Inverse Trigonometric Functions Class 12th Notes - Free NCERT Class 12 Maths Chapter 2 Notes - Download PDF

NCERT Notes for Class 12 Chapter 2 Inverse Trigonometric Functions: Free PDF Download

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NCERT Notes for Class 12 Chapter 2 Inverse Trigonometric Functions

The inverse of any function exists if the function 'f' is one-one and onto. The inverse of function 'f' is denoted by f ^-1. The trigonometric functions are neither one-one nor onto over their domain and natural ranges.
So, the domains and ranges of trigonometric functions are restricted to ensure the existence of their inverse. The inverse trigonometric functions, denoted by sin ^-1 x or (arc sin x), cos ^-1x, etc., denote the angles whose sine, cosine, etc, is equal to x.

Note: Do not confuse inverse function $f^{-1}$ with $\frac{1}{f}$ . The inverse trigonometric functions like $\sin^{-1}x$ is not the same as $\frac{1}{\sin x}$.

Domain & Range of Inverse Trigonometric Functions

$
\begin{array}{|l|l|l|l|}
\hline \text { S.No } &{f}(\mathrm{x}) & \text { Domain } & \text { Range } \\
\hline(1) & \sin ^{-1} \mathrm{x} & |\mathrm{x}| \leq 1 & {\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]} \\
(2) & \cos ^{-1} \mathrm{x} & |\mathrm{x}| \leq 1 & {[0, \pi]} \\
(3) & \tan ^{-1} \mathrm{x} & \mathrm{x} \in \mathrm{R} & \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \\
(4) & \sec ^{-1} \mathrm{x} & |\mathrm{x}| \geq 1 & {[0, \pi] -\left\{\frac{\pi}{2}\right\} \text { or, }\left[0, \frac{\pi}{2}\right) \cup\left(\frac{\pi}{2}, \pi\right]} \\
(5) & \operatorname{cosec}^{-1} \mathrm{x} & |\mathrm{x}| \geq 1 & {\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\}} \\
(6) & \cot ^{-1} \mathrm{x} & \mathrm{x} \in \mathrm{R} & (0, \pi) \\
\hline
\end{array}
$

Graphs Of Inverse Trigonometric Functions

(i) Graphs of sin x and sin^-1 x

$\begin{aligned} &\mathrm{f}:\left[\frac{-\pi}{2}, \frac{\pi}{2}\right] \rightarrow[-1,1]\\ &\mathrm{f}(\mathrm{x})=\sin \mathrm{x} \end{aligned}$

$\begin{aligned} &\mathrm{f}^{-1}:[-1,1] \rightarrow[-\frac{\pi}2, \frac{\pi}2]\\ &\mathrm{f}^{-1}(\mathrm{x})=\sin ^{-1}(\mathrm{x}) \end{aligned}$

The domain of the inverse sine function is $[-1,1]$ and the range is $[-\frac{\pi}2, \frac{\pi}2]$.

(ii) Graphs of cos x and cos^-1 x

$\begin{aligned} &f:[0, \pi] \rightarrow[-1,1] \\ &f(x)=\cos x \end{aligned}$

$\begin{aligned} &\mathrm{f}^{-1}:[-1,1] \rightarrow[0, \pi] \\ &\mathrm{f}^{-1}(\mathrm{x})=\cos ^{-1} \mathrm{x} \end{aligned}$

The domain of the inverse cosine function is $[-1,1]$, and the range is $[0,\pi]$.

(iii) Graphs of cosec x and cosec^-1 x

$\begin{aligned} &f:[-\frac{\pi}2,0) \cup(0, \frac{\pi}2] \rightarrow(-\infty,-1] \cup[1, \infty) \\ &f(x)=\operatorname{cosec} x \end{aligned}$

$\begin{aligned} &\mathrm{f}^{-1}:(-\infty,-1] \cup[1, \infty) \rightarrow[-\frac{\pi}2,0) \cup(0, \frac{\pi}2] \\ &\mathrm{f}^{-1}(\mathrm{x})=\operatorname{cosec}^{-1} \mathrm{x} \end{aligned}$

The domain of the inverse cosec function is $(-\infty,-1] \cup[1, \infty)$ and the range is $\left[-\frac{\pi}{2}, 0\right) \cup\left(0, \frac{\pi}{2}\right]$.

(iv) Graphs of sec x and sec^-1 x

$\begin{aligned} &\mathrm{f}:[0,\frac{\pi}2) \cup(\frac{\pi}2, \pi] \rightarrow(-\infty,-1] \cup[1, \infty) \\ &\mathrm{f}(\mathrm{x})=\sec \mathrm{x}
\end{aligned}$

$\begin{aligned} &\mathrm{f}^{-1}:(-\infty,-1] \cup[1, \infty) \rightarrow[0, \frac{\pi}2) \cup(\frac{\pi}2, \pi] \\ &\mathrm{f}^{-1}(\mathrm{x})=\sec ^{-1} \mathrm{x} \end{aligned}$

The domain of the inverse secant function is $(-\infty,-1] \cup[1, \infty)$ and the range is $\left[0, \frac{\pi}{2}\right) \cup\left(\frac{\pi}{2}, \pi\right]$.

(v) Graphs of tan x and tan^-1 x

$\begin{aligned} &\mathrm{f}:(-\frac{\pi}2, \frac{\pi}2) \rightarrow \mathrm{R} \\ &\mathrm{f}(\mathrm{x})=\tan \mathrm{x} \end{aligned}$

$\begin{aligned} &\mathrm{f}^{-1}: \mathrm{R} \rightarrow(-\frac{\pi}2, \frac{\pi}2) \\ &\mathrm{f}^{-1}(\mathrm{x})=\tan ^{-1} \mathrm{x} \end{aligned}$

The domain of the inverse tangent function is $(-\infty, \infty)$ and the range is $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.

