Inverse Trigonometric Functions Class 12 Maths Chapter 2 CBSE Notes - 2025-26

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, ${\sin }^{-1}x\ne {\displaystyle \frac{1}{\sin x}}$.

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

• The functions $${\sin }^{-1}x\mathrm{\&}{\tan }^{-1}x$$ are increasing functions in their domain.

• The functions $${\cos }^{-1}x\mathrm{\&}{\cot }^{-1}x$$ 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). $${\sin }^{-1}\left({\displaystyle \frac{1}{x}}\right)=\cos e{c}^{-1}x$$, for all $$x\in (-\mathrm{\infty },1]\cup [1,\mathrm{\infty })$$

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,

$$\Rightarrow {\displaystyle \frac{1}{x}}=\sin \theta$$

$$\{\because x\in (-\mathrm{\infty },-1]\cup [1,\mathrm{\infty })\}\Rightarrow {\displaystyle \frac{1}{x}}\in [-1,1]\left\{0\right\}$$

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

$$\Rightarrow \theta ={\sin }^{-1}\left({\displaystyle \frac{1}{x}}\right)$$ ……(ii)

From (i) and (ii), we get

$${\sin }^{-1}\left({\displaystyle \frac{1}{x}}\right)=\cos e{c}^{-1}x$$

Hence proved.

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

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

Taking $$\sec$$ on both sides,

$$\Rightarrow x=\sec \theta$$

Using reciprocal identity,

$$\Rightarrow {\displaystyle \frac{1}{x}}=\cos \theta$$

$$\Rightarrow \theta ={\cos }^{-1}\left({\displaystyle \frac{1}{x}}\right)$$ ……(ii)

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

$$\{\begin{array}{rl}& \because x=(-\mathrm{\infty },-1]\cup [1,\mathrm{\infty })\\ & \Rightarrow {\displaystyle \frac{1}{x}}\in [-1,1]-\left\{0\right\}\text{ and }\theta \in [0,\pi ]\end{array}$$

From (i) and (ii), we get

$${\cos }^{-1}\left({\displaystyle \frac{1}{x}}\right)={\sec }^{-1}\left(x\right)$$

Hence proved.

c. $${\tan }^{-1}\left({\displaystyle \frac{1}{x}}\right)=\{\begin{array}{rl}& {\cot }^{-1}x,\text{ for }x>0\\ & -\pi +{\cot }^{-1}x,\text{ for }x<0\end{array}$$ 

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

Now there are two cases that arise:

Case I: When $$x>0$$

In this case, we have $$\theta \in (0,{\displaystyle \frac{\pi }{2}})$$

Considering $${\cot }^{-1}x=\theta$$ 

Taking $$\cot$$ on both sides,

$$\Rightarrow x=\cot \theta$$

Using reciprocal property,                               

$$\Rightarrow {\displaystyle \frac{1}{x}}=\tan \theta$$

$$\theta ={\tan }^{-1}\left({\displaystyle \frac{1}{x}}\right)$$  ……(ii)

From (i) and (ii), we get   $$\{\because \theta \in (0,{\displaystyle \frac{\pi }{2}})\}$$

$${\tan }^{-1}\left({\displaystyle \frac{1}{x}}\right)={\cot }^{-1}x$$, for all $$x>0$$

Case II: When $$x<0$$

In this case, we have $$\theta \in ({\displaystyle \frac{\pi }{2}},\pi )\{\because x=\cot \theta <0\}$$

Now, $${\displaystyle \frac{\pi }{2}}<\theta <\pi$$

$$\Rightarrow -{\displaystyle \frac{\pi }{2}}<\theta -\pi <0$$

$$\Rightarrow \theta -\pi \in (-{\displaystyle \frac{\pi }{2}},0)$$

$$\therefore {\cot }^{-1}x=\theta$$

Taking $$\cot$$ on both sides,

$$\Rightarrow x=\cot \theta$$

Using reciprocal property,

$$\Rightarrow {\displaystyle \frac{1}{x}}=\tan \theta$$

$$\Rightarrow {\displaystyle \frac{1}{x}}=-\tan (\pi -\theta )$$

$$\Rightarrow {\displaystyle \frac{1}{x}}=\tan (\theta -\pi )\{\because \tan (\pi -\theta )=-\tan \theta \}$$

$$\Rightarrow \theta -\pi ={\tan }^{-1}\left({\displaystyle \frac{1}{x}}\right)\{\because \theta -\pi \in (-{\displaystyle \frac{\pi }{2}},0)\}$$

$$\Rightarrow {\tan }^{-1}\left({\displaystyle \frac{1}{x}}\right)=-\pi +\theta$$   ……(iii)

From (i) and (iii), we get

$${\tan }^{-1}\left({\displaystyle \frac{1}{x}}\right)=-\pi +{\cot }^{-1}x$$, if $$x<0$$

Hence it is proved that $${\tan }^{-1}\left({\displaystyle \frac{1}{x}}\right)=\{\begin{array}{rl}& {\cot }^{-1}x,\text{ for }x>0\\ & -\pi +{\cot }^{-1}x,\text{ for }x<0\end{array}$$.

