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

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

Class 12 Mathematics chapter Inverse Trigonometric functions dealing with inverse of trigonometric ratios, domain, range of inverse trigonometric functions have weightage not only in class 12th boards but its concept is used in JEE and other competitive exams. Thus keeping the overall importance of the chapter in our mind Vedantu has prepared step by step revision notes of all topics in a descriptive manner under the vision of our subject experts with an idea that basic concepts should be clear and students should get a sufficient amount of practise for each class 12 maths inverse trigonometry solution through problem-solving.

CBSE Class 12 Maths Revision Notes 2023-24 - Chapter Wise PDF Notes for Free

In the table below we have provided the PDF links of all the chapters of CBSE Class 12 Maths whereby the students are required to revise the chapters by downloading the PDF. 


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Inverse Trigonometric Functions Class 12 Notes Maths - Basic Subjective Questions


Section–A (1 Mark Questions)

1. Find the principal value of $tan^{-1}\sqrt{3}$ .

Ans. $\because -\frac{\pi }{2}< tan^{-1}x< \dfrac{\pi }{2}$ 

$tan^{-1}\sqrt{3}= \dfrac{\pi }{3}$


2. Find the principal value of the $sin^{-1}\left ( -\frac{\sqrt{3}}{2} \right )$ .

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

Then $sin\;y=\dfrac{\sqrt{3}}{2}$ 

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

[$\because sin(\Theta )=-sin\Theta$]

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


3. The value of $sin(tan^{-1}x+tan^{-1}1/x)$   for x>0 is……………

Ans. $\begin{array}{l} \sin \left(\tan ^{-1}(x)+\cot ^{-1}(x)\right) \\ {\left[\because \tan ^{-1} \theta=\cot ^{-1}(1 / \theta)\right]} \\ =\sin \left(\frac{\pi}{2}\right)=1\left[\because \tan ^{-1} \theta+\cot ^{-1} \theta=\frac{\pi}{2}\right] .\end{array}$


4. Evaluate $tan^{-1}(sin\left ( \frac{-\pi }{2} \right ))$. 

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

[$\because sin(\Theta )=sin\Theta$ ]

$tan^{-1}(-1)=-\dfrac{\pi }{4}$.


5. Find the domain of $sin^{-1}2x$ .

Ans. Let $sin^{-1}2x =\Theta$  so that $2x=sin\Theta$

Now $-1\leq sin\Theta \leq -1,\;i.e\; -1\leq 2x\leq 1$ which gives $-\dfrac{1}{2}\leq x\leq \dfrac{1}{2}$ .


Section–B (2 Marks Questions)

6. If $3tan^{-1}x+cot^{-1}x=\pi$  , then find x.

Ans.  Given that $3tan^{-1}x+cot^{-1}x=\pi$ 

$\begin{array}{l} \Rightarrow 2 \tan ^{-1} x+\tan ^{-1} x+\cot ^{-1} x=\pi \\ \Rightarrow 2 \tan ^{-1} x+\dfrac{\pi}{2}=\pi \quad\left[\because \tan ^{-1} x+\cot ^{-1} x=\dfrac{\pi}{2}\right] \\ \Rightarrow 2 \tan ^{-1} x=\pi-\dfrac{\pi}{2} \\ \Rightarrow 2 \tan ^{-1} x=\dfrac{\pi}{2} \\ \Rightarrow \tan ^{-1} x=\dfrac{\pi}{4} \\ \Rightarrow \tan ^{-1} x=\tan ^{-1}(1) \\ \therefore x=1 . \end{array}$


7.  Find the value of $sin^{-1}\left [ cos\left ( \dfrac{33\pi }{5} \right ) \right ]$ .

Ans. $\begin{array}{l} \sin ^{-1}\left[\cos \left(\dfrac{33 \pi}{5}\right)\right]=\sin ^{-1}\left[\cos \left(6 \pi+\dfrac{3 \pi}{5}\right)\right] \\ =\sin ^{-1}\left[\cos \dfrac{3 \pi}{5}\right] \quad[\because \cos (2 n \pi+x)=\cos x] \\ =\sin ^{-1}\left[\cos \left(\dfrac{\pi}{2}+\dfrac{\pi}{10}\right)\right] \\ =\sin ^{-1}\left[-\sin \left(\dfrac{\pi}{10}\right)\right] \quad\left[\because \cos \left(\dfrac{\pi}{2}+\theta\right)=-\sin \theta\right] \\ =\sin ^{-1}\left[\sin \left(\dfrac{-\pi}{10}\right)\right] \quad[\because \sin (-\theta)-\sin \theta] \\ =\dfrac{-\pi}{10} \text {. } \\\end{array}$


