Question

If $\sin ^{-1}\left(\frac{3}{x}\right)+\sin ^{-1}\left(\frac{4}{x}\right)=\frac{\pi}{2}$, then $x$ is equal to

• A. 3
• B. 5
• C. 7
• D. 11

Answer 1

Answer: (B)

Given that,
$\therefore \sin ^{-1}\left(\frac{3}{x}\right)+\sin ^{-1}\left(\frac{4}{x}\right)=\frac{\pi}{2}$
$\sin ^{-1}\left(\frac{3}{x}\right)=\frac{\pi}{2}-\sin ^{-1}\left(\frac{4}{x}\right)$
$\Rightarrow \sin ^{-1}\left(\frac{3}{x}\right)=\cos ^{-1}\left(\frac{4}{x}\right)$
$ \Rightarrow \sin ^{-1}\left(\frac{3}{x}\right) =\sin ^{-1}\left(\frac{\sqrt{x^{2}-16}}{x}\right)$
$\Rightarrow \frac{3}{x} =\frac{\sqrt{x^{2}-16}}{x} $
$ \Rightarrow 9=x^{2}-16 $
$ \Rightarrow x^{2}=25 $
$ \Rightarrow x =\pm 5 $
$\Rightarrow x =5 $
( $\because-5$ is not satisfied the given equation)