Question

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

Answer 1

Answer: 3

$si{n}^{-1}\left(\frac{x}{5}\right)+cose{c}^{-1}\left(\frac{5}{4}\right)=\frac{\pi }{2}$
$\Rightarrow si{n}^{-1}\left(\frac{x}{5}\right)=\frac{\pi }{2}-cose{c}^{-1}\left(\frac{5}{4}\right)$
$\Rightarrow si{n}^{-1}\left(\frac{x}{5}\right)=\frac{\pi }{2}-si{n}^{-1}\left(\frac{4}{5}\right)$
$\left[\because si{n}^{-1}x+co{s}^{-1}x=\pi 2\right]$
$\Rightarrow si{n}^{-1}\left(\frac{x}{5}\right)=co{s}^{-1}\left(\frac{4}{5}\right)$ ___$\left(i\right)$
Let $co{s}^{-1}\frac{4}{5}=A$
$\Rightarrow cosA=\frac{4}{5}$
$\Rightarrow A=co{s}^{-1}\left(4/5\right)$

$\Rightarrow sinA=\frac{3}{5}$
$\Rightarrow A=si{n}^{-1}\frac{3}{5}$
$\therefore co{s}^{-1}\left(4/5\right)=si{n}^{-1}\left(3/5\right)$
$\therefore$ equation $\left(i\right)$ become,
$si{n}^{-1}\frac{x}{5}=si{n}^{-1}\frac{3}{5}$
$\Rightarrow \frac{x}{5}=\frac{3}{5}$
$\Rightarrow x=3$