Question

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

• A. $3$
• B. $2$
• C. $1$
• D. $0$

Answer 1

Answer: (A)

$3$

We know, ${\csc }^{-1}\frac{5}{4}={\sin }^{-1}\frac{4}{5}$

$\therefore$ we have, ${\sin }^{-1}\frac{x}{5}+{\sin }^{-1}\frac{4}{5}=\frac{\pi }{2}$

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

$\Rightarrow {\cos }^{-1}\frac{4}{5}={\sin }^{-1}\frac{x}{5}$

We know, ${\cos }^{-1}\frac{4}{5}={\sin }^{-1}\frac{3}{5}$

(because $co{s}^{-1}x=si{n}^{-1}\sqrt{1-x^2}$)

$\therefore x=3$