Question

Solve the equation for $x:{\sin }^{-1}\frac{5}{x}+{\sin }^{-1}\frac{12}{x}=\frac{\pi }{2},x\ne 0$.

Answer 1

The given equation is
${\sin }^{-1}\frac{5}{x}+{\sin }^{-1}\frac{12}{x}=\frac{\pi }{2}$

$\Rightarrow \frac{12}{x}=\sin \left(\begin{array}{c}\frac{\pi }{2}-{\sin }^{-1}\frac{5}{x}\end{array}\right)$

$\Rightarrow \frac{12}{x}=\cos \left(\begin{array}{c}{\sin }^{-1}\frac{5}{x}\end{array}\right)$

$\Rightarrow \frac{12}{x}=\sqrt{1-{\left(\begin{array}{c}\frac{5}{x}\end{array}\right)}^2}$ $\because \cos \left({\sin }^{-1}x\right)=\sqrt{1-x^2}$

${\left(\begin{array}{c}\frac{12}{x}\end{array}\right)}^2=1-\frac{25}{x^2}$

$\Rightarrow \frac{144}{x^2}+\frac{25}{x^2}=1$

$\Rightarrow \frac{169}{x^2}=1$

$x^2=169$

$\Rightarrow x=\pm \sqrt{169}$

$\Rightarrow x=13,-13$

But $x=-13$ doesn't satisfy the given equation.
Hence the solution of the given equation is $x=13$.