Practice Exercise 2.2

Revise Chapter Notes

Overview

This page provides an Overview of Chapter 2 Inverse Trigonometric Functions for CBSE Class 12. Key topics covered: Q3 - Q7. Simplest Form, Q8 - Q12. Finding Values, Q13 - Q15. MCQs. All content is aligned with the NCERT syllabus for Board Exam preparation.

Inverse Trigonometric Functions: New problems to test your understanding.

Chapter Index Start Solving
Q1 Prove Q2 Prove Q3 Simplify Q4 Simplify Q5 Simplify Q6 Simplify Q7 Simplify Q8 Find Q9 Find Q10 Find Q11 Find Q12 Find Q13 MCQ Q14 MCQ Q15 MCQ

Q1 - Q2. Proving Properties

Practice Question 1
Prove: \( \tan^{-1} x + \tan^{-1} \frac{2x}{1 - x^2} = \tan^{-1} \frac{3x - x^3}{1 - 3x^2} \) for \( |x| < \frac{1}{\sqrt{3}} \)
View Step-by-Step Solution
Step 1: Simplify LHS
We know that \( \tan^{-1} \frac{2x}{1 - x^2} = 2 \tan^{-1} x \).
So, LHS = \( \tan^{-1} x + 2 \tan^{-1} x = 3 \tan^{-1} x \).
Step 2: Simplify RHS
Let \( x = \tan \theta \).
RHS = \( \tan^{-1} \frac{3 \tan \theta - \tan^3 \theta}{1 - 3 \tan^2 \theta} \)
= \( \tan^{-1} (\tan 3\theta) \) = \( 3\theta \).
Step 3: Conclusion
Since \( \theta = \tan^{-1} x \), RHS = \( 3 \tan^{-1} x \).
Thus, LHS = RHS.
Property Proved.
Practice Question 2
Prove: \( \cos^{-1}(1 - 2x^2) = 2 \sin^{-1} x \) for \( 0 \le x \le 1 \)
View Solution
Step 1: Substitution
Let \( x = \sin \theta \). Then \( \theta = \sin^{-1} x \).
LHS = \( \cos^{-1}(1 - 2 \sin^2 \theta) \)
Step 2: Identity
We know \( \cos 2\theta = 1 - 2 \sin^2 \theta \).
So, LHS = \( \cos^{-1}(\cos 2\theta) \).
Step 3: Result
LHS = \( 2\theta = 2 \sin^{-1} x = \text{RHS} \).
Property Proved.

Q3 - Q7. Simplest Form

Practice Question 3
Simplify: \( \tan^{-1} \left( \frac{\cos x}{1 + \sin x} \right) \)
View Solution
Step 1: Convert
Use \( \cos x = \sin(\frac{\pi}{2} - x) \) and \( \sin x = \cos(\frac{\pi}{2} - x) \).
Exp = \( \tan^{-1} \frac{\sin(\frac{\pi}{2} - x)}{1 + \cos(\frac{\pi}{2} - x)} \)
Step 2: Half-Angle Formula
Numerator: \( 2 \sin(\frac{\pi}{4} - \frac{x}{2}) \cos(\frac{\pi}{4} - \frac{x}{2}) \)
Denominator: \( 2 \cos^2(\frac{\pi}{4} - \frac{x}{2}) \)
Result = \( \tan^{-1} \tan(\frac{\pi}{4} - \frac{x}{2}) \).
Simplest Form: \( \frac{\pi}{4} - \frac{x}{2} \)
Practice Question 4
Simplify: \( \tan^{-1} \left( \frac{1 - \cos x}{\sin x} \right) \)
View Solution
Step 1: Identities
\( 1 - \cos x = 2 \sin^2 \frac{x}{2} \)
\( \sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2} \)
Step 2: Simplify
Ratio = \( \frac{2 \sin^2 (x/2)}{2 \sin (x/2) \cos (x/2)} = \tan \frac{x}{2} \).
\( \tan^{-1} (\tan \frac{x}{2}) = \frac{x}{2} \).
Simplest Form: \( \frac{x}{2} \)
Practice Question 5
Simplify: \( \tan^{-1} \left( \frac{a \cos x - b \sin x}{b \cos x + a \sin x} \right) \)
View Solution
Step 1: Divide
Divide numerator and denominator by \( b \cos x \):
\( \tan^{-1} \left( \frac{\frac{a}{b} - \tan x}{1 + \frac{a}{b} \tan x} \right) \)
Step 2: Identity
This matches formula \( \tan^{-1} x - \tan^{-1} y \).
Here \( x = a/b \) and \( y = \tan x \).
Result = \( \tan^{-1} \frac{a}{b} - \tan^{-1}(\tan x) \).
Simplest Form: \( \tan^{-1} \frac{a}{b} - x \)
Practice Question 6
Simplify: \( \sin^{-1} \left( \frac{x}{\sqrt{x^2 + a^2}} \right) \)
View Solution
Step 1: Substitution
Put \( x = a \tan \theta \). Then \( \sqrt{x^2 + a^2} = a \sec \theta \).
Step 2: Simplify
\( \sin^{-1} \left( \frac{a \tan \theta}{a \sec \theta} \right) = \sin^{-1} (\sin \theta) = \theta \).
Since \( \theta = \tan^{-1} \frac{x}{a} \).
Simplest Form: \( \tan^{-1} \frac{x}{a} \)
Practice Question 7
Simplify: \( \tan^{-1} \left( \frac{3x - x^3}{1 - 3x^2} \right) \)
View Solution
Step 1: Substitution
Put \( x = \tan \theta \).
Expression = \( \tan^{-1} \left( \frac{3 \tan \theta - \tan^3 \theta}{1 - 3 \tan^2 \theta} \right) \).
Step 2: Formula
This is the formula for \( \tan 3\theta \).
So, \( \tan^{-1}( \tan 3\theta ) = 3\theta \).
Simplest Form: \( 3 \tan^{-1} x \)

