Question

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

• A. 4
• B. 5
• C. 1
• D. 3

Answer 1

Answer: (D)

${\sin }^{-1}\left(\frac{x}{5}\right)+cose{c}^{-1}\left(\frac{5}{4}\right)=\frac{\pi }{2}$
$\Rightarrow {\sin }^{-1}\left(\frac{x}{5}\right)=\frac{\pi }{2}-cose{c}^{-1}\left(\frac{5}{4}\right)$
$\Rightarrow {\sin }^{-1}\left(\frac{x}{5}\right)=\frac{\pi }{2}-{\sin }^{-1}\left(\frac{4}{5}\right)$
$\left[\because {\sin }^{-1}x+{\cos }^{-1}x=\pi /2\right]$
$\Rightarrow {\sin }^{-1}\left(\frac{x}{5}\right)={\cos }^{-1}\left(\frac{4}{5}\right)$ ...(1)
Let ${\cos }^{-1}\frac{4}{5}=A\Rightarrow \cos A=\frac{4}{5}$
$\Rightarrow A={\cos }^{-1}\left(4/5\right)$
$\Rightarrow \sin A=\frac{3}{5}$
$\Rightarrow {\sin }^{-1}\frac{3}{5}$
$\therefore {\cos }^{-1}\left(4/5\right)={\sin }^{-1}\left(3/5\right)$
$\therefore$ equation (1) become,
${\sin }^{-1}\frac{x}{5}={\sin }^{-1}\frac{3}{5}\Rightarrow \frac{x}{5}=\frac{3}{5}\Rightarrow x=3$