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)