NCERT Solutions Class 12 Maths Chapter 2 Miscellaneous | LearnCBSEHub.in

NCERT Solutions

Class 12 • Maths • Miscellaneous Ex
Questions 1 & 2
1. Find the value of $\cos^{-1}(\cos \frac{13\pi}{6})$
2. Find the value of $\tan^{-1}(\tan \frac{7\pi}{6})$

1. $\cos^{-1}(\cos \frac{13\pi}{6})$

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

$\frac{13\pi}{6} = 2\pi + \frac{\pi}{6}$. It is not in range.

$$ \cos \frac{13\pi}{6} = \cos(2\pi + \frac{\pi}{6}) = \cos \frac{\pi}{6} $$ $$ \cos^{-1}(\cos \frac{\pi}{6}) = \frac{\pi}{6} $$

2. $\tan^{-1}(\tan \frac{7\pi}{6})$

Range of principal value of $\tan^{-1}$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$.

$$ \tan \frac{7\pi}{6} = \tan(\pi + \frac{\pi}{6}) = \tan \frac{\pi}{6} $$ $$ \tan^{-1}(\tan \frac{\pi}{6}) = \frac{\pi}{6} $$
Answers: 1) $\frac{\pi}{6}$   2) $\frac{\pi}{6}$
Question 3
Prove that $2\sin^{-1}\frac{3}{5} = \tan^{-1}\frac{24}{7}$
Proof:

Let $\sin^{-1}\frac{3}{5} = \theta \implies \sin\theta = \frac{3}{5}$.

Then $\cos\theta = \sqrt{1 – (\frac{3}{5})^2} = \frac{4}{5}$, so $\tan\theta = \frac{3}{4}$.

Thus, $\theta = \tan^{-1}\frac{3}{4}$.

LHS $= 2\tan^{-1}\frac{3}{4}$

Using $2\tan^{-1}x = \tan^{-1}\frac{2x}{1-x^2}$:

$$ = \tan^{-1} \left( \frac{2 \times \frac{3}{4}}{1 – (\frac{3}{4})^2} \right) = \tan^{-1} \left( \frac{3/2}{1 – 9/16} \right) $$ $$ = \tan^{-1} \left( \frac{3/2}{7/16} \right) = \tan^{-1} \left( \frac{3}{2} \times \frac{16}{7} \right) = \tan^{-1} \frac{24}{7} $$
LHS = RHS
Question 4
Prove that $\sin^{-1}\frac{8}{17} + \sin^{-1}\frac{3}{5} = \tan^{-1}\frac{77}{36}$

Convert both to $\tan^{-1}$:

  • $\sin^{-1}\frac{8}{17} \to \tan^{-1}\frac{8}{15}$ (since $\sqrt{17^2-8^2}=15$)
  • $\sin^{-1}\frac{3}{5} \to \tan^{-1}\frac{3}{4}$ (since $\sqrt{5^2-3^2}=4$)

LHS $= \tan^{-1}\frac{8}{15} + \tan^{-1}\frac{3}{4}$

$$ = \tan^{-1} \left( \frac{\frac{8}{15} + \frac{3}{4}}{1 – \frac{8}{15}\cdot\frac{3}{4}} \right) $$ $$ = \tan^{-1} \left( \frac{\frac{32+45}{60}}{\frac{60-24}{60}} \right) = \tan^{-1} \frac{77}{36} $$
LHS = RHS
Question 8
Prove that $\tan^{-1}\frac{1}{5} + \tan^{-1}\frac{1}{7} + \tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{8} = \frac{\pi}{4}$

Group terms: $(\tan^{-1}\frac{1}{5} + \tan^{-1}\frac{1}{7}) + (\tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{8})$

First part: $\tan^{-1} \frac{\frac{1}{5}+\frac{1}{7}}{1-\frac{1}{35}} = \tan^{-1} \frac{12/35}{34/35} = \tan^{-1} \frac{6}{17}$

Second part: $\tan^{-1} \frac{\frac{1}{3}+\frac{1}{8}}{1-\frac{1}{24}} = \tan^{-1} \frac{11/24}{23/24} = \tan^{-1} \frac{11}{23}$

Now add results: $\tan^{-1}\frac{6}{17} + \tan^{-1}\frac{11}{23}$

$$ = \tan^{-1} \frac{\frac{6}{17}+\frac{11}{23}}{1-\frac{6}{17}\cdot\frac{11}{23}} = \tan^{-1} \frac{138+187}{391-66} = \tan^{-1} \frac{325}{325} $$ $$ = \tan^{-1}(1) = \frac{\pi}{4} $$
LHS = RHS
Question 9
Prove $\tan^{-1}\sqrt{x} = \frac{1}{2}\cos^{-1}\left(\frac{1-x}{1+x}\right), x \in [0, 1]$
Proof:

Let $\sqrt{x} = \tan\theta \implies x = \tan^2\theta$. So $\theta = \tan^{-1}\sqrt{x}$.

