If sec ^{-1}left(frac{5}{x}right)+sin ^{-1} l | vedclass

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(D) Given the equation: $\sec ^{-1}\left(\frac{5}{x}\right)+\sin ^{-1} \left(\frac{4}{5}\right)=\frac{\pi}{2}$.We know the identity $\sin ^{-1}(y) + \

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$4$

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(D) Given the equation: $\sec ^{-1}\left(\frac{5}{x}\right)+\sin ^{-1} \left(\frac{4}{5}\right)=\frac{\pi}{2}$.We know the identity $\sin ^{-1}(y) + \cos ^{-1}(y) = \frac{\pi}{2}$ for $y \in [-1, 1]$.Thus,$\frac{\pi}{2} - \sin ^{-1} \left(\frac{4}{5}\right) = \cos ^{-1} \left(\frac{4}{5}\right)$.Substituting this into the original equation,we get: $\sec ^{-1}\left(\frac{5}{x}\right) = \cos ^{-1} \left(\frac{4}{5}\right)$.Since $\sec ^{-1}(z) = \cos ^{-1} \left(\frac{1}{z}\right)$,we have $\cos ^{-1} \left(\frac{x}{5}\right) = \cos ^{-1} \left(\frac{4}{5}\right)$.Comparing the arguments,we get $\frac{x}{5} = \frac{4}{5}$,which implies $x = 4$.

If $\sec ^{-1}\left(\frac{5}{x}\right)+\sin ^{-1} \left(\frac{4}{5}\right)=\frac{\pi}{2}$,where $x \neq 0$,then $x=$ . . . . . . .

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