If sin ^{-1}left(frac{3}{x}right)+sin ^{-1}le | vedclass

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(B) Given that, $\sin ^{-1}\left(\frac{3}{x}\right)+\sin ^{-1}\left(\frac{4}{x}\right)=\frac{\pi}{2}$.We know that $\sin ^{-1}(A) + \cos ^{-1}(A) = \f

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$5$

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(B) Given that, $\sin ^{-1}\left(\frac{3}{x}ight)+\sin ^{-1}\left(\frac{4}{x}ight)=\frac{\pi}{2}$.We know that $\sin ^{-1}(A) + \cos ^{-1}(A) = \frac{\pi}{2}$, so $\sin ^{-1}\left(\frac{4}{x}ight) = \frac{\pi}{2} - \cos ^{-1}\left(\frac{4}{x}ight)$.Rearranging the given equation: $\sin ^{-1}\left(\frac{3}{x}ight) = \frac{\pi}{2} - \sin ^{-1}\left(\frac{4}{x}ight)$.Using the identity $\cos ^{-1}(A) = \frac{\pi}{2} - \sin ^{-1}(A)$, we get $\sin ^{-1}\left(\frac{3}{x}ight) = \cos ^{-1}\left(\frac{4}{x}ight)$.Let $\sin ^{-1}\left(\frac{3}{x}ight) = 	heta$, then $\sin 	heta = \frac{3}{x}$, which implies $\cos 	heta = \sqrt{1 - \left(\frac{3}{x}ight)^2} = \frac{\sqrt{x^2-9}}{x}$.Thus, $	heta = \cos ^{-1}\left(\frac{\sqrt{x^2-9}}{x}ight)$.Equating the two expressions: $\frac{\sqrt{x^2-9}}{x} = \frac{4}{x}$.Squaring both sides: $x^2 - 9 = 16$, which gives $x^2 = 25$.Since the domain of $\sin ^{-1}$ requires $|\frac{3}{x}| \le 1$ and $|\frac{4}{x}| \le 1$, we have $|x| \ge 4$. Thus, $x = 5$.

If $\sin ^{-1}\left(\frac{3}{x}\right)+\sin ^{-1}\left(\frac{4}{x}\right)=\frac{\pi}{2}$, then $x$ is equal to

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