Q8 - Q12. Finding Values

Practice Question 8
Find the value of: \( \tan^{-1} \left[ 2 \sin \left( 4 \cos^{-1} \frac{\sqrt{3}}{2} \right) \right] \)
View Solution
Step 1
\( \cos^{-1}(\frac{\sqrt{3}}{2}) = \frac{\pi}{6} \).
Step 2
\( 4 \times (\frac{\pi}{6}) = \frac{2\pi}{3} \). Then \( \sin(\frac{2\pi}{3}) = \frac{\sqrt{3}}{2} \).
Step 3
\( \tan^{-1}[ 2 \times \frac{\sqrt{3}}{2} ] = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \).
Value = \( \frac{\pi}{3} \)
Practice Question 9
Find the value of: \( \cos \left( \sin^{-1} \frac{3}{5} + \cos^{-1} \frac{4}{5} \right) \)
View Solution
Step 1: Convert
Note that \( \cos^{-1} \frac{4}{5} \) corresponds to the same triangle as \( \sin^{-1} \frac{3}{5} \).
So, \( \cos^{-1} \frac{4}{5} = \sin^{-1} \frac{3}{5} \).
Expression = \( \cos( \sin^{-1} \frac{3}{5} + \sin^{-1} \frac{3}{5} ) = \cos( 2 \sin^{-1} \frac{3}{5} ) \).
Step 2: Formula
Let \( \theta = \sin^{-1} \frac{3}{5} \). We need \( \cos 2\theta \).
\( \cos 2\theta = 1 - 2 \sin^2 \theta = 1 - 2(3/5)^2 \)
\( = 1 - 18/25 = 7/25 \).
Value = \( \frac{7}{25} \)
Practice Question 10
Find value of: \( \cos^{-1}( \cos \frac{5\pi}{4} ) \)
View Solution
Principal Range
\( [0, \pi] \). \( \frac{5\pi}{4} \) is outside.
Adjust
\( \cos(5\pi/4) = \cos(2\pi - 3\pi/4) = \cos(3\pi/4) \).
Alternatively \( \cos(5\pi/4) = \cos(\pi + \pi/4) = -\cos(\pi/4) \).
\( \cos^{-1}(-\frac{1}{\sqrt{2}}) = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \).
Value = \( \frac{3\pi}{4} \)
Practice Question 11
Find value of: \( \sin^{-1}( \sin \frac{7\pi}{6} ) \)
View Solution
Principal Range
\( [-\pi/2, \pi/2] \).
Adjust
\( \sin(7\pi/6) = \sin(\pi + \pi/6) = -\sin(\pi/6) \).
\( \sin^{-1}(-1/2) = -\pi/6 \).
Value = \( -\frac{\pi}{6} \)
Practice Question 12
Find value of: \( \cos( \tan^{-1} \frac{3}{4} + \sin^{-1} \frac{3}{5} ) \)
View Solution
Convert
Let \( A = \tan^{-1}(3/4) \) so \( \tan A = 3/4 \), \( \sin A=3/5, \cos A=4/5 \).
Let \( B = \sin^{-1}(3/5) \) so \( \sin B = 3/5, \cos B=4/5 \).
Formula
\( \cos(A+B) = \cos A \cos B - \sin A \sin B \)
\( = (4/5)(4/5) - (3/5)(3/5) = 16/25 - 9/25 \).
Value = \( \frac{7}{25} \)

Q13 - Q15. MCQs

Practice Question 13
\( \tan^{-1}( \tan \frac{5\pi}{6} ) \) is equal to:
(A) \( 5\pi/6 \) (B) \( -\pi/6 \) (C) \( \pi/6 \) (D) \( 7\pi/6 \)
View Solution
Range Check
\( (-\pi/2, \pi/2) \). \( 5\pi/6 \) is outside.
Adjust
\( \tan(5\pi/6) = \tan(\pi - \pi/6) = -\tan(\pi/6) \).
Inverse is \( -\pi/6 \).
Correct Option: (B) \( -\pi/6 \)
Practice Question 14
\( \cos( \pi - \cos^{-1}(-\frac{\sqrt{3}}{2}) ) \) is equal to:
(A) \( \sqrt{3}/2 \) (B) \( -1/2 \) (C) \( 1/2 \) (D) \( -\sqrt{3}/2 \)
View Solution
Simplify Inverse
\( \cos^{-1}(-\frac{\sqrt{3}}{2}) = \frac{5\pi}{6} \).
Calculate
\( \pi - \frac{5\pi}{6} = \frac{\pi}{6} \).
\( \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2} \).
Correct Option: (A) \( \frac{\sqrt{3}}{2} \)
Practice Question 15
\( \sin^{-1}(-\frac{1}{2}) + \cos^{-1}(-\frac{1}{2}) \) is equal to:
(A) \( \pi \) (B) \( \pi/2 \) (C) 0 (D) \( -\pi/2 \)
View Solution
Identity
We know \( \sin^{-1} x + \cos^{-1} x = \pi/2 \) for all \( x \in [-1, 1] \).
Here \( x = -1/2 \), which is valid.
Correct Option: (B) \( \pi/2 \)
Exercise 2.1 Chapter 2 Miscellaneous