#### Need solution for RD Sharma Maths Class 12 Chapter 3 Inverse Trigonometric Functions Excercise Fill in the Blanks Question 15

$-1$

Hint:

You must know the rules of inverse trigonometric function.

Given:

$\sin^{-1}x+\sin^{-1}y+\sin^{-1}z=\frac{-3\pi}{2}$

Solution:

$\sin^{-1}x+\sin^{-1}y+\sin^{-1}z=\frac{-3\pi}{2}$

$\because \frac{-\pi}{2}< \sin^{-1}x\leq \frac{\pi}{2}$

$\because \sin^{-1}x=\frac{-\pi}{2},\sin^{-1}y=\frac{-\pi}{2},\sin^{-1}z=\frac{-\pi}{2}$

$\because x=y=z=-1$

$\because xyz=-1$

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$\frac{\pi}{6}$

Hint:

You must know the rules of inverse trigonometric function.

Given:

$\cos^{-1}\left \{\sin\left (\cos^{-1}\frac{1}{2} \right ) \right \}$

Solution:

$\cos^{-1}\left \{\sin\left (\cos^{-1}\frac{1}{2} \right ) \right \}$

$\cos^{-1}\left \{\sin\left (\frac{\pi}{3} \right ) \right \}$                                                                    $\left [ cos^{-1}\frac{1}{2}=cos^{-1}\left ( cos\frac{\pi}{3} \right )=\frac{\pi}{3} \right ]$

$\cos^{-1}\left \{ \frac{\sqrt{3}}{2} \right \}$                                                                                  $\left [\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2} \right ]$

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