Solve for x:


Solve for x:

$\sin ^{-1} \frac{8}{x}+\sin ^{-1} \frac{15}{x}=\frac{\pi}{2}$



To find: value of x

Given: $\sin ^{-1} \frac{8}{x}+\sin ^{-1} \frac{15}{x}=\frac{\pi}{2}$

We know $\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$

Let $\sin ^{-1} \frac{8}{x}=\mathrm{P}$

$\Rightarrow \sin P=\frac{8}{x}$

Therefore, $\cos \mathrm{P}=\frac{\sqrt{\mathrm{x}^{2}-64}}{\mathrm{x}}$

$P=\cos ^{-1} \frac{\sqrt{x^{2}-64}}{x}$

$\cos ^{-1} \frac{\sqrt{x^{2}-64}}{x}+\sin ^{-1} \frac{15}{x}=\frac{\pi}{2}$

Therefore, $\frac{\sqrt{x^{2}-64}}{x}=\frac{15}{x}$

$\Rightarrow \sqrt{x^{2}-64}=15$

Squaring both sides,

$\Rightarrow x^{2}-64=225$

$\Rightarrow x^{2}=289$

$\Rightarrow x=\pm 17$

Therefore, $x=\pm 17$ are the required values of $x$. 

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