Question

If $sin^{-1} \left(\frac{x}{5}\right)+cosec^{-1}\left(\frac{5}{4}\right)=\frac{\pi}{2}$, then the value of $x$ is

Answer 1

$sin^{-1} \left(\frac{x}{5}\right)+cosec^{-1}\left(\frac{5}{4}\right)=\frac{\pi}{2}$
$\Rightarrow sin^{-1} \left(\frac{x}{5}\right)=\frac{\pi}{2}-cosec^{-1}\left(\frac{5}{4}\right)$
$\Rightarrow sin^{-1}\left(\frac{x}{5}\right)=\frac{\pi}{2}-sin^{-1}\left(\frac{4}{5}\right)$
$\left[\because sin^{-1}\,x+cos^{-1}\,x=\pi 2\right]$
$\Rightarrow sin^{-1} \left(\frac{x}{5}\right) = cos^{-1} \left(\frac{4}{5}\right)$ ___$\left(i\right)$
Let $cos^{-1} \frac{4}{5} = A $
$\Rightarrow cos\,A =\frac{4}{5}$
$\Rightarrow A=cos^{-1}\left(4 /5\right)$

$\Rightarrow sin\,A =\frac{3}{5}$
$\Rightarrow A=sin^{-1}\frac{3}{5}$
$\therefore cos^{-1}\left(4 /5\right)=sin^{-1} \left(3 /5\right)$
$\therefore $ equation $\left(i\right)$ become,
$sin^{-1} \frac{x}{5} = sin^{-1} \frac{3}{5} $
$\Rightarrow \frac{x}{5}=\frac{3}{5} $
$\Rightarrow x=3$