Question

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

• A. $1$
• B. $3$
• C. $4$
• D. $5$

Answer 1

Answer: (B)

$3$

${\sin }^{-1}\frac{x}{5}=\frac{\pi }{2}-cose{c}^{-1}\frac{5}{4}$

take sin on both sides,

$\frac{x}{5}=\cos \left(cose{c}^{-1}\frac{5}{4}\right)\left[\sin \right({\sin }^{-1}x)=x]$

$\because cose{c}^{-1}\frac{5}{4}={\cos }^{-1}\frac{3}{5}$

$\frac{x}{5}=\cos \left({\cos }^{-1}\frac{3}{5}\right)$

$\frac{x}{5}=\frac{3}{5}$

$x=3$