[{{\sin }^{-1}}\left( \sin \theta  \right)\ne \theta \], if \[\notin \left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]\].

Let us understand this better. The function \[y={{\sin }^{-1}}\left( \sin x \right)\] is periodic and has period \[2\pi \].

To draw this graph, we should draw the graph for one interval of length \[2\pi \] and repeat the entire values of x.

As we know,

\[{{\sin }^{-1}}\left( \sin x \right)=\left\{ \begin{align} & x;\text{ }-\dfrac{\pi }{2}\le x\le \dfrac{\pi }{2} \\ & \left( \pi -x \right);\text{ }-\dfrac{\pi }{2}\le \pi -x < \dfrac{\pi }{2}\left( \text{i}\text{.e}\text{.,}\dfrac{\pi }{2}\le x\le \dfrac{3\pi }{2} \right) \\ \end{align} \right.\]

\[\Rightarrow {{\sin }^{-1}}\left( \sin x \right)=\left\{ \begin{align} & x,\text{ }-\dfrac{\pi }{2}\le x\le \dfrac{\pi }{2} \\ & \pi -x,\text{ }\dfrac{\pi }{2}\le x\le \dfrac{3\pi }{2}, \\ \end{align} \right.\]

This is plotted as

Thus, we can note that the graph for \[y={{\sin }^{-1}}\left( \sin x \right)\] is a straight line up and a straight line down with slopes \[1\] and \[-1\] respectively lying between \[\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]\].

The below result for the definition of \[{{\sin }^{-1}}\left( \sin x \right)\] must be kept in mind. \[y={{\sin }^{-1}}\left( \sin x \right)=\left\{ \begin{align} & x+2\pi ;\text{ }-\dfrac{5\pi }{2}\le x\le -\dfrac{3\pi }{2} \\ & -\pi -x;\text{ }-\dfrac{3\pi }{2}\le x\le -\dfrac{\pi }{2} \\ & x;\text{ }-\dfrac{\pi }{2}\le x\le \dfrac{\pi }{2} \\ & \pi -x;\text{ }\dfrac{\pi }{2}\le x\le \dfrac{3\pi }{2} \\ & x-2\pi ;\text{ }\dfrac{3\pi }{2}\le x\le \dfrac{5\pi }{2}\text{ }...\text{and so on} \\ \end{align} \right.\]

Now we consider \[y={{\cos }^{-1}}\left( \cos x \right)\] which is periodic and has period \[2\pi \].

To draw this graph, we should draw the graph for one interval of length \[2\pi \] and repeat the entire values of \[x\] of length \[2\pi \]        

As we know,

\[{{\cos }^{-1}}\left( \cos x \right)=\left\{ \begin{align} & x;\text{ }0\le x\le \pi \\ & 2\pi -x;\text{ }0\le 2\pi -x\le \pi , \\ \end{align} \right.\] \[\Rightarrow {{\cos }^{-1}}\left( \cos x \right)=\left\{ \begin{align} & x;\text{ }0\le x\le \pi \\ & 2\pi -x;\text{ }\pi \le x\le 2\pi , \\ \end{align} \right.\]

Thus, it has been defined for \[0 < x < 2\pi \] that has length \[2\pi \].

So, its graph could be plotted as;

Thus, the curve \[y={{\cos }^{-1}}\left( \cos x \right)\] and we can not the results as \[{{\cos }^{-1}}\left( \cos x \right)=\left\{ \begin{align} & -x,\text{ if x}\in \left[ -\pi ,0 \right] \\ & x,\text{ if x}\in \left[ 0,\pi \right] \\ & 2\pi -x,\text{ if x}\in \left[ \pi ,2\pi \right] \\ & -2\pi +x,\text{ if x}\in \left[ 2\pi ,3\pi \right]\text{ and so on}\text{.} \\ \end{align} \right.\]

Next, we consider \[y={{\tan }^{-1}}\left( \tan x \right)\] which is periodic and has period \[\pi \].

To draw this graph, we should draw the graph for one interval of length \[\pi \] and repeat the entire values of x.

We know \[{{\tan }^{-1}}\left( \tan x \right)=\left\{ x;-\dfrac{\pi }{2} < x < \dfrac{\pi }{2} \right\}\]. Thus, it has been defined for \[-\dfrac{\pi }{2} < x < \dfrac{\pi }{2}\] that has length \[\pi \].

The graph is plotted as

Thus, the curve for \[y={{\tan }^{-1}}\left( \tan x \right)\], where \[y\] is not defined for \[x\in \left( 2n+1 \right)\dfrac{\pi }{2}\]. The below result can be kept in mind.

\[{{\tan }^{-1}}\left( \tan x \right)=\left\{ \begin{align} & -\pi -x,\text{ if x}\in \left[ -\dfrac{3\pi }{2},-\dfrac{\pi }{2} \right] \\ & x,\text{ if x}\in \left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right] \\ & x-\pi ,\text{ if x}\in \left[ \dfrac{\pi }{2},\dfrac{3\pi }{2} \right] \\ & x-2\pi ,\text{ if x}\in \left[ \dfrac{3\pi }{2},\dfrac{5\pi }{2} \right]\text{ and so on}\text{.} \\ \end{align} \right.\]

Additional Formulas

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

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

3. \[{{\cos }^{-1}}x+{{\cos }^{-1}}y={{\cos }^{-1}}\left( xy-\sqrt{1-{{x}^{2}}}\sqrt{1-{{y}^{2}}} \right)\]

4. \[{{\cos }^{-1}}x-{{\cos }^{-1}}y={{\cos }^{-1}}\left( xy+\sqrt{1-{{x}^{2}}}\sqrt{1-{{y}^{2}}} \right)\]

5. \[{{\tan }^{-1}}x+{{\tan }^{-1}}y+{{\tan }^{-1}}z={{\tan }^{-1}}\left[ \dfrac{x+y+z-xyz}{1-xy-yz-zx} \right]\], if \[x>0,y>0,z>0\And xy+yz+zx < 1\]

6. \[{{\tan }^{-1}}x+{{\tan }^{-1}}y+{{\tan }^{-1}}z=\pi \] when \[x+y+z=xyz\]

7. \[{{\tan }^{-1}}x+{{\tan }^{-1}}y+{{\tan }^{-1}}z=\dfrac{\pi }{2}\] when \[xy+yz+zx=1\]

8. \[{{\sin }^{-1}}x+{{\sin }^{-1}}y+{{\sin }^{-1}}z=\dfrac{3\pi }{2}\text{; }x=y=z=1\]

9. \[{{\cos }^{-1}}x+{{\cos }^{-1}}y+{{\cos }^{-1}}z=3\pi ;\text{ }x=y=z=-1\]

10. \[{{\tan }^{-1}}1+{{\tan }^{-1}}2+2{{\tan }^{-1}}3={{\tan }^{-1}}1+{{\tan }^{-1}}\dfrac{1}{2}+{{\tan }^{-1}}\dfrac{1}{3}=\dfrac{\pi }{2}\]

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Importance of Maths Chapter 2 Notes Inverse Trigonometric Functions Class 12

• These notes simplify the understanding of inverse trigonometric functions, which are essential for solving complex maths problems.

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