NCERT Solutions Class 12 Maths Chapter 2 Ex 2.2 | LearnCBSEHub.in

NCERT Solutions

Class 12 • Maths • Ex 2.2
Question 1
Prove the following: $3\sin^{-1} x = \sin^{-1} (3x – 4x^3), x \in [-\frac{1}{2}, \frac{1}{2}]$
Proof:

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

RHS: $\sin^{-1}(3x – 4x^3)$

$$ = \sin^{-1}(3\sin\theta – 4\sin^3\theta) $$

Using the identity $\sin 3\theta = 3\sin\theta – 4\sin^3\theta$:

$$ = \sin^{-1}(\sin 3\theta) $$ $$ = 3\theta $$

Substitute $\theta = \sin^{-1}x$ back:

$$ = 3\sin^{-1}x $$ $$ = \text{LHS} $$
LHS = RHS
Question 2
Prove the following: $3\cos^{-1} x = \cos^{-1} (4x^3 – 3x), x \in [\frac{1}{2}, 1]$
Proof:

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

RHS: $\cos^{-1}(4x^3 – 3x)$

$$ = \cos^{-1}(4\cos^3\theta – 3\cos\theta) $$

Using identity $\cos 3\theta = 4\cos^3\theta – 3\cos\theta$:

$$ = \cos^{-1}(\cos 3\theta) $$ $$ = 3\theta $$

Substitute $\theta = \cos^{-1}x$:

$$ = 3\cos^{-1}x = \text{LHS} $$
LHS = RHS
Question 3
Write the function in the simplest form: $\tan^{-1} \frac{\sqrt{1+x^2}-1}{x}, x \ne 0$

Put $x = \tan\theta \implies \theta = \tan^{-1}x$.

$$ \tan^{-1} \left( \frac{\sqrt{1+\tan^2\theta}-1}{\tan\theta} \right) = \tan^{-1} \left( \frac{\sec\theta – 1}{\tan\theta} \right) $$

Convert to sin/cos:

$$ = \tan^{-1} \left( \frac{1-\cos\theta}{\sin\theta} \right) $$

Using Half-angle formulas:

$$ = \tan^{-1} \left( \frac{2\sin^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}} \right) = \tan^{-1} \left( \tan\frac{\theta}{2} \right) $$ $$ = \frac{\theta}{2} $$
Simplest Form: $\frac{1}{2}\tan^{-1}x$
Question 4
Write in simplest form: $\tan^{-1} \sqrt{\frac{1-\cos x}{1+\cos x}}, 0 < x < \pi$

Using $1-\cos x = 2\sin^2\frac{x}{2}$ and $1+\cos x = 2\cos^2\frac{x}{2}$:

$$ \tan^{-1} \sqrt{\frac{2\sin^2\frac{x}{2}}{2\cos^2\frac{x}{2}}} = \tan^{-1} \sqrt{\tan^2\frac{x}{2}} $$ $$ = \tan^{-1} \left| \tan\frac{x}{2} \right| $$

Since $x \in (0, \pi)$, $\tan\frac{x}{2}$ is positive.

$$ = \frac{x}{2} $$
Simplest Form: $\frac{x}{2}$
Question 5
Write in simplest form: $\tan^{-1} \left(\frac{\cos x – \sin x}{\cos x + \sin x}\right)$

Divide numerator and denominator by $\cos x$:

$$ \tan^{-1} \left(\frac{1 – \tan x}{1 + \tan x}\right) $$

Using $\tan(\frac{\pi}{4} – x)$ formula:

$$ = \tan^{-1} \left( \tan(\frac{\pi}{4} – x) \right) $$ $$ = \frac{\pi}{4} – x $$
Simplest Form: $\frac{\pi}{4} – x$
Question 6
Write in simplest form: $\tan^{-1} \frac{x}{\sqrt{a^2-x^2}}, |x| < a$

Put $x = a\sin\theta \implies \theta = \sin^{-1}\frac{x}{a}$.

$$ \tan^{-1} \frac{a\sin\theta}{\sqrt{a^2 – a^2\sin^2\theta}} = \tan^{-1} \frac{a\sin\theta}{a\cos\theta} $$ $$ = \tan^{-1} (\tan\theta) = \theta $$
Simplest Form: $\sin^{-1}\frac{x}{a}$
Question 7
Write in simplest form: $\tan^{-1} \left(\frac{3a^2x – x^3}{a^3 – 3ax^2}\right)$

Put $x = a\tan\theta \implies \theta = \tan^{-1}\frac{x}{a}$.

$$ \tan^{-1} \left(\frac{3a^2(a\tan\theta) – (a\tan\theta)^3}{a^3 – 3a(a\tan\theta)^2}\right) $$ $$ = \tan^{-1} \left(\frac{a^3(3\tan\theta – \tan^3\theta)}{a^3(1 – 3\tan^2\theta)}\right) $$ $$ = \tan^{-1} (\tan 3\theta) = 3\theta $$
Simplest Form: $3\tan^{-1}\frac{x}{a}$
Question 8
Find the value of: $\tan^{-1} [2\cos (2\sin^{-1} \frac{1}{2})]$

1. Inner term: $\sin^{-1}\frac{1}{2} = \frac{\pi}{6}$.

2. Next: $2 \times \frac{\pi}{6} = \frac{\pi}{3}$.

3. Next: $2\cos(\frac{\pi}{3}) = 2(\frac{1}{2}) = 1$.

4. Final: $\tan^{-1}(1)$.

Answer: $\frac{\pi}{4}$
Question 9
Find the value of: $\tan \frac{1}{2} [\sin^{-1}\frac{2x}{1+x^2} + \cos^{-1}\frac{1-y^2}{1+y^2}], |x| < 1, y > 0, xy < 1$

Using identities $\sin^{-1}\frac{2x}{1+x^2} = 2\tan^{-1}x$ and $\cos^{-1}\frac{1-y^2}{1+y^2} = 2\tan^{-1}y$.

$$ \tan \frac{1}{2} [2\tan^{-1}x + 2\tan^{-1}y] $$ $$ = \tan [\tan^{-1}x + \tan^{-1}y] $$ $$ = \tan \left[ \tan^{-1} \left(\frac{x+y}{1-xy}\right) \right] $$
Answer: $\frac{x+y}{1-xy}$
Questions 10-15
Evaluate Q10 to Q15 (Principal Values & Expressions)

Q10: $\sin^{-1}(\sin \frac{2\pi}{3}) = \sin^{-1}(\sin \frac{\pi}{3}) = \frac{\pi}{3}$.

Q11: $\tan^{-1}(\tan \frac{3\pi}{4}) = \tan^{-1}(-\tan \frac{\pi}{4}) = -\frac{\pi}{4}$.

Q12: $\tan(\sin^{-1}\frac{3}{5} + \cot^{-1}\frac{3}{2}) = \tan(\tan^{-1}\frac{3}{4} + \tan^{-1}\frac{2}{3}) = \frac{17}{6}$.

Q13: $\cos^{-1}(\cos \frac{7\pi}{6}) = \frac{5\pi}{6}$. (Option B)

Q14: $\sin(\frac{\pi}{3} – \sin^{-1}(-\frac{1}{2})) = 1$. (Option D)

Q15: $\tan^{-1}\sqrt{3} – \cot^{-1}(-\sqrt{3}) = -\frac{\pi}{2}$. (Option B)

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