8.  Find the domain of the function $cos^{-1}(2x-1)$ .

Ans. Let $\Theta =cos^{-1}(2x-1)$ 

$\therefore cos\Theta = cos(cos^{-1}(2x-1))$

$-1\leq cos\Theta \leq 1$

$-1\leq cos(cos^{-1}(2x-1)\leq 1$

$-1\leq 2x-1\leq 1\Rightarrow -1+1\leq 2x\leq 1+1$

$-0\leq 2x\leq 2\Rightarrow 0\leq x\leq 1$

$\therefore$ main of the given function is [0-,1] .



9. If $tan^{-1}x+tan^{-1}y=\frac{4\pi }{5}$ then find $cot^{-1}x+cot^{-1}y$ .

Ans. Given that $\begin{array}{l} \tan ^{-1} x+\tan ^{-1} y=\dfrac{4 \pi}{5} \\ \Rightarrow \dfrac{\pi}{2}-\cot ^{-1} x+\dfrac{\pi}{2}-\cot ^{-1} y=\dfrac{4 \pi}{5} \\ {\left[\because \tan ^{-1} x+\cot ^{-1} x=\dfrac{\pi}{2}\right]} \\ \Rightarrow \pi-\left(\cot ^{-1} x+\cot ^{-1} y\right)=\dfrac{4 \pi}{5} \\ \Rightarrow \cot ^{-1} x+\cot ^{-1} y=\pi-\dfrac{4 \pi}{5} \\ \Rightarrow \cot ^{-1} x+\cot ^{-1} y=\dfrac{\pi}{5} \\\end{array}$


10. Find the value of $tan^{-1}\left ( 2cos\left ( \frac{2\pi }{3} \right ) \right )$ .


Ans. Given that $tan^{-1}\left ( 2cos\left ( \frac{2\pi }{3} \right ) \right )$ 

But we know that cos $\dfrac{\pi }{3}=\dfrac{1}{2}$ 

Therefore $tan^{-1}$ $tan^{-1}\left ( 2cos\left ( \dfrac{2\pi }{3} \right ) \right )=tan^{-1}\left ( 2cos\left (\pi - \dfrac{\pi }{3} \right ) \right )$  $cos(\pi -\Theta )=-cos\Theta$

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

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

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

Hence the principal value of $tan^{-1}\left ( 2cos\left ( \frac{2\pi }{3} \right ) \right )R  is $-\dfrac{\pi }{4}$ .


11. Evaluate: $cot\left ( sin^{-1}\left ( \dfrac{3}{4}\right )+sec^{-1}\left ( \dfrac{4}{3} \right ) \right  )$ .


Ans. Given, $cot\left ( sin^{-1}\left ( \dfrac{3}{4}\right )+sec^{-1}\left ( \dfrac{4}{3} \right ) \right  )$ 

$\begin{array}{l} =\cot \left(\sin ^{-1} \dfrac{3}{4}+\cos ^{-1} \dfrac{3}{4}\right) \\ \left(\because \sec ^{-1} x=\cos ^{-1} \dfrac{1}{x}\right) \end{array}$

We know that \sin ^{-1} x+\cos ^{-1} x=\dfrac{\pi}{2}

By substituting these values in given questions,

$\cot \left(\sin ^{-1} \dfrac{3}{4}+\cos ^{-1} \dfrac{3}{4}\right)=\cot \dfrac{\pi}{2}=0 \text {. }$



12. Prove that: $tan^{-1}\left ( \dfrac{1}{2} \right )+tan^{-1}\left ( \dfrac{1}{13} \right )=tan^{-1}\left ( \dfrac{2}{9} \right )$ .