(vi) Graphs of cot x and cot^-1 x

$\begin{aligned} &\mathrm{f}:(0, \pi) \rightarrow \mathrm{R} \\ &\mathrm{f}(\mathrm{x})=\cot \mathrm{x} \end{aligned}$

$\begin{aligned} &\mathrm{f}^{-1}: \mathrm{R} \rightarrow(0, \pi) \\ &\mathrm{f}^{-1}(\mathrm{x})=\cot ^{-1} \mathrm{x} \end{aligned}$

The domain of the inverse cotangent function is $(-\infty, \infty)$ and the range is $(0, \pi)$.

Note

• All the inverse trigonometric functions represent an angle.
• If $x\geq 0$, then all six inverse trigonometric functions viz sin^-1 x, cos^-1 x, tan^-1 x, sec^-1x, cosec^-1x, and cot^-1x represent an acute angle.
• If x < 0, then sin^-1x, tan^-1x & cosec^-1x represent an angle from $-\frac{\pi}2$ to 0.
• If x < 0, then cos^-1 x, cot^-1 x & sec^-1 x represent an obtuse angle.
• (III)rd quadrant is never used in inverse trigonometric functions.

Important Trigonometric Functions Formula

There are some important formulas in the NCERT Class 12 Maths chapter 2, Inverse trigonometric functions, which will be useful while solving NCERT problems. These formulas are conditional and can be used for a certain range of values of x.

Inverse Trigonometric Functions: Previous Year Question and Answer

Question 1:
The value of $\sin ^{-1}\left ( \cos \frac{3\pi }{5} \right )$ is: [CBSE 2020]

Solution:
$\\ \sin ^{-1}\left ( \cos \frac{3\pi }{5} \right ) \\\\ = \sin ^{-1}\left ( \sin (\frac{\pi}{2} - \frac{3\pi }{5}) \right ) \\\\ = \sin ^{-1}\left ( \sin (\frac{5\pi- 6\pi}{10}) \right ) \\\\ = \sin ^{-1}\left ( \sin \frac{-\pi}{10} \right ) \\\\ =- \frac{\pi}{10}$

Hence, the correct answer is $-\frac{\pi}{10}$.

Question 2:
Evaluate : $\sin ^{-1}\left(\sin \frac{3 \pi}{5}\right)$ [CBSE 2025]

Solution:
The inverse sine function $\sin ^{-1} x$ has a principal value range of:
$
\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] .
$
This means:

$
\sin ^{-1}(\sin \theta)=\theta, \text { only if } \theta \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] .
$

If $\theta$ is outside this interval, $\sin ^{-1}(\sin \theta)$ will be the principal value of the sine function within $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.
Now, $
\frac{3 \pi}{5}=\frac{3 \times 180^{\circ}}{5}=108^{\circ}
$
which is not in the principal range $\left[-90^{\circ}, 90^{\circ}\right]$.
Since $\sin (\pi-x)=\sin x$,
and $\pi-\frac{3 \pi}{5}=\frac{2 \pi}{5}$, and $\frac{2 \pi}{5}=\frac{2 \times 180^{\circ}}{5}=72^{\circ}$, which is in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$,
we can write:

$
\sin \frac{3 \pi}{5}=\sin \left(\pi-\frac{3 \pi}{5}\right)=\sin \frac{2 \pi}{5} .
$
Therefore,

$
\sin ^{-1}\left(\sin \frac{3 \pi}{5}\right)=\sin ^{-1}\left(\sin \frac{2 \pi}{5}\right)=\frac{2 \pi}{5} .
$

Hence, the correct answer is ($\frac{2\pi}{5}$).

Question 3:
Prove that $\tan \left[2 \tan ^{-1}\left(\frac{1}{2}\right)-\cot ^{-1} 3\right]=\frac{9}{13}$
[CBSE 2020]

Solution:

To prove:

$\begin{aligned} & \tan \left[2 \tan ^{-1}\left(\frac{1}{2}\right)-\cot ^{-1} 3\right]=\frac{9}{13} \\ & \text { LHS }=\tan \left[2 \tan ^{-1} \frac{1}{2}-\cot ^{-1} 3\right] \\ & =\tan \left[\tan ^{-1}\left(\frac{2 \times \frac{1}{2}}{1-\left(\frac{1}{2}\right)^2}\right)-\cot ^{-1} 3\right] \\ & =\tan \left[\tan ^{-1}\left\{\frac{1}{1-\frac{1}{4}}\right\}-\tan ^{-1} \frac{1}{3}\right] \\ & =\tan \left[\tan ^{-1} \frac{4}{3}-\tan ^{-1} \frac{1}{3}\right] \\ & =\tan \left[\tan ^{-1}\left(\frac{\frac{4}{3}-\frac{1}{3}}{1+\frac{4}{9}}\right)\right] \\ & =\tan \left[\tan ^{-1}\left(\frac{\frac{4-1}{3}}{\frac{9+4}{9}}\right)\right] \\ & =\tan \left[\tan ^{-1}\left(\frac{1}{\frac{13}{9}}\right)\right] \\ & =\tan \left[\tan ^{-1}\left(\frac{9}{13}\right)\right] \\ & =\frac{9}{13}=R H S\end{aligned}$

Hence, the given statement is proved.

NCERT Class 12 Notes Chapter Wise

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NCERT Exemplar Solutions for Class 12 Subject-wise

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• NCERT Exemplar Class 12 Solutions
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• NCERT Solutions for Class 12 Mathematics
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NCERT Books and Syllabus

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