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If $\sin ^{-1} \frac{5}{x}+\sin ^{-1} \frac{12}{x}=\frac{\pi}{2}$, then what is the value of $x$ ?
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The correct answer is:
13
$\sin ^{-1} \frac{5}{x}+\sin ^{-1} \frac{12}{x}=\frac{\pi}{2}$
$\Rightarrow \sin ^{-1} \frac{5}{x}+\cos ^{-1} \frac{\sqrt{x^{2}-144}}{x}=\frac{\pi}{2}$
$\left(\because \sin ^{-1} x=\cos ^{-1} \sqrt{1-x^{2}}\right)$
But we know $\sin ^{-1} y+\cos ^{-1} y=\frac{\pi}{2}$
$\therefore \quad \frac{5}{x}=\frac{\sqrt{x^{2}-144}}{x}$
$\Rightarrow 5=\sqrt{x^{2}-144}$
$\Rightarrow 25=x^{2}-144 \Rightarrow x^{2}=169$
$\Rightarrow \quad x=13$
$\Rightarrow \sin ^{-1} \frac{5}{x}+\cos ^{-1} \frac{\sqrt{x^{2}-144}}{x}=\frac{\pi}{2}$
$\left(\because \sin ^{-1} x=\cos ^{-1} \sqrt{1-x^{2}}\right)$
But we know $\sin ^{-1} y+\cos ^{-1} y=\frac{\pi}{2}$
$\therefore \quad \frac{5}{x}=\frac{\sqrt{x^{2}-144}}{x}$
$\Rightarrow 5=\sqrt{x^{2}-144}$
$\Rightarrow 25=x^{2}-144 \Rightarrow x^{2}=169$
$\Rightarrow \quad x=13$
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