Question

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

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (C)

2

Answer 2

We know that $\left|\sin ^{-1} x\right| \leq \frac{\pi}{2}$
Hence, from the given relation we observe that each of
$\sin ^{-1} x, \sin ^{-1} y$ and $\sin ^{-1} z$ will be $\frac{\pi}{2}$
so that $x=y=z=\sin \frac{\pi}{2}=1$
$\therefore x^{9}+y^{9}+z^{9}-\frac{1}{x^{9} y^{9} z^{9}}$
$=(1)^{9}+(1)^{9}+(1)^{9}-\frac{1}{(1)^{9}(1)^{9}(1)^{9}}$
$=1+1+1-\frac{1}{1 \times 1 \times 1}$
$=3-1=2$