Question

The number of solutions of the equation ${\sin }^{-1}\left(\frac{1+x^2}{2x}\right)=\frac{\pi }{2}\sec (x-1)$ is

Answer 1

Since, domain of ${\sin }^{-1}x$ is $[-1,1]$
Therefore, $-1\le \frac{1+x^2}{2x}\le 1$
$1+2x+x^2\ge 0$ and $1-2x+x^2\le 0$
$(1+x{)}^2\ge 0$ and $(1-x{)}^2\le 0$
$-1\le x\le 1$
However, only $x=1$ satisfies the above equation.
Therefore, only one solution exists.