Inverse Trigonometric Function Notes for Class 12 and IIT JEE

Inverse Trigonometric Function Notes: Definition and Basics

The functions $\sin x$, $\cos x$ and $\tan x$ are many-one on their natural domains, so their inverses do not exist until the domain is restricted. Once each function is restricted to a suitable interval, called the principal value branch, it becomes one-one and onto, and its inverse is well defined. These inverses are the inverse trigonometric functions: $\sin^{-1}x$, $\cos^{-1}x$, $\tan^{-1}x$, $\cot^{-1}x$, $\sec^{-1}x$ and $\csc^{-1}x$. They are also written $\arcsin x$, $\arccos x$, $\arctan x$ and so on.

By definition, $y=\sin^{-1}x \iff x=\sin y$ with $y\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$.

Notation warning: $\sin^{-1}x$ is the angle whose sine is $x$. It is not $\dfrac{1}{\sin x}$, which is $\csc x$. This is one of the most common errors at the start of the chapter.

These notes assume you already know the basic ratios and identities. If needed, revise the trigonometric formulas class 11 page first.

Domain and Range in Inverse Trigonometric Function Notes (Principal Value Branches)

The principal value of an inverse trigonometric function is the value that lies in its principal value branch. Unless a question says otherwise, only principal values are taken.

Quick memory aid: $\sin^{-1}$, $\tan^{-1}$ and $\csc^{-1}$ give angles in the first and fourth quadrants, so negative inputs give negative angles. $\cos^{-1}$, $\cot^{-1}$ and $\sec^{-1}$ give angles in the first and second quadrants.

Why Are These Ranges Chosen?

Each range is an interval on which the original trigonometric function is one-one and covers its full range of values. For example, $\cos x$ is strictly decreasing on $[0,\pi]$, so it has an inverse there. On $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ it is not one-one, since $\cos\frac{\pi}{3}=\cos\left(-\frac{\pi}{3}\right)$, so that interval cannot be used for $\cos^{-1}x$.

Graphs of Inverse Trigonometric Functions

The graph of each inverse function is the mirror image of the restricted trigonometric function about the line $y=x$. Three shapes to remember:

• $\sin^{-1}x$ increases from $\left(-1,-\frac{\pi}{2}\right)$ to $\left(1,\frac{\pi}{2}\right)$ and passes through the origin.
• $\cos^{-1}x$ decreases from $(-1,\pi)$ to $(1,0)$ and passes through $\left(0,\frac{\pi}{2}\right)$.
• $\tan^{-1}x$ increases for all real $x$ and has horizontal asymptotes $y=\pm\frac{\pi}{2}$.

The visual notes at the end of this page show these graphs.

How to Find the Principal Value of Inverse Trigonometric Functions

1. Write $y=f^{-1}(x)$, so $x=f(y)$.
2. Find the reference angle whose trigonometric value matches $|x|$.
3. Place the angle in the principal value branch of that function, using the sign of $x$.

Formulas in Inverse Trigonometric Function Notes for JEE and Boards

CBSE Class 12 mainly asks for definitions, domain, range, principal values and graphs. The identities below are mostly used in JEE-level simplification and in calculus.

Inverse of a Function (Cancellation) Properties

$$\sin(\sin^{-1}x)=x,\quad x\in[-1,1]$$ $$\sin^{-1}(\sin x)=x,\quad x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$$ $$\cos^{-1}(\cos x)=x,\quad x\in[0,\pi]$$ $$\tan^{-1}(\tan x)=x,\quad x\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$$

Outside these intervals, first reduce the angle into the principal range, then apply the inverse.

Negative Argument Properties

$$\sin^{-1}(-x)=-\sin^{-1}x,\quad \tan^{-1}(-x)=-\tan^{-1}x,\quad \csc^{-1}(-x)=-\csc^{-1}x$$ $$\cos^{-1}(-x)=\pi-\cos^{-1}x,\quad \cot^{-1}(-x)=\pi-\cot^{-1}x,\quad \sec^{-1}(-x)=\pi-\sec^{-1}x$$

Reciprocal Properties

$$\sin^{-1}\frac{1}{x}=\csc^{-1}x,\qquad \cos^{-1}\frac{1}{x}=\sec^{-1}x,\qquad |x|\ge 1$$ $$\tan^{-1}\frac{1}{x}=\cot^{-1}x\ \ (x>0),\qquad \tan^{-1}\frac{1}{x}=\cot^{-1}x-\pi\ \ (x<0)$$

Complementary Properties

$$\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2},\quad x\in[-1,1]$$ $$\tan^{-1}x+\cot^{-1}x=\frac{\pi}{2},\quad x\in\mathbb{R}$$ $$\sec^{-1}x+\csc^{-1}x=\frac{\pi}{2},\quad |x|\ge 1$$

