cos left( sin^{-1} frac{5}{13} right) = | vedclass

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Question

(A) Let $\sin^{-1} \frac{5}{13} = x$. Then,$\sin x = \frac{5}{13}$. Since $\cos^2 x = 1 - \sin^2 x$,we have $\cos x = \sqrt{1 - \left( \frac{5}{13} \r

Vedclass · Accepted Answer

$\frac{12}{13}$

Vedclass · Answer

(A) Let $\sin^{-1} \frac{5}{13} = x$. Then,$\sin x = \frac{5}{13}$. Since $\cos^2 x = 1 - \sin^2 x$,we have $\cos x = \sqrt{1 - \left( \frac{5}{13} \right)^2} = \sqrt{1 - \frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13}$. Therefore,$\cos \left( \sin^{-1} \frac{5}{13} \right) = \cos x = \frac{12}{13}$. Since the value $\frac{12}{13}$ is present in option $A$,the correct answer is $A$.

$\cos \left( \sin^{-1} \frac{5}{13} \right) = $

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