Chapter 2 - Properties of Inverse Trigonometric Functions

Overview

This page provides comprehensive Class 12 Maths Chapter 2 Properties of Inverse Trigonometric Functions. Chapter 2 - Properties 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: $\sin(\sin^{-1}x)$, where $-1 \le x \le 1$.

Let $y = \sin^{-1}x$.

Then $\sin y = x$.

Since the principal value range of $\sin^{-1}x$ is $[-\pi/2, \pi/2]$, the sine function is one-one in this interval.

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

Q2 2020

00:00

Evaluate: $\cos(\cos^{-1}x)$, where $-1 \le x \le 1$.

Let $y = \cos^{-1}x$.

Then $\cos y = x$.

The principal value range of $\cos^{-1}x$ is $[0, \pi]$, where cosine is one-one.

Hence, $\cos(\cos^{-1}x) = x$.

Q3 2021

00:00

Evaluate: $\tan(\tan^{-1}x)$, for all real $x$.

Let $y = \tan^{-1}x$.

Then $\tan y = x$.

The principal value range of $\tan^{-1}x$ is $(-\pi/2, \pi/2)$, where tangent is one-one.

Therefore, $\tan(\tan^{-1}x) = x$.

Q4 2021

00:00

Find the value of $\sin^{-1}(\sin \frac{5\pi}{6})$.

(a)$\frac{\pi}{6}$

(b)$\frac{5\pi}{6}$

(c)$-\frac{\pi}{6}$

(d)$-\frac{5\pi}{6}$

We know that $\sin(5\pi/6) = 1/2$.

So $\sin^{-1}(\sin 5\pi/6) = \sin^{-1}(1/2)$.

The principal value of $\sin^{-1}(1/2)$ in $[-\pi/2, \pi/2]$ is $\pi/6$.

$\pi/6$

Q5 2022

00:00

Find the value of $\cos^{-1}(\cos \frac{5\pi}{6})$.

(a)$\frac{\pi}{6}$

(b)$\frac{5\pi}{6}$

(c)$\frac{7\pi}{6}$

(d)$\frac{11\pi}{6}$

We know that $\cos(5\pi/6) = -\sqrt{3}/2$.

The principal value range of $\cos^{-1}x$ is $[0, \pi]$.

Since $5\pi/6$ lies in $[0, \pi]$, $\cos^{-1}(\cos 5\pi/6) = 5\pi/6$.

$\frac{5\pi}{6}$

Q6 2022

00:00

Find the value of $\tan^{-1}(\tan \frac{5\pi}{3})$.

(a)$\frac{5\pi}{3}$

(b)$\frac{\pi}{3}$

(c)$-\frac{\pi}{3}$

(d)$\frac{2\pi}{3}$

The principal value range of $\tan^{-1}x$ is $(-\pi/2, \pi/2)$.

$\tan(5\pi/3) = \tan(-\pi/3) = -\sqrt{3}$.

So $\tan^{-1}(\tan 5\pi/3) = \tan^{-1}(-\sqrt{3})$.

The principal value of $\tan^{-1}(-\sqrt{3})$ is $-\pi/3$.

$-\frac{\pi}{3}$

Q7 2021

00:00

Evaluate: $\sin^{-1}x + \cos^{-1}x$, for $-1 \le x \le 1$.

(a)$0$

(b)$\pi/2$

(c)$\pi$

(d)$2\pi$

Let $\sin^{-1}x = y$, then $\sin y = x$, where $y \in [-\pi/2, \pi/2]$.

We know the identity: $\sin^{-1}x + \cos^{-1}x = \pi/2$ for all $x$ in $[-1,1]$.

$\pi/2$

Q8 2021

00:00

Evaluate: $\tan^{-1}x + \cot^{-1}x$, for $x > 0$.

(a)$0$

(b)$\pi/4$

(c)$\pi/2$

(d)$\pi$

For $x > 0$, we use the identity: $\tan^{-1}x + \cot^{-1}x = \pi/2$.

This follows from the complementary angle property of inverse functions.

$\pi/2$

Q9 2023

00:00

Assertion (A): $\sin^{-1}x + \cos^{-1}x = \pi/2$ for all $x$ in $[-1,1]$. Reason (R): The ranges of $\sin^{-1}x$ and $\cos^{-1}x$ are $[-\pi/2, \pi/2]$ and $[0, \pi]$ respectively.

(a)Both A and R are true and R is the correct explanation of A.

(b)Both A and R are true but R is not the correct explanation of A.

(c)A is true but R is false.

(d)A is false but R is true.

The identity $\sin^{-1}x + \cos^{-1}x = \pi/2$ is true for all $x$ in $[-1,1]$.

The given reason correctly states the principal value ranges which ensure complementary angles.

Thus, both the Assertion and Reason are true and the Reason explains the Assertion.

Both A and R are true and R is the correct explanation of A.

Q10 2024

00:00

Find the value of $\cos^{-1}(-x)$ in terms of $\cos^{-1}x$, where $-1 \le x \le 1$.

(a)$\pi - \cos^{-1}x$

(b)$\cos^{-1}x - \pi$

(c)$\pi + \cos^{-1}x$

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

Let $y = \cos^{-1}x$, so $\cos y = x$, where $y \in [0, \pi]$.

Then $\cos(\pi - y) = -\cos y = -x$.

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

$\pi - \cos^{-1}x$