Question

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

• A. $0$
• B. $3$
• C. $-3$
• D. $9$

Answer 1

The correct option is A

$0$

Explanation for the correct option :

Given that, ${\sin }^{-1}x+{\sin }^{-1}y+{\sin }^{-1}z=\frac{3\pi }{2}$.

${\sin }^{-1}x$ cannot be greater than $\frac{\mathrm{\pi }}{2},{\sin }^{-1}x={\sin }^{-1}y={\sin }^{-1}z=\frac{\mathrm{\pi }}{2}$.
$x=y=z=1$
Substitute values in $x^{100}+y^{100}+z^{100}-\frac{9}{\left[x^{101}+y^{101}+z^{101}\right]}$ we get:

$\begin{array}{rcl}\therefore \left(1+1+1\right)-\frac{9}{\left(1+1+1\right)}& =& 3-\frac{9}{3}\\ & =& 3-3\\ & =& 0\end{array}$

Hence, option A is correct.