RHS $= \frac{1}{2}\cos^{-1}\left(\frac{1-\tan^2\theta}{1+\tan^2\theta}\right)$

We know $\cos 2\theta = \frac{1-\tan^2\theta}{1+\tan^2\theta}$.

$$ = \frac{1}{2}\cos^{-1}(\cos 2\theta) $$ $$ = \frac{1}{2}(2\theta) = \theta $$ $$ = \tan^{-1}\sqrt{x} = \text{LHS} $$
LHS = RHS
Question 10
Prove $\cot^{-1}\left( \frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} – \sqrt{1-\sin x}} \right) = \frac{x}{2}, x \in (0, \frac{\pi}{4})$

We use identities:

  • $1+\sin x = (\cos\frac{x}{2} + \sin\frac{x}{2})^2$
  • $1-\sin x = (\cos\frac{x}{2} – \sin\frac{x}{2})^2$

So, $\sqrt{1+\sin x} = \cos\frac{x}{2} + \sin\frac{x}{2}$ and $\sqrt{1-\sin x} = \cos\frac{x}{2} – \sin\frac{x}{2}$.

Numerator: $(\cos\frac{x}{2} + \sin\frac{x}{2}) + (\cos\frac{x}{2} – \sin\frac{x}{2}) = 2\cos\frac{x}{2}$

Denominator: $(\cos\frac{x}{2} + \sin\frac{x}{2}) – (\cos\frac{x}{2} – \sin\frac{x}{2}) = 2\sin\frac{x}{2}$

$$ \cot^{-1} \left( \frac{2\cos\frac{x}{2}}{2\sin\frac{x}{2}} \right) = \cot^{-1}(\cot\frac{x}{2}) = \frac{x}{2} $$
LHS = RHS
Question 12
Prove $\frac{9\pi}{8} – \frac{9}{4}\sin^{-1}\frac{1}{3} = \frac{9}{4}\sin^{-1}\frac{2\sqrt{2}}{3}$

LHS $= \frac{9}{4} (\frac{\pi}{2} – \sin^{-1}\frac{1}{3})$

Using $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$, we get $\frac{\pi}{2} – \sin^{-1}\frac{1}{3} = \cos^{-1}\frac{1}{3}$.

So LHS $= \frac{9}{4} \cos^{-1}\frac{1}{3}$.

Convert $\cos^{-1}\frac{1}{3}$ to sine:

Let $\cos^{-1}\frac{1}{3} = \theta \implies \cos\theta = \frac{1}{3}$.

$\sin\theta = \sqrt{1 – (1/3)^2} = \sqrt{8/9} = \frac{2\sqrt{2}}{3}$.

So LHS $= \frac{9}{4} \sin^{-1}\frac{2\sqrt{2}}{3} = \text{RHS}$.

LHS = RHS
Questions 13 & 14
13. Solve $2\tan^{-1}(\cos x) = \tan^{-1}(2\text{cosec } x)$
14. Solve $\tan^{-1}\frac{1-x}{1+x} = \frac{1}{2}\tan^{-1}x$

13. Solution:

LHS $= \tan^{-1}\frac{2\cos x}{1-\cos^2 x} = \tan^{-1}\frac{2\cos x}{\sin^2 x}$.

So, $\frac{2\cos x}{\sin^2 x} = 2\text{cosec } x = \frac{2}{\sin x}$.

$\frac{\cos x}{\sin x} = 1 \implies \cot x = 1 \implies x = \frac{\pi}{4}$.

14. Solution:

LHS $= \tan^{-1}1 – \tan^{-1}x = \frac{\pi}{4} – \tan^{-1}x$.

Equation: $\frac{\pi}{4} – \tan^{-1}x = \frac{1}{2}\tan^{-1}x$.

$\frac{3}{2}\tan^{-1}x = \frac{\pi}{4} \implies \tan^{-1}x = \frac{\pi}{6}$.

$x = \tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}$.

MCQs
15. $\sin(\tan^{-1} x)$ equals?
16. $\sin^{-1}(1-x) – 2\sin^{-1}x = \frac{\pi}{2}$, find x?

15. Let $\tan^{-1}x = \theta \implies \tan\theta = \frac{x}{1}$.

Hypotenuse $= \sqrt{1+x^2}$. So $\sin\theta = \frac{x}{\sqrt{1+x^2}}$.

Answer: (D) $\frac{x}{\sqrt{1+x^2}}$

16. Put values from options.

If $x=0$: $\sin^{-1}(1) – 0 = \frac{\pi}{2}$. Correct.

If $x=\frac{1}{2}$: $\sin^{-1}(\frac{1}{2}) – 2\sin^{-1}(\frac{1}{2}) = \frac{\pi}{6} – \frac{\pi}{3} \ne \frac{\pi}{2}$.

Answer: (C) 0

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