Question

The number of triplets $(x,y,z)$ satisfies the equation $si{n}^{-1}x+si{n}^{-1}y+si{n}^{-1}z=\frac{3\pi }{2}$ is

• A. $1$
• B. $2$
• C. $0$
• D. infinite

Answer 1

Answer: (A)

$si{n}^{-1}x+si{n}^{-1}y+si{n}^{-1}z=\frac{3\pi }{2}$
$\therefore -\frac{\pi }{2}<si{n}^{-1}x<\frac{\pi }{2}$,
$\frac{-\pi }{2}<si{n}^{-1}y<\frac{\pi }{2}$
and $\frac{-\pi }{2}<si{n}^{-1}z<\frac{\pi }{2}$
Hence the above condition satisfies if
$si{n}^{-1}x=si{n}^{-1}y=si{n}^{-1}z=\frac{\pi }{2}$
$\Rightarrow x=y=z=1$