Question

Number of triplets $(x,y,z)$ satisfying ${\sin }^{-1}x+{\sin }^{-1}y+{\cos }^{-1}z=2\pi$ is

• A. $1$
• B. $0$
• C. $2$
• D. $\infty$

Answer 1

The correct option is A $1$

Let $f(x,y,z)={\sin }^{-1}x+{\sin }^{-1}y+{\cos }^{-1}z$
It will attain the value $2\pi$ only if ${\sin }^{-1}x={\sin }^{-1}y=\frac{\pi }{2}$ and ${\cos }^{-1}z=\pi$.
This is possible only if $x=y=1$ and $z=-1$.
Hence there is only one solution $f(1,1,-1)=2\pi$