Class 12 Maths Chapter 2: Inverse Trigonometry Miscellaneous Exercise

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Overview

Review the Chapter 2 miscellaneous exercise through worked problems on principal values, inverse-trigonometric properties, expression simplification, and equations.

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
Exercise 2.2 Exercise 3.1