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

The correct option is C $2$

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