Practice Miscellaneous Exercise

Overview

This page provides an Overview of Chapter 2 Inverse Trigonometric Functions for CBSE Class 12. Key topics covered: Q3 - Q7. Proving Properties, Q8 - Q10. Simplest Form, Q11 - Q14. Solve & Select. All content is aligned with the NCERT syllabus for Board Exam preparation.

Inverse Trigonometric Functions: Advanced problems on principal values, properties, and equations.

Chapter Index Start Solving

Q1 Val Q2 Val Q3 Prove Q4 Prove Q5 Prove Q6 Prove Q7 Prove Q8 Simp Q9 Simp Q10 Simp Q11 Solve Q12 Solve Q13 MCQ Q14 MCQ

Q1 - Q2. Principal Values

Practice Question 1

Find the value of: \( \cos^{-1} (\cos \frac{5\pi}{3}) \)

View Step-by-Step Solution

Step 1: Check Range

Range of \( \cos^{-1} \) is \( [0, \pi] \).
\( \frac{5\pi}{3} \) is approx \( 300^\circ \), which is outside the range.

Step 2: Adjust Angle

Use periodicity: \( \cos(2\pi - \theta) = \cos \theta \).
\( \cos(\frac{5\pi}{3}) = \cos(2\pi - \frac{\pi}{3}) = \cos(\frac{\pi}{3}) \).

Step 3: Result

\( \cos^{-1}(\cos \frac{\pi}{3}) = \frac{\pi}{3} \) (Since \( \frac{\pi}{3} \in [0, \pi] \)).

Value = \( \frac{\pi}{3} \)

Practice Question 2

Find the value of: \( \tan^{-1} (\tan \frac{5\pi}{4}) \)

View Solution

Range Check

Range of \( \tan^{-1} \) is \( (-\frac{\pi}{2}, \frac{\pi}{2}) \). \( \frac{5\pi}{4} \) is outside.

Adjustment

\( \tan(\frac{5\pi}{4}) = \tan(\pi + \frac{\pi}{4}) = \tan(\frac{\pi}{4}) \).

Value = \( \frac{\pi}{4} \)

Q3 - Q7. Proving Properties

Practice Question 3

Prove that: \( 2\tan^{-1} \frac{1}{2} = \tan^{-1} \frac{4}{3} \)

View Solution

Formula

Use \( 2\tan^{-1}x = \tan^{-1}(\frac{2x}{1-x^2}) \).

Calculation

LHS = \( \tan^{-1}(\frac{2(1/2)}{1-(1/2)^2}) = \tan^{-1}(\frac{1}{1-1/4}) \)
= \( \tan^{-1}(\frac{1}{3/4}) = \tan^{-1}(\frac{4}{3}) \).

LHS = RHS. Proved.

Practice Question 4

Prove: \( \sin^{-1} \frac{5}{13} + \sin^{-1} \frac{12}{13} = \frac{\pi}{2} \)

View Solution

Conversion

Let \( \sin^{-1} \frac{12}{13} = \theta \Rightarrow \sin \theta = \frac{12}{13} \).
Then \( \cos \theta = \frac{5}{13} \Rightarrow \theta = \cos^{-1} \frac{5}{13} \).

Identity

LHS = \( \sin^{-1} \frac{5}{13} + \cos^{-1} \frac{5}{13} \).
We know \( \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \).

LHS = \( \pi/2 \). Proved.

Practice Question 5

Prove: \( \cos^{-1} \frac{3}{5} + \cos^{-1} \frac{4}{5} = \frac{\pi}{2} \)

View Solution

Conversion

Convert one term to sine. Let \( \cos^{-1} \frac{4}{5} = x \).
\( \cos x = 4/5 \Rightarrow \sin x = 3/5 \). So \( x = \sin^{-1} \frac{3}{5} \).

Result

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

Proved.

Practice Question 6

Prove: \( \tan^{-1} \frac{1}{2} + \tan^{-1} \frac{1}{3} = \frac{\pi}{4} \)

View Solution

Formula

\( \tan^{-1} x + \tan^{-1} y = \tan^{-1} (\frac{x+y}{1-xy}) \) if \( xy < 1 \).
Here \( (1/2)(1/3) = 1/6 < 1 \).

Calculation

LHS = \( \tan^{-1} (\frac{1/2 + 1/3}{1 - 1/6}) = \tan^{-1} (\frac{5/6}{5/6}) = \tan^{-1}(1) \).

LHS = \( \pi/4 \). Proved.

Practice Question 7

Prove: \( \tan^{-1} x + \tan^{-1} y + \tan^{-1} z = \pi \) if \( x+y+z = xyz \)

View Solution

Formula

\( \tan^{-1} (\frac{x+y+z-xyz}{1-xy-yz-zx}) \).

