cos left(sin ^{-1} frac{1}{5}+cos ^{-1} xrigh | vedclass

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Question

(B) Given the equation: $\cos \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} xight)=0$. We know that $\cos 	heta = 0$ implies $	heta = \frac{\pi}{2}$ (co

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$\frac{1}{5}$

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(B) Given the equation: $\cos \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} xight)=0$. We know that $\cos 	heta = 0$ implies $	heta = \frac{\pi}{2}$ (considering the principal value branch). Therefore,$\sin ^{-1} \frac{1}{5}+\cos ^{-1} x = \frac{\pi}{2}$. Rearranging the terms,we get $\cos ^{-1} x = \frac{\pi}{2} - \sin ^{-1} \frac{1}{5}$. Using the identity $\sin ^{-1} y + \cos ^{-1} y = \frac{\pi}{2}$,we have $\cos ^{-1} y = \frac{\pi}{2} - \sin ^{-1} y$. Thus,$\cos ^{-1} x = \cos ^{-1} \frac{1}{5}$. Comparing both sides,we get $x = \frac{1}{5}$.

$\cos \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=0$,then $x=$ . . . . . . .

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