2. Property II

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

2. $${\tan }^{-1}(-x)=-{\tan }^{-1}x$$, for all $$x\in R$$

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

Clearly, $$-x\in [-1,1]$$ for all $$x\in [-1,1]$$

Let us prove a) by taking $${\sin }^{-1}(-x)=\theta$$

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

$$-x=\sin \theta$$   ……(i)

$$\Rightarrow x=-\sin \theta$$

$$\Rightarrow x=\sin (-\theta )$$

$$\Rightarrow -\theta ={\sin }^{-1}x$$

$$\{\because x\in [-1,1]\text{ and }-\theta \in [-{\displaystyle \frac{\pi }{2}},{\displaystyle \frac{\pi }{2}}]\text{ for all }\theta \in [-{\displaystyle \frac{\pi }{2}},{\displaystyle \frac{\pi }{2}}]\}$$

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

From (i) and (ii), we get

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

Hence proved.

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

3. Property III

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

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

3. $${\cot }^{-1}(-x)=\pi -{\cot }^{-1}x$$, for all $$x\in R$$

Clearly, $$-x\in [-1,1]$$ for all $$x\in [-1,1]$$

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

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

$$-x=\cos \theta$$

$$\Rightarrow x=-\cos \theta$$

$$\Rightarrow x=\cos (\pi -\theta )$$

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

$${\cos }^{-1}x=\pi -\theta$$

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

From (i) and (ii), we get

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

Hence Proved.

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

4.  Property IV

a) $${\sin }^{-1}x+{\cos }^{-1}x={\displaystyle \frac{\pi }{2}}$$, for all $$x\in [-1,1]$$

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

Then, $$\theta \in [-{\displaystyle \frac{\pi }{2}},{\displaystyle \frac{\pi }{2}}][\because x\in [-1,1]]$$

$$\Rightarrow -{\displaystyle \frac{\pi }{2}}\le \theta \le {\displaystyle \frac{\pi }{2}}$$

$$\Rightarrow -{\displaystyle \frac{\pi }{2}}\le -\theta \le {\displaystyle \frac{\pi }{2}}$$

$$\Rightarrow 0\le {\displaystyle \frac{\pi }{2}}-\theta \le \pi$$

$$\Rightarrow {\displaystyle \frac{\pi }{2}}-\theta \in [0,\pi ]$$

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

Taking $\sin$ on both sides, we get

$$\Rightarrow x=\sin \theta$$

Changing functions, we get

$$\Rightarrow x=\cos ({\displaystyle \frac{\pi }{2}}-\theta )$$

$$\Rightarrow {\cos }^{-1}x={\displaystyle \frac{\pi }{2}}-\theta$$

$$\{\because x\in [-1,1]\text{ and }({\displaystyle \frac{\pi }{2}}-\theta )\in [0,\pi ]\}$$                                                

$$\Rightarrow \theta +{\cos }^{-1}x={\displaystyle \frac{\pi }{2}}$$ ……(ii)

From (i) and (ii), we get

$${\sin }^{-1}x+{\cos }^{-1}x={\displaystyle \frac{\pi }{2}}$$

Hence proved.

b) $${\tan }^{-1}x+{\cot }^{-1}x={\displaystyle \frac{\pi }{2}}$$, for all $$x\in R$$

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

Then, $$\theta \in (-{\displaystyle \frac{\pi }{2}},{\displaystyle \frac{\pi }{2}})\{\because x\in R\}$$

$$\Rightarrow -{\displaystyle \frac{\pi }{2}}<\theta <{\displaystyle \frac{\pi }{2}}$$

$$\Rightarrow -{\displaystyle \frac{\pi }{2}}<-\theta <{\displaystyle \frac{\pi }{2}}$$

$$\Rightarrow 0<{\displaystyle \frac{\pi }{2}}-\theta <\pi$$

$$\Rightarrow ({\displaystyle \frac{\pi }{2}}-\theta )\in (0,\pi )$$

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

Taking $\tan$ on both sides, we get

$$\Rightarrow x=\tan \theta$$

$$\Rightarrow x=\cot ({\displaystyle \frac{\pi }{2}}-\theta )$$

$$\Rightarrow {\cot }^{-1}x={\displaystyle \frac{\pi }{2}}-\theta \{\because {\displaystyle \frac{\pi }{2}}-\theta \in (0,\pi )\}$$

$$\Rightarrow \theta +{\cot }^{-1}x={\displaystyle \frac{\pi }{2}}$$   ……(ii)

