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Need solution for RD Sharma maths class 12 chapter 3 Inverse Trigonometric Functions exercise  Very short answer question 3 math

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Answer: 3

Given: \sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z=\frac{3 \pi}{2}, x+y+z=?

Hint: The maximum value in the range of \sin ^{-1}  is \frac{\pi}{2}


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

Here sum of three inverse of \sin x  is 3  terms \frac{\pi}{2}, i.e. every \sin  inverse function is equal to \frac{\pi}{2}.            

\begin{array}{l} \sin ^{-1} x=\frac{\pi}{2}, \sin ^{-1} y=\frac{\pi}{2}, \sin ^{-1} z=\frac{\pi}{2} \\\\ x=\frac{\sin \pi}{2}, y=\sin \frac{\pi}{2}, z=\sin \frac{\pi}{2} \\\\ x=1, y=1, z=1 \\\\ x+y+z=1+1+1 \\\\ \quad=3 \end{array}

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