If \sin^{-1}\left ( \frac{3}{x} \right )+\sin^{-1}\left ( \frac{4}{x} \right )=\frac{\pi }{2} then find the value of x.

 

 

 

 
 
 
 
 

Answers (1)

Let \sin^{-1}\left ( \frac{3}{x} \right )=A and \sin^{-1}\left ( \frac{4}{x} \right )=B

A+B=\frac{\pi}{2}\Rightarrow A=\frac{\pi}{2}-B

\sin A=\sin \left (\frac{\pi}{2}-B \right )=\cos B

\cos B = \sqrt{1-\sin^2 B}     

    \frac{3}{x}=\sqrt{1-\frac{16}{x^2}}

Taking square on both sides

\frac{9}{x^2}={1-\frac{16}{x^2}}

\frac{9}{x^2} + \frac{16}{x^2}=1\Rightarrow \frac{25}{x^2}=1

x^2=25\Rightarrow x=\pm 5\: \: \: x=5

 

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