Chapter 2 - Simplification of Inverse Trigonometric Functions
Overview
This page provides comprehensive Class 12 Maths Chapter 2 Simplification of Inverse Trigonometric Functions. Chapter 2 - Simplification of Inverse Trigonometric Functions - Free study material for Class 12 Maths. NCERT Solutions, Notes, and PYQs.
Class 12 Mathematics | CBSE Previous Year Questions
Q1
2020
00:00
Evaluate: $\sin(\tan^{-1}x)$, where $x > 0$.
Let $\theta = \tan^{-1}x$, so $\tan\theta = x$ and $\theta \in (-\pi/2, \pi/2)$.
Consider a right triangle with opposite side = $x$, adjacent side = $1$.
Then hypotenuse = $\sqrt{1 + x^2}$.
So $\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{x}{\sqrt{1 + x^2}}$.
Hence, $\sin(\tan^{-1}x) = \frac{x}{\sqrt{1 + x^2}}$.
$\dfrac{x}{\sqrt{1 + x^2}}$
Q2
2020
00:00
Find the value of $\cos(\sin^{-1}x)$, where $-1 \le x \le 1$.
Let $\theta = \sin^{-1}x$, so $\sin\theta = x$ and $\theta \in [-\pi/2, \pi/2]$.
Consider a right triangle with opposite side = $x$ and hypotenuse = $1$.
Then adjacent side = $\sqrt{1 - x^2}$.
So $\cos\theta = \frac{\sqrt{1 - x^2}}{1} = \sqrt{1 - x^2}$.
Hence, $\cos(\sin^{-1}x) = \sqrt{1 - x^2}$.
$\sqrt{1 - x^2}$
Q3
2021
00:00
Evaluate: $\tan(\cos^{-1}x)$, where $-1 < x < 1$.
Let $\theta = \cos^{-1}x$, so $\cos\theta = x$ and $\theta \in [0, \pi]$.
In a right triangle, take adjacent side = $x$ and hypotenuse = $1$.
Then opposite side = $\sqrt{1 - x^2}$.
So $\tan\theta = \frac{\sqrt{1 - x^2}}{x}$.
Hence, $\tan(\cos^{-1}x) = \frac{\sqrt{1 - x^2}}{x}$.
$\dfrac{\sqrt{1 - x^2}}{x}$
Q4
2022
00:00
Evaluate: $\sin^{-1}(\cos x)$, where $0 \le x \le \pi$.
We know that $\cos x = \sin(\frac{\pi}{2} - x)$.
So $\sin^{-1}(\cos x) = \sin^{-1}(\sin(\frac{\pi}{2} - x))$.
For $0 \le x \le \pi$, the value of $(\frac{\pi}{2} - x)$ lies in $[-\pi/2, \pi/2]$.
Hence, $\sin^{-1}(\sin(\frac{\pi}{2} - x)) = \frac{\pi}{2} - x$.
$\frac{\pi}{2} - x$
Q5
2023
00:00
Simplify: $\tan(\sin^{-1}x)$, where $-1 < x < 1$.
Let $\theta = \sin^{-1}x$, so $\sin\theta = x$ and $\theta \in [-\pi/2, \pi/2]$.
In a right triangle, opposite side = $x$, hypotenuse = $1$.
Then adjacent side = $\sqrt{1 - x^2}$.
So $\tan\theta = \frac{x}{\sqrt{1 - x^2}}$.
Hence, $\tan(\sin^{-1}x) = \frac{x}{\sqrt{1 - x^2}}$.
$\dfrac{x}{\sqrt{1 - x^2}}$
Q6
2023
00:00
Evaluate: $\cos(\tan^{-1}x)$, where $x > 0$.
Let $\theta = \tan^{-1}x$, so $\tan\theta = x$ and $\theta \in (-\pi/2, \pi/2)$.
Take a right triangle with opposite side = $x$ and adjacent side = $1$.
Then hypotenuse = $\sqrt{1 + x^2}$.
So $\cos\theta = \frac{1}{\sqrt{1 + x^2}}$.
Hence, $\cos(\tan^{-1}x) = \frac{1}{\sqrt{1 + x^2}}$.
$\dfrac{1}{\sqrt{1 + x^2}}$
Q7
2022
00:00
Find the value of $\sec(\sin^{-1}x)$, where $0 < x < 1$.
Let $\theta = \sin^{-1}x$, so $\sin\theta = x$.
In a right triangle, opposite side = $x$, hypotenuse = $1$.
Adjacent side = $\sqrt{1 - x^2}$.
So $\cos\theta = \sqrt{1 - x^2}$.
Hence, $\sec\theta = \frac{1}{\cos\theta} = \frac{1}{\sqrt{1 - x^2}}$.
Therefore, $\sec(\sin^{-1}x) = \frac{1}{\sqrt{1 - x^2}}$.
$\dfrac{1}{\sqrt{1 - x^2}}$
Q8
2024
00:00
Simplify: $\sin^{-1}(\sin x)$, where $-\frac{\pi}{2} \le x \le \frac{\pi}{2}$.
The principal value range of $\sin^{-1}x$ is $[-\pi/2, \pi/2]$.
Given that $x$ itself lies in this interval,
Therefore, $\sin^{-1}(\sin x) = x$.
$x$
Q9
2024
00:00
Evaluate: $\tan^{-1}(\sin(\cos^{-1}x))$, where $-1 \le x \le 1$.
Let $\theta = \cos^{-1}x$, so $\cos\theta = x$ and $\theta \in [0, \pi]$.
Then $\sin\theta = \sqrt{1 - x^2}$.
So the expression becomes $\tan^{-1}(\sqrt{1 - x^2})$.
Hence, the simplified form is $\tan^{-1}(\sqrt{1 - x^2})$.
$\tan^{-1}(\sqrt{1 - x^2})$
Q10
2025
00:00
Assertion (A): $\tan(\sin^{-1}x) = \frac{x}{\sqrt{1 - x^2}}$ for $|x|<1$. Reason (R): If $\sin\theta = x$, then $\cos\theta = \sqrt{1 - x^2}$.
Let $\theta = \sin^{-1}x$, then $\sin\theta = x$.
In a right triangle, hypotenuse = 1, opposite side = x, so adjacent side = $\sqrt{1 - x^2}$.
Thus, $\tan\theta = \frac{x}{\sqrt{1 - x^2}}$.
Hence, the Assertion is true and the Reason correctly explains it.
Both A and R are true and R is the correct explanation of A.