Ans. LHS = $tan^{-1}\left ( \dfrac{1}{7} \right )+tan^{-1}\left ( \dfrac{1}{13} \right )$

$=tan^{-1}\left ( \dfrac{\dfrac{1}{3}+\dfrac{1}{13}}{1-\dfrac{1}{7}\times\dfrac{1}{13}} \right ) $                                  

$=tan^{-1}\left ( \dfrac{\dfrac{13+7}{91}}{\dfrac{91-1}{91}} \right )$

$=tan^{-1}\left ( \dfrac{20}{90} \right )$

$=tan^{-1}\left ( \dfrac{2}{9} \right )$ = RHS

Hence proved.




PDF Summary - Class 12 Maths Inverse Trigonometric Functions Notes (Chapter 2)


Domain and Range of all Inverse Trigonometric Functions

Function

Domain

Range

1. \[y={{\sin }^{-1}}x\text{ if }x=\sin y\]

\[-1\le x\le 1\]

\[\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]\]

2. \[y={{\cos }^{-1}}x\text{if }x=\cos y\]

\[-1\le x\le 1\]

\[[0,\pi ]\]

3. \[y={{\tan }^{-1}}x\text{if }x=\tan y\]

\[-\infty  < x < \infty \]

\[\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)\]

4. \[y={{\cot }^{-1}}x\text{if }x=\cot y\]

\[-\infty  < x < \infty \]

\[(0,\pi )\]

5.\[y=\text{cose}{{\text{c}}^{-1}}x\text{if }x=\text{cosec}y\]

\[(-\infty ,-1]\cup [1,\infty )\]

\[\left[ -\dfrac{\pi }{2},0 \right)\cup \left( 0,\dfrac{\pi }{2} \right]\]

6. \[y={{\sec }^{-1}}x\text{if }x=\sec y\]

\[(-\infty ,-1]\cup [1,\infty )\]

\[\left[ 0,\dfrac{\pi }{2} \right)\cup \left( \dfrac{\pi }{2},\pi  \right]\]

  • 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

(image will be uploaded soon)

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;

(image will be uploaded soon)

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

(image will be uploaded soon)

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.


The Inverse Trigonometric Functions

The Formulas

Arcsine

$\sin^{-1} (-x) = -\sin^{-1} (x), x \epsilon -1, 1$

Arccosine

$\cos^{-1} (x) = \pi - \cos^{-1} (x), x \epsilon -1, 1$

Arctangent

$\tan^{-1} (-x), x \epsilon \mathbb{R}$

Arccotangent

$\cot^{-1} (x) = \pi - \cot^{-1} x, x \epsilon \mathbb{R}$

Arcsecant

$\sec^{-1} (x) = \pi - \sec^{-1}x, x \geq 1$

Arccosecant

$\text{cosec}^{-1} (x) = -\text{cosec}^{-1} x, x \geq 1$


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.


Function Name

Notation

Definition

Domain of x

Range

Arcsine or inverse sine

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

X = sin y

-1 ≤ x ≤ 1

$-\dfrac{\pi}{2} \leq y \leq \pi$ or - 90 degree ≤ y ≤ 90 degree

Inverse cosine or arccosine

$y = \cos^{-1} (x)$

X = cos y

-1 ≤ x ≤ 1

0 ≤ y ≤ π or 0 degree ≤ y ≤ 180 degrees

Inverse tangent or arctangent

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

X = tan y

For all real numbers

$-\dfrac{\pi}{2} < y < \dfrac{\pi}{2}$ or - 90 degrees < y  < 90 degrees

Inverse cot or arccotangent

$y = \cot^{-1} (x)$

X = cot y

For all real numbers

0 < y < π or 0 degree < y < 180 degrees

Inverse secant or arcsecant

$y = \sec^{-1} (x)$

X = sec y

X ≤ -1 or 1 ≤ x

$0 \leq y < \dfrac{\pi}{2}$ or $\dfrac{\pi}{2} < y \leq \pi$ or 0 degree ≤ y < 90 degrees or 90 degrees < y ≤ 180 degrees

Arccosecant

$y = \text{cosec}^{-1} (x)$

X = csc y

X ≤ -1 or 1 ≤ x

$-\dfrac{\pi}{2} \leq y < 0$ or $0 < y \leq \dfrac{\pi}{2}$ or -90 degrees ≤ y < 0 degree or 0 degree < y ≤ 90 degrees


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.


