Question

If ${\sin }^{-1}x+{\sin }^{-1}y+{\sin }^{-1}z=\frac{3\pi }{2}$, the value of $x^{100}+y^{100}+z^{100}-\frac{9}{x^{101}+y^{101}+z^{101}}$ is-

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is A $0$

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 ${\sin }^{-1}x\in [-\frac{\pi }{2},\frac{\pi }{2}]$
then $=\sin \frac{\pi }{2}=1\Rightarrow x=y=z=1$
Hence desired value is $=1+1+1+-\frac{9}{1+1+1}=3-\frac{9}{3}=0$