From (i) and (ii), we get

$${\tan }^{-1}x+{\cot }^{-1}x={\displaystyle \frac{\pi }{2}}$$

c) $${\sec }^{-1}x+\cos e{c}^{-1}x={\displaystyle \frac{\pi }{2}}$$, for all $$x\in (-\mathrm{\infty },-1]\cup [1,\mathrm{\infty })$$

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

Then, $$\theta \in [0,\pi ]-\left\{{\displaystyle \frac{\pi }{2}}\right\}\{\because x\in (-\mathrm{\infty },-1]\cup [1,\mathrm{\infty })\}$$

$$\Rightarrow 0\le \theta \le \pi ,\theta \ne {\displaystyle \frac{\pi }{2}}$$

$$\Rightarrow -\pi \le -\theta \le 0,\theta \ne {\displaystyle \frac{\pi }{2}}$$

$$\Rightarrow -{\displaystyle \frac{\pi }{2}}\le {\displaystyle \frac{\pi }{2}}-\theta \le {\displaystyle \frac{\pi }{2}},{\displaystyle \frac{\pi }{2}}-\theta \ne 0$$

$$\Rightarrow ({\displaystyle \frac{\pi }{2}}-\theta )\in [-{\displaystyle \frac{\pi }{2}},{\displaystyle \frac{\pi }{2}}],{\displaystyle \frac{\pi }{2}}-\theta \ne 0$$

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

Taking $\sec$ on both sides, we get

$$\Rightarrow x=\sec \theta$$

$$\Rightarrow x=\cos ec({\displaystyle \frac{\pi }{2}}-\theta )$$

$$\Rightarrow \cos e{c}^{-1}x={\displaystyle \frac{\pi }{2}}-\theta$$

$$\{\because ({\displaystyle \frac{\pi }{2}}-\theta )\in [-{\displaystyle \frac{\pi }{2}},{\displaystyle \frac{\pi }{2}}],{\displaystyle \frac{\pi }{2}}-\theta \ne 0\}$$

$$\Rightarrow \theta +\cos e{c}^{-1}x={\displaystyle \frac{\pi }{2}}$$   ….…(ii)

From (i) and (ii), we get

$${\sec }^{-1}x+\cos e{c}^{-1}x={\displaystyle \frac{\pi }{2}}$$

5. Property V

1. $${\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}{\displaystyle \frac{x+y}{1-xy}},xy<1$$

2. $${\tan }^{-1}x-{\tan }^{-1}y={\tan }^{-1}{\displaystyle \frac{x-y}{1+xy}},xy>-1$$

3. $${\tan }^{-1}x+{\tan }^{-1}y=\pi +{\tan }^{-1}\left({\displaystyle \frac{x+y}{1-xy}}\right),xy>1;x,y=0$$

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

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

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

$\tan (\theta +\varphi )={\displaystyle \frac{\tan \theta +tan\varphi }{1-\tan \theta \tan \varphi }}$

Writing in terms of $xandy$,

$\tan (\theta +\varphi )={\displaystyle \frac{x+y}{1-xy}}$

$\Rightarrow \theta +\varphi ={\tan }^{-1}\left({\displaystyle \frac{x+y}{1-xy}}\right)$

Therefore $${\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}{\displaystyle \frac{x+y}{1-xy}},xy<1$$.

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. $$2{\tan }^{-1}x={\sin }^{-1}{\displaystyle \frac{2x}{1+x^2}},\left|x\right|\le 1$$

2. $$2{\tan }^{-1}x={\cos }^{-1}{\displaystyle \frac{1-x^2}{1+x^2}},x\ge 0$$

3. $$2{\tan }^{-1}x={\tan }^{-1}{\displaystyle \frac{2x}{1-x^2}},-1<x<1$$

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

Taking $\tan$ on both sides, we get

$x=\tan y$

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

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

${\sin }^{-1}{\displaystyle \frac{2x}{1+x^2}}={\sin }^{-1}(\sin 2y)$

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

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

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

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 $$x\in [-1,1]$$

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

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

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

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

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 $$\theta \in [-{\displaystyle \frac{\pi }{2}},{\displaystyle \frac{\pi }{2}}]$$ and $$x\in [-1,1]$$ 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 $$x\in [-1,1]$$ and we proved it.