Inverse Trigonometric Function

$\dfrac{dy}{dx}$

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

$\dfrac{1}{\sqrt{1 - x^2}}$

$y = \cos^{-1} (x)$

$-\dfrac{1}{\sqrt{1 - x^2}}$

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

$\dfrac{1}{{1 + x^2}}$

$y = \cot^{-1} (x)$

$\dfrac{1}{{1 + x^2}}$

$y = \sec^{-1} (x)$

$\dfrac{x}{\sqrt{x^2 - 1}}$

$y = \text{cosec}^{-1} (x)$

$-\dfrac{x}{\sqrt{x^2 - 1}}$


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.

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

1. Find the value of x, if sin (x) = 2.

We know that sin x = 2,

X = sin-1 (2), this is not possible

This means that there is no value of x for which sin x = 2. This is true since the domain of sin-1x is -1 to 1 for the value of x.

2. What is the value of sin-1 (sin (π / 6))?

sin-1 (sin (π / 6)) = π / 6

We have arrived at the above answer by using the sin-1 (sin (x) ) = x identity.

3. What is the value of sin (cos-1 3 / 5)?

Let’s suppose that cos-1 3 /5 = x

This means that cos x = 3 / 5

We also know that sin x = √1 - cos2 x

So, sin x = √1 - 9 / 25 = 4 / 5

This means that sin x = sin (cos-1 3 / 5) = 4 / 5.

4. Find the value of sin(cot - 1x)

Let’s assume that cot-1 x = θ

X = cotθ

Now, cosec θ = √1 + cot2θ  = √1 + x2

Hence, sinθ = 1 / cos ec θ = 1 / √1 + x2

Θ = sin-1 1 / √1 + x2

Hence, sin (cot - 1x) = sin (sin-1 x 1 / √1 + x2 = 1 /√ 1 + x2 = (1 + x2)-1 / 2.

5. What is the value of sec-1 [sec (-30 degrees)]

sec-1 [sec (-30 degrees)] = sec-1 (sec 30 degrees) = 30 degrees.

6. Where can I download the latest Chapter 2 Inverse Trigonometric functions of Class 12 Maths notes?

You can download the Chapter 2 Inverse Trigonometric functions of Class 12 Maths notes from the Vedantu website. Here at Vedantu, you will find all revision notes from which you can study.

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  • Once the revision notes pdf is saved on your device, you can use it to study later offline.

7. What are the various functions discussed in Chapter 2 Inverse Trigonometric functions of Class 12 Maths?

As per the syllabus of Class 12 Maths CBSE, the second chapter in the Maths textbook of Class 12 is inverse trigonometric functions. Previously, you have learnt about trigonometric functions and keeping that knowledge and concepts in mind, you will easily be able to understand this particular chapter. You will learn about the inverse functions of trigonometry like arcsine, arctangent, arccosine, arccotangent, arcsecant, arccosecant functions. You will need to understand these topics to solve the sums from the chapter.


These solutions are available on Vedantu's official website(vedantu.com) and mobile app free of cost.

8. What are the topics in Chapter 2 Inverse Trigonometric functions of Class 12 Maths?

Chapter 2 Inverse Trigonometric functions of Class 12 Maths deals with the Inverse Trigonometric Functions. You will have to study various subtopics like inverse functions, equations using these inverse functions, and the graphs pertaining to these. You will also have to memorize the important formulas given in the chapter to be able to solve the questions. You must be acquainted with the notations and symbols for the various functions so that it becomes easier to move forward.

9. How can I get all trigonometry questions right in the Class 12 Maths exam?

It is not a very big deal to be able to get all the sums correct from the trigonometry section in your board exam. The key to getting full marks is practice. You must be consistent with your practice and solve as many sums as possible from the trigonometry chapter in order to be confident and accurate. This will enhance your concepts and problem solving skills as well. Apart from this, you must not forget to refer to mock papers and previous year papers to understand the pattern and prepare accordingly.

10. What is meant by inverse trigonometric functions according to Chapter 2 Inverse Trigonometric functions of Class 12 Maths?

In your previous chapter, you gathered ideas about the normal trigonometry, its functions, derivatives etc. Now when you study inverse trigonometry, it is nothing but the inverse or the reciprocal of the actual one. That is, all the functions, angles, derivatives, formulas etc. everything will be written as the reciprocal or opposite of the corresponding trigonometric functions, angles, derivatives , formulas etc. You must refer to the Vedantu explanation of this chapter in order to understand clearly.