Class 12 ncert solution math exercise 2.1

23/01/2023 by Dileep namdev

EXERCISE 2.1

Find the principal values of the following:(Class 12 ncert solution math exercise 2.1)

Question 1: $\sin^{-1}(\frac{-1}{2})$

Solution : Principal value of

$\sin^{-1}(\frac{-1}{2})= -\frac{\pi}{6}\in[-\pi/2,\pi/2]$

Question 2: $\cos^{-1}(\frac{\sqrt{3}}{2})$

Solution : Principal value of

$\cos^{-1}(\frac{\sqrt{3}}{2})= \pi/6 \in[0, \pi]$

Question 3: $\operatorname{cosec}^{-1}(2)$

Solution: Principal value of

$\operatorname{cosec}^{-1}(2)= \pi/6\in[-pi/2, \pi/2]-\{0\}$

Question 4: $\tan^{-1}(-\sqrt{3})$

Solution: Principal value of

$\tan^{-1}(-\sqrt{3}) = -\pi/3\in(-\pi/2,\pi/2)$

Question 5: $\cos^{-1}(-\frac{1}{2})$

Solution :Principal value of

$\cos^{-1}(-\frac{1}{2})= \pi-\frac{\pi}{3}$

$= 2\pi/3 \in [0,\pi]$

Question 6: $\tan^{-1}(-1)$

Solution: Principal value of

$\tan^{-1}(-1) = -\pi/4\in(-\pi/2, \pi/2)$

Question 7: $\sec^{-1}(\frac{2}{\sqrt{3}})$

Solution : Principal value of

$\sec^{-1}(\frac{2}{\sqrt{3}})=\pi/6\in[0, \pi]-\{\pi/2\}$

Question 8: $\cot^{-1}(\sqrt{3})$

Solution: Principal value of

$\cot^{-1}(\sqrt{3}) = \pi/6\in(0,\pi)$

Question 9: $\cos^{-1}(-\frac{1}{\sqrt{2}})$

Solution: Principal value of

$\cos^{-1}(-\frac{1}{\sqrt{2}}) = \pi - \pi/4$

$= 3\pi/4\in (0, \pi)$

Question 10: $\operatorname{cosec}^{-1}(-\sqrt{2})$

Solution: Principal value of

$\operatorname{cosec}^{-1}(-\sqrt{2})=-\pi/4 \in [-\pi/2,\pi/2]-\{0\}$

Find the values of the following:

Question 11: $\tan^{-1}(1)+\cos^{-1}(-1/2)+\sin^{-1}(-1/2)$

Solution: Since
$\tan^{-1}(1) = \pi/4\in(-\pi/2,\pi/2)$

$\cos^{-1}(-1/2)= \pi - \pi/3$

$= 2\pi/3 \in[0,\pi]$

$\sin^{-1}(-1/2)=-\pi/6\in[-\pi/2,\pi/2]$

Hence

$\tan^{-1}(1) = \pi/4+2\pi/3-\pi/6$

$= \frac{3\pi+8\pi-2\pi}{12}$

$= \frac{9\pi}{12}$

$=\frac{3\pi}{12}$

Question 12: $\cos^{-1}(1/2)+2\sin^{-1}(1/2)$

Solution : Since,

$\cos^{-1}(1/2)= \pi/3\in[0, \pi]$

$\sin^{-1}(1/2)= \pi/6\in[-\pi/2, \pi/2]$

Hence,
$\cos^{-1}(1/2)+2\sin^{-1}(1/2) = \pi/3+2\pi/6$

$= 2\pi/3$

Question 13: Find the value of $\sin ^{-1} x=y$ , then

(A) $0 \leq y \leq \pi$

(B) $-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$

(C) $0 \leq y \leq \pi$

(D) $-\frac{\pi}{2}<y<\frac{\pi}{2}$

Answer: (B).

Question 14: Find the value of $\tan ^{-1} \sqrt{3}-\sec ^{-1}(-2)$ is equal to

(A) 0

(B) $-\frac{\pi}{3}$

(C) $\frac{\pi}{3}$

(D) $\frac{2 \pi}{3}$

Answer (B)

Since, $\tan ^{-1}(\sqrt{3})=\pi/3\in (-\pi/2, \pi/2)$
Hence,

$\sec^{-1}(-2)=2\pi/3\in[0,\pi]-\{\pi/2\}$

Hence,

$\tan ^{-1} \sqrt{3}-\sec ^{-1}(-2)= \pi/3-2\pi/3$

$= -\pi/3$ .

Case study inverse trigonometry 2 — Two men on either side of a temple 30 metres high observes

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