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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