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

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

Let us understand this better. The function $$y={\sin }^{-1}(\sin x)$$ 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,

$${\sin }^{-1}(\sin x)=\{\begin{array}{rl}& x;-{\displaystyle \frac{\pi }{2}}\le x\le {\displaystyle \frac{\pi }{2}}\\ & (\pi -x);-{\displaystyle \frac{\pi }{2}}\le \pi -x<{\displaystyle \frac{\pi }{2}}(\text{i}\text{.e}\text{.,}{\displaystyle \frac{\pi }{2}}\le x\le {\displaystyle \frac{3\pi }{2}})\end{array}$$

$$\Rightarrow {\sin }^{-1}(\sin x)=\{\begin{array}{rl}& x,-{\displaystyle \frac{\pi }{2}}\le x\le {\displaystyle \frac{\pi }{2}}\\ & \pi -x,{\displaystyle \frac{\pi }{2}}\le x\le {\displaystyle \frac{3\pi }{2}},\end{array}$$

This is plotted as

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

The below result for the definition of $${\sin }^{-1}(\sin x)$$ must be kept in mind. $$y={\sin }^{-1}(\sin x)=\{\begin{array}{rl}& x+2\pi ;-{\displaystyle \frac{5\pi }{2}}\le x\le -{\displaystyle \frac{3\pi }{2}}\\ & -\pi -x;-{\displaystyle \frac{3\pi }{2}}\le x\le -{\displaystyle \frac{\pi }{2}}\\ & x;-{\displaystyle \frac{\pi }{2}}\le x\le {\displaystyle \frac{\pi }{2}}\\ & \pi -x;{\displaystyle \frac{\pi }{2}}\le x\le {\displaystyle \frac{3\pi }{2}}\\ & x-2\pi ;{\displaystyle \frac{3\pi }{2}}\le x\le {\displaystyle \frac{5\pi }{2}}...\text{and so on}\end{array}$$

Now we consider $$y={\cos }^{-1}(\cos x)$$ 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,

$${\cos }^{-1}(\cos x)=\{\begin{array}{rl}& x;0\le x\le \pi \\ & 2\pi -x;0\le 2\pi -x\le \pi ,\end{array}$$ $$\Rightarrow {\cos }^{-1}(\cos x)=\{\begin{array}{rl}& x;0\le x\le \pi \\ & 2\pi -x;\pi \le x\le 2\pi ,\end{array}$$

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

Next, we consider $$y={\tan }^{-1}(\tan x)$$ 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 $${\tan }^{-1}(\tan x)=\{x;-{\displaystyle \frac{\pi }{2}}<x<{\displaystyle \frac{\pi }{2}}\}$$. Thus, it has been defined for $$-{\displaystyle \frac{\pi }{2}}<x<{\displaystyle \frac{\pi }{2}}$$ that has length $$\pi$$.

The graph is plotted as

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

$${\tan }^{-1}(\tan x)=\{\begin{array}{rl}& -\pi -x,\text{ if x}\in [-{\displaystyle \frac{3\pi }{2}},-{\displaystyle \frac{\pi }{2}}]\\ & x,\text{ if x}\in [-{\displaystyle \frac{\pi }{2}},{\displaystyle \frac{\pi }{2}}]\\ & x-\pi ,\text{ if x}\in [{\displaystyle \frac{\pi }{2}},{\displaystyle \frac{3\pi }{2}}]\\ & x-2\pi ,\text{ if x}\in [{\displaystyle \frac{3\pi }{2}},{\displaystyle \frac{5\pi }{2}}]\text{ and so on}\text{.}\end{array}$$

Additional Formulas

1. $${\sin }^{-1}x+{\sin }^{-1}y={\sin }^{-1}(x\sqrt{1-y^2}+y\sqrt{1-x^2})$$

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

3. $${\cos }^{-1}x+{\cos }^{-1}y={\cos }^{-1}(xy-\sqrt{1-x^2}\sqrt{1-y^2})$$

4. $${\cos }^{-1}x-{\cos }^{-1}y={\cos }^{-1}(xy+\sqrt{1-x^2}\sqrt{1-y^2})$$

5. $${\tan }^{-1}x+{\tan }^{-1}y+{\tan }^{-1}z={\tan }^{-1}\left[{\displaystyle \frac{x+y+z-xyz}{1-xy-yz-zx}}\right]$$, if $$x>0,y>0,z>0\mathrm{\&}xy+yz+zx<1$$

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

7. $${\tan }^{-1}x+{\tan }^{-1}y+{\tan }^{-1}z={\displaystyle \frac{\pi }{2}}$$ when $$xy+yz+zx=1$$

8. $${\sin }^{-1}x+{\sin }^{-1}y+{\sin }^{-1}z={\displaystyle \frac{3\pi }{2}}\text{; }x=y=z=1$$

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

10. $${\tan }^{-1}1+{\tan }^{-1}2+2{\tan }^{-1}3={\tan }^{-1}1+{\tan }^{-1}{\displaystyle \frac{1}{2}}+{\tan }^{-1}{\displaystyle \frac{1}{3}}={\displaystyle \frac{\pi }{2}}$$

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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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