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

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

Answers (2)

Answer:

                -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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Answer:

                \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}

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