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need solution for rd sharma maths class 12 chapter inverse trigonometric functions exercise 3.2 question 5 sub question (iii)

Answers (1)

Answer: ${\frac{3\pi }{2}}$

Hint: The range of the principal value branch of $\cos ^{-1}$ is $\left [ 0,\pi \right ]$

Given: $\sin ^{-1}\left ( \frac{-1}{2} \right )+2\cos ^{-1}\left ( \frac{-\sqrt{3}}{2} \right )$

Solution:

Let $\sin ^{-1}\left ( \frac{-1}{2} \right )=x$

$\sin\; \sin x=\frac{-1}{2}=-\sin\; \sin \left ( \frac{\pi }{6} \right )$

$\sin ^{-1}\left ( \frac{-1}{2} \right )=\frac{-\pi }{6}$

Let $\cos ^{-1}\left ( \frac{-\sqrt{3}}{2} \right )=y$

$\cos y=\left ( \frac{-\sqrt{3}}{2} \right )=\cos \left ( \pi -\frac{\pi }{6} \right )$

$\cos ^{-1}\left ( \frac{-\sqrt{3}}{2} \right )=\frac{5\pi }{6}$

Hence, $\sin ^{-1}\left ( \frac{-1}{2} \right )+2\cos ^{-1}\left ( \frac{-\sqrt{3}}{2} \right )=\frac{-\pi }{6}+2\left ( \frac{5\pi }{6} \right )$

$=\frac{-\pi }{6}+\frac{10\pi }{6}$

$=\frac{9\pi }{6}$

$=\frac{3\pi }{2}$

Principal Value of $\sin ^{-1}\left ( \frac{-1}{2} \right )+2\cos ^{-1}\left ( \frac{-\sqrt{3}}{2} \right )$ is ${\frac{3\pi }{2}}$ .

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