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

In the language of trigonometry, inverse functions let angles speak for themselves. Trigonometry tells you where you’re going; inverse trigonometric functions tell you where you started. Imagine you have moved to a new town and need to visit the nearest bank urgently. You know the distance, but you don't know the angles which you should turn to reach there quickly. This is where inverse trigonometric functions are very helpful. In the Inverse Trigonometric Functions Class 12th notes PDF, students will learn about the domain and range of all trigonometric functions, along with their graphs. The main purpose of these NCERT notes of Inverse Trigonometry class 12 PDF is to provide students with an efficient study material from which they can revise the entire chapter.

Inverse trigonometric functions are the detectives of mathematics, finding the angles hidden behind sine, cosine, and tangent. Like trigonometric functions, these are also used in geometry, navigation, science, and engineering. Students should complete the inverse trigonometric functions class 12 questions and answers from the textbook to strengthen all the basic concepts. These Inverse Trigonometric Functions Class 12 Handwritten Notes are prepared by experienced Careers360 teachers, keeping the needs of students in mind. These NCERT Maths notes for class 12 will also clarify some important questions of students related to this chapter. For your syllabus, solutions, and chapter-wise PDFs, head over to this link: NCERT.

Also, read,

• NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions
• NCERT Exemplar Class 12 Maths Solutions Chapter 2 Inverse Trigonometric Functions

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:-

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

Graphs Of Inverse Trigonometric Functions:-

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

$\begin{array}{rl}& \mathrm{f}:[\frac{-\pi }{2},\frac{\pi }{2}]\to [-1,1]\\ & \mathrm{f}(\mathrm{x})=\sin \mathrm{x}\end{array}$

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

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:

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

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

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

$\begin{array}{rl}& f:[-\frac{\pi }{2},0)\cup (0,\frac{\pi }{2}]\to (-\mathrm{\infty },-1]\cup [1,\mathrm{\infty })\\ & f(x)=\mathrm{cosec}x\end{array}$

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

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

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

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

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

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

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

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

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

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

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

$\begin{array}{rl}& \mathrm{f}:(0,\pi )\to \mathrm{R}\\ & \mathrm{f}(\mathrm{x})=\cot \mathrm{x}\end{array}$

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

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

Note:-

• All the inverse trigonometric functions represent an angle.
• If $x\ge 0$, 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 ${\sin }^{-1}(\cos \frac{3\pi }{5})$ is: [CBSE 2020]

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

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

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

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

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

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

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

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

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

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

Solution:

To prove:

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

Hence, the given statement is proved.

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