If sin ^{-1}left(frac{12}{x}right)+sin ^{-1}l | vedclass

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(C) Given equation is $\sin ^{-1}\left(\frac{12}{x}\right)+\sin ^{-1}\left(\frac{5}{x}\right)=\frac{\pi}{2}$.We know that $\sin ^{-1}(u) + \cos ^{-1}(

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$13$

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(C) Given equation is $\sin ^{-1}\left(\frac{12}{x}ight)+\sin ^{-1}\left(\frac{5}{x}ight)=\frac{\pi}{2}$.We know that $\sin ^{-1}(u) + \cos ^{-1}(u) = \frac{\pi}{2}$, so $\sin ^{-1}(u) = \frac{\pi}{2} - \cos ^{-1}(u)$.Rearranging the given equation: $\sin ^{-1}\left(\frac{12}{x}ight) = \frac{\pi}{2} - \sin ^{-1}\left(\frac{5}{x}ight)$.Using the identity $\frac{\pi}{2} - \sin ^{-1}(u) = \cos ^{-1}(u)$, we get $\sin ^{-1}\left(\frac{12}{x}ight) = \cos ^{-1}\left(\frac{5}{x}ight)$.Let $\sin ^{-1}\left(\frac{12}{x}ight) = 	heta$, then $\sin 	heta = \frac{12}{x}$.Since $\sin^2 	heta + \cos^2 	heta = 1$, we have $\cos 	heta = \sqrt{1 - \left(\frac{12}{x}ight)^2} = \frac{\sqrt{x^2 - 144}}{x}$.Thus, $	heta = \cos ^{-1}\left(\frac{\sqrt{x^2 - 144}}{x}ight)$.Equating the arguments: $\frac{\sqrt{x^2 - 144}}{x} = \frac{5}{x}$.Squaring both sides: $x^2 - 144 = 25$.$x^2 = 169$, which gives $x = 13$ (since $x$ must be positive for the domain of $\sin^{-1}$ here).

If $\sin ^{-1}\left(\frac{12}{x}\right)+\sin ^{-1}\left(\frac{5}{x}\right)=\frac{\pi}{2}$, then $x=$

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