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 $=1+1+1+-\frac{9}{1+1+1}=3-\frac{9}{3}=0$