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

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

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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 $\sin ^{-1}(	heta) + \cos ^{-1}(	heta) = \frac{\pi}{2}$,so $\sin ^{-1}\left(\frac{3}{x}ight) = \frac{\pi}{2} - \sin ^{-1}\left(\frac{4}{x}ight) = \cos ^{-1}\left(\frac{4}{x}ight)$.Let $\sin ^{-1}\left(\frac{3}{x}ight) = 	heta$,then $\sin 	heta = \frac{3}{x}$,which implies $\cos 	heta = \sqrt{1 - \left(\frac{3}{x}ight)^2} = \frac{\sqrt{x^2 - 9}}{x}$.Equating $\cos 	heta = \frac{4}{x}$,we get $\frac{\sqrt{x^2 - 9}}{x} = \frac{4}{x}$.Squaring both sides,$x^2 - 9 = 16$,which gives $x^2 = 25$.Thus,$x = 5$ (since $x = -5$ does not satisfy the domain of the inverse sine function in the original equation).

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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