Apply Condition

Since \( x+y+z = xyz \), numerator is 0.
\( \tan^{-1}(0) = \pi \) (Given general context of positive reals sum).

Proved.

Q8 - Q10. Simplest Form

Practice Question 8

Prove: \( \tan^{-1} x = \frac{1}{2} \cos^{-1} (\frac{1-x^2}{1+x^2}) \) for \( x \ge 0 \)

View Solution

Substitution

Let \( x = \tan \theta \).
RHS = \( \frac{1}{2} \cos^{-1} (\frac{1-\tan^2\theta}{1+\tan^2\theta}) \).

Identity

RHS = \( \frac{1}{2} \cos^{-1} (\cos 2\theta) = \frac{1}{2}(2\theta) = \theta \).

RHS = \( \tan^{-1} x \). Proved.

Practice Question 9

Simplify: \( \tan^{-1} \left( \frac{\sqrt{1+\cos x} - \sqrt{1-\cos x}}{\sqrt{1+\cos x} + \sqrt{1-\cos x}} \right) \)

View Solution

Identities

\( 1+\cos x = 2\cos^2(x/2) \), \( 1-\cos x = 2\sin^2(x/2) \).

Substitute

Numerator: \( \sqrt{2}\cos(x/2) - \sqrt{2}\sin(x/2) \)
Denominator: \( \sqrt{2}\cos(x/2) + \sqrt{2}\sin(x/2) \)

Divide by Cos

\( \tan^{-1} (\frac{1-\tan(x/2)}{1+\tan(x/2)}) = \tan^{-1}(\tan(\pi/4 - x/2)) \).

Simplest Form: \( \frac{\pi}{4} - \frac{x}{2} \)

Practice Question 10

Simplify: \( \tan^{-1} \left( \frac{\sqrt{1+x^2}-1}{x} \right) \)

View Solution

Substitution

Put \( x = \tan \theta \). Exp = \( \frac{\sec \theta - 1}{\tan \theta} \).

Simplification

\( \frac{1-\cos \theta}{\sin \theta} = \frac{2\sin^2(\theta/2)}{2\sin(\theta/2)\cos(\theta/2)} = \tan(\theta/2) \).

Simplest Form: \( \frac{1}{2} \tan^{-1} x \)

Q11 - Q14. Solve & Select

Practice Question 11

Solve for x: \( 2\tan^{-1}(\sin x) = \tan^{-1}(2\sec x) \)

View Solution

Apply Formula

LHS: \( \tan^{-1}(\frac{2\sin x}{1-\sin^2 x}) = \tan^{-1}(\frac{2\sin x}{\cos^2 x}) \).
Equate arguments: \( \frac{2\sin x}{\cos^2 x} = \frac{2}{\cos x} \).

Solve

\( \tan x = 1 \).

\( x = \frac{\pi}{4} \)

Practice Question 12

Solve for x: \( \tan^{-1}(2x) + \tan^{-1}(3x) = \frac{\pi}{4} \)

View Solution

Formula

\( \tan^{-1}(\frac{2x+3x}{1-6x^2}) = \frac{\pi}{4} \).
\( \frac{5x}{1-6x^2} = 1 \).

Quadratic Equation

\( 6x^2 + 5x - 1 = 0 \). Factors: \( (6x-1)(x+1) = 0 \).
\( x = 1/6 \) or \( x = -1 \).

Check

\( x = -1 \) gives LHS as negative, but RHS is positive. Reject -1.

\( x = \frac{1}{6} \)

Practice Question 13 (MCQ)

The value of \( \cos(\tan^{-1} x) \) is equal to:
(A) \( \sqrt{1+x^2} \) (B) \( \frac{1}{\sqrt{1+x^2}} \) (C) \( 1+x^2 \) (D) None

View Solution

Triangle Method

Let \( \tan^{-1} x = \theta \Rightarrow \tan \theta = x/1 \).
Opp = x, Adj = 1. Hyp = \( \sqrt{1+x^2} \).
\( \cos \theta = \text{Adj}/\text{Hyp} = 1/\sqrt{1+x^2} \).

Correct Option: (B)

Practice Question 14 (MCQ)

If \( \sin^{-1}(1-x) - 2\sin^{-1}x = \frac{\pi}{2} \), then x is equal to:
(A) 0, 1/2 (B) 1, 1/2 (C) 0 (D) 1/2

View Solution

Substitution Check

Try x = 0: \( \sin^{-1}(1) - 0 = \pi/2 \). (Works).
Try x = 1/2: \( \sin^{-1}(1/2) - 2\sin^{-1}(1/2) = -\sin^{-1}(1/2) = -\pi/6 \ne \pi/2 \).

Correct Option: (C) 0