Question

If $\sin^{-1} x + \sin^{-1}y + \sin^{-1} z = \frac{3 \pi}{2} $ then the value of $x^{100} + y^{100} + z^{100} - \frac{ 3}{x^{101} + y^{101} + z^{101}} $ is

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (A)

Given, $\sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z=\frac{3 \pi}{2}$
This will happen only when $\sin ^{-1} x =\sin ^{-1} y =\sin ^{-1} z =\frac{\pi}{2}$
Since $\sin ^{-1} x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
then $=\sin \frac{\pi}{2}=1 \Rightarrow x=y=z=1$
Hence desired value is $=1+1+1+-\frac{9}{1+1+1}=3-\frac{9}{3}=0$