If sin ^{-1}left(frac{x}{13}right)+operatorna | vedclass

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(C) Given the equation: $\sin ^{-1}\left(\frac{x}{13}\right)+\operatorname{cosec}^{-1}\left(\frac{13}{12}\right)=\frac{\pi}{2}$ We know that $\operato

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) Given the equation: $\sin ^{-1}\left(\frac{x}{13}ight)+\operatorname{cosec}^{-1}\left(\frac{13}{12}ight)=\frac{\pi}{2}$ We know that $\operatorname{cosec}^{-1}(y) = \sec^{-1}(\frac{1}{y})$. Therefore,$\operatorname{cosec}^{-1}\left(\frac{13}{12}ight) = \sec^{-1}\left(\frac{12}{13}ight)$. Substituting this into the equation: $\sin ^{-1}\left(\frac{x}{13}ight) + \sec^{-1}\left(\frac{12}{13}ight) = \frac{\pi}{2}$. We also know the identity $\sin^{-1}(	heta) + \cos^{-1}(	heta) = \frac{\pi}{2}$,which implies $\cos^{-1}(	heta) = \frac{\pi}{2} - \sin^{-1}(	heta)$. Since $\sec^{-1}\left(\frac{12}{13}ight) = \cos^{-1}\left(\frac{13}{12}ight)$ is not correct,let us use $\sec^{-1}(y) = \cos^{-1}(\frac{1}{y})$. Thus,$\sec^{-1}\left(\frac{12}{13}ight) = \cos^{-1}\left(\frac{13}{12}ight)$ is invalid as the domain of $\cos^{-1}$ is $[-1, 1]$. Correct approach: $\operatorname{cosec}^{-1}\left(\frac{13}{12}ight) = \sin^{-1}\left(\frac{12}{13}ight)$. So,$\sin ^{-1}\left(\frac{x}{13}ight) + \sin^{-1}\left(\frac{12}{13}ight) = \frac{\pi}{2}$. $\sin ^{-1}\left(\frac{x}{13}ight) = \frac{\pi}{2} - \sin^{-1}\left(\frac{12}{13}ight) = \cos^{-1}\left(\frac{12}{13}ight)$. Since $\cos^{-1}\left(\frac{12}{13}ight) = \sin^{-1}\left(\sqrt{1 - (\frac{12}{13})^2}ight) = \sin^{-1}\left(\sqrt{\frac{169-144}{169}}ight) = \sin^{-1}\left(\frac{5}{13}ight)$. Comparing both sides,we get $\frac{x}{13} = \frac{5}{13}$,which implies $x = 5$.

If $\sin ^{-1}\left(\frac{x}{13}\right)+\operatorname{cosec}^{-1}\left(\frac{13}{12}\right)=\frac{\pi}{2}$,then the value of $x$ is

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