Question

lf ${\sin }^{-1}x+{\sin }^{-1}y+{\sin }^{-1}z=\frac{3\pi }{2}$, then $x^{100}+y^{100}+z^{100}-\frac{9}{x^{101}+y^{101}+z^{101}}=$

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (A)

$0$

${\sin }^{-1}x+{\sin }^{-1}y+{\sin }^{-1}z=\frac{3\pi }{2}$
Considering the range of sin inverse function, the only possible solution is
${\sin }^{-1}x=\frac{\pi }{2}={\sin }^{-1}y={\sin }^{-1}z$
$x=y=z=1$
$\therefore x^{100}+y^{100}+z^{100}-\frac{9}{x^{100}+y^{100}+z^{100}}$
$=3-\frac{9}{3}$
$=0$