Sum and Difference of Inverse Tangents

$$\tan^{-1}x+\tan^{-1}y=\begin{cases}\tan^{-1}\dfrac{x+y}{1-xy}, & xy<1\\[2mm] \pi+\tan^{-1}\dfrac{x+y}{1-xy}, & x,y>0,\ xy>1\\[2mm] -\pi+\tan^{-1}\dfrac{x+y}{1-xy}, & x,y<0,\ xy>1\end{cases}$$ $$\tan^{-1}x-\tan^{-1}y=\tan^{-1}\frac{x-y}{1+xy},\quad xy>-1$$

Double-Angle (Conversion) Forms

$$2\tan^{-1}x=\sin^{-1}\frac{2x}{1+x^{2}},\quad |x|\le 1$$ $$2\tan^{-1}x=\cos^{-1}\frac{1-x^{2}}{1+x^{2}},\quad x\ge 0$$ $$2\tan^{-1}x=\tan^{-1}\frac{2x}{1-x^{2}},\quad -1<x<1$$

Solved Examples from the Inverse Trigonometric Function Notes

Example 1. Find $\sin^{-1}\left(\sin\frac{5\pi}{6}\right)$.

Since $\frac{5\pi}{6}\notin\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$, we cannot cancel directly. Now $\sin\frac{5\pi}{6}=\sin\left(\pi-\frac{\pi}{6}\right)=\frac{1}{2}$, so $\sin^{-1}\frac{1}{2}=\frac{\pi}{6}$.

Example 2. Find $\cos^{-1}\left(\cos\frac{7\pi}{6}\right)$.

$\cos\frac{7\pi}{6}=\cos\left(2\pi-\frac{5\pi}{6}\right)=\cos\frac{5\pi}{6}$, and $\frac{5\pi}{6}\in[0,\pi]$. So the answer is $\frac{5\pi}{6}$.

Example 3. Evaluate $\sin\left(\cos^{-1}\frac{3}{5}\right)$.

Let $\theta=\cos^{-1}\frac{3}{5}$ with $\theta\in[0,\pi]$, so $\sin\theta\ge 0$. Then $\sin\theta=\sqrt{1-\frac{9}{25}}=\frac{4}{5}$.

Example 4. Show that $\tan^{-1}1+\tan^{-1}2+\tan^{-1}3=\pi$.

For $\tan^{-1}1+\tan^{-1}2$: $x=1$ and $y=2$ are positive and $xy=2>1$, so the sum is $\pi+\tan^{-1}\frac{3}{1-2}=\pi-\tan^{-1}3$. Adding $\tan^{-1}3$ gives $\pi$.

Example 5. Simplify $\tan^{-1}\dfrac{\sqrt{1+x^{2}}-1}{x}$ for $x\ne 0$.

Put $x=\tan\theta$ with $\theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$. The expression becomes $\dfrac{\sec\theta-1}{\tan\theta}=\dfrac{1-\cos\theta}{\sin\theta}=\tan\frac{\theta}{2}$. Since $\frac{\theta}{2}$ lies in the principal range, the result is $\frac{\theta}{2}=\frac{1}{2}\tan^{-1}x$.

Common Mistakes in Inverse Trigonometric Functions

These are the slips that most inverse trigonometric function notes warn about:

• Reading $\sin^{-1}x$ as $\frac{1}{\sin x}$.
• Writing $\sin^{-1}(\sin x)=x$ for every $x$. It holds only inside the principal range.
• Giving an answer outside the principal value branch, such as $\frac{7\pi}{6}$ for $\sin^{-1}\left(-\frac{1}{2}\right)$.
• Using the $\tan^{-1}x+\tan^{-1}y$ formula without checking the sign of $1-xy$.
• Using the wrong range for $\cot^{-1}x$. It is $(0,\pi)$, not $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$.

Inverse Trigonometric Function Notes for the CBSE Class 12 and JEE Syllabus

• CBSE Class 12: The topic is Chapter 2 of NCERT Maths Part 1. In the 2026-27 course structure, Relations and Functions (Chapters 1 and 2 together) carries 8 marks out of 80 theory marks. Check the official curriculum for any change.
• JEE Main: These functions are used in functions, trigonometry and calculus. Practise with JEE Main chapterwise PYQs after finishing the theory.
• Later chapters: The derivatives appear in Continuity and Differentiability:

$$\frac{d}{dx}\sin^{-1}x=\frac{1}{\sqrt{1-x^{2}}},\quad \frac{d}{dx}\cos^{-1}x=-\frac{1}{\sqrt{1-x^{2}}},\quad \frac{d}{dx}\tan^{-1}x=\frac{1}{1+x^{2}}$$

NEET UG has no Mathematics section, so this chapter matters for boards and JEE-type exams.

Inverse Trigonometric Function Notes