Inverse Trigonometric Functions Class 12 Notes CBSE Maths Chapter 2 (Free PDF Download)

PDF Summary - Class 12 Maths Inverse Trigonometric Functions Notes (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,

(image will be uploaded soon)

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

(image will be uploaded soon)

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

(image will be uploaded soon)

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

(image will be uploaded soon)

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

(image will be uploaded soon)

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

(image will be uploaded soon)

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

Revision Notes for Class 12 Maths 2 - Inverse Trigonometric Functions

In this chapter, students will learn the functions which are developed by finding the inverse of trigonometric ratios, graphs of inverse trigonometric functions, their domains, range, etc. The chapters hold their relevance, as well as importance in 12th grade as well and the concepts that we study here will help us to easily grasp the upcoming concepts that we will be studying for your competitive exams like JEE mains, JEE advanced, State Engineering Entrance Exams, etc.

Also, its weightage in the 12th board exam makes it one of the most important chapters for those who want to score really well in the board exams. Thus Vedantu has prepared revision notes for Inverse trigonometric functions so that students can easily revise the concepts before the exams and are able to score well.

Class 12 Inverse Trigonometry chapter 2 notes have been prepared with an objective of an overall evolution of student’s concepts in a manner that the students understand all the class 12 maths inverse trigonometry solution, theorems, formulas, and derivations quite effectively by linking them with their practical applications. Revision notes will benefit the students in various ways particularly it is going to speed up the problem-solving skills of the students and for this, we suggest practising these solutions on a regular basis. The revision notes provided by Vedantu is in a very illustrative and descriptive way. We keep every important point in their mind while preparing for a book like your exam pattern, important concepts which need to be highlighted again and again, etc. 

Vedantu Revision notes play a very crucial role in preparation for class 12th board exam as it gives you a path to give your best during exam preparation in every aspect as it develops a better understanding of different concepts particularly with our precise revision notes which will help students in the preparation of different competitive exams like NTSE and various olympiads class 12th students.

The revision notes are prepared after extensive research by our experts which aims to provide easy, precise and accurate compiling of concepts used throughout the chapter. Students can easily download these revision notes which will contribute not only to their process of learning but also with better output in the examination.

The Important Formulas

The next topic that a student needs to focus on while going over Class 12 maths ch 2 notes is that of the functions and their formulas. We have formulated a table of all the formulas. And that table is mentioned below.

Graphs Related to Inverse Trigonometric Functions

According to most inverse trigonometric functions notes Class 12, there are mainly six inverse trigonometric functions for every trigonometric ratio. We have already talked about the basic formulas of those six inverse trigonometric functions in this chapter 12 maths Class 12 notes. So, we will directly move to the graphs and properties.

• Arcsine Function

As mentioned above, the arcsine function is the inverse of the sine function. It is denoted by $\sin^{-1}x$ . The graph for representing this function is shown in the image attached below.

The domain of the function is -1 ≤ x ≤ 1. And the range is $-\dfrac{\pi}{2} \leq y \leq \dfrac{\pi}{2}$ .

• Arccosine Function

As mentioned in this Class 12 inverse trigonometric functions notes, the arccosine function is defined as the inverse of the cosine function. It is denoted by $\cos^{-1}x$ . The graph for representing this function is attached below.

If you look at this graph carefully, then you will realize that the inverse of the cos function can also be expressed as $y = \cos^{-1}x$ (arccosine x). The domain of the function is -1 ≤ x ≤ 1, and the range is 0 ≤ y ≤ π.

• Arctangent Function

The arctangent function can be explained as the inverse of the tangent function. It is denoted by tan-1x. 

The inverse of the tangent function can also be expressed as $y = \tan^{-1}x$ (arctangent x). The domain of this function is -∞ < x < ∞. The value for range can also be expressed as $-\dfrac{\pi}{2} < y < \dfrac{\pi}{2}$ .

• Arccotangent (Arccot) Function

The arccotangent or arccot function is the inverse of the cotangent function, and it is denoted by by $\cot^{-1}x$ . The inverse of the cotangent function can be said to be $y = \cot^{-1}x$ (arccotangent x). The value of the domain can be expressed by -∞ < x < ∞, and the range can be depicted by 0 < y < π.

• Arcsecant Function

One can explain the arcsecant function as being the inverse of the secant function. It can be denoted by sec-1x. The inverse of the secant function can be denoted by y = sec-1x (arcsecant x). The domain can be explained as -∞ ≤ x ≤ -1 or 1 ≤ x ≤ ∞, and for range, the value can be denoted by $-\dfrac{\pi}{2} < y < \dfrac{\pi}{2}$; y ≠ 0 .

• Arccosecant Function

According to experts, the arccosecant function can be defined as the inverse of the cosecant function. It can be defined as cosec-1x. Readers should be familiar with the graph for this function. On the basis of the graph, the function can be expressed as y = cosec-1x (arccosecant x). The domain can be said to be -∞ ≤ x ≤ -1 or 1 ≤ x ≤ ∞ and the range is $-\dfrac{\pi}{2} < y < \dfrac{\pi}{2}$ ; y ≠ 0 .

Inverse Trigonometric Functions and Its Derivatives

In this section of maths Class 12 Chapter 2 notes, readers will be able to learn about all inverse trigonometric functions along with their definition, notations, domains, and ranges. We have formulated a table that contains all the information. And that table is mentioned below.

It should be noted here that the derivatives of inverse trigonometric functions are classified as being first-order derivatives. Do you wish to know what those derivatives are? If yes, then go through the table that is mentioned below.

Fun Facts About Inverse Trigonometric Functions

Did you know that inverse trigonometric functions can also be defined in certain intervals or under restricted domains? We have looked at the inverse trigonometric functions but not at the basic functions. However, the basic trigonometric functions are sin x, cos x, tan x, cosec x, sec x, and cot x.

Conclusion

Vedantu offers concise and comprehensive revision notes for Class 12 Maths Chapter 2 - Inverse Trigonometric Functions. These revision notes are meticulously prepared by expert educators to help students revise and understand the key concepts of the chapter effectively. The revision notes provided by Vedantu cover all the important topics, formulas, and key points in a clear and concise manner. They serve as a handy resource for last-minute revision, quick recaps, and exam preparation. Students can access these revision notes on the Vedantu website or app, making it convenient for them to review the chapter and enhance their understanding of Inverse Trigonometric Functions.