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

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(A) Given,$\sin ^{-1}\left(\frac{x}{13}\right)+\operatorname{cosec}^{-1}\left(\frac{13}{12}\right)=\frac{\pi}{2} \quad ...(i)$ We know that $\operator

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$5$

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(A) Given,$\sin ^{-1}\left(\frac{x}{13}ight)+\operatorname{cosec}^{-1}\left(\frac{13}{12}ight)=\frac{\pi}{2} \quad ...(i)$ We know that $\operatorname{cosec}^{-1}(z) = \sin^{-1}(\frac{1}{z})$. Therefore,$\operatorname{cosec}^{-1}\left(\frac{13}{12}ight) = \sin^{-1}\left(\frac{12}{13}ight)$. Substituting this into equation $(i)$,we get: $\sin ^{-1}\left(\frac{x}{13}ight)+\sin ^{-1}\left(\frac{12}{13}ight)=\frac{\pi}{2}$. We know the identity $\sin^{-1}(	heta) + \cos^{-1}(	heta) = \frac{\pi}{2}$. Also,$\sin^{-1}(\frac{12}{13}) = \cos^{-1}(\sqrt{1 - (\frac{12}{13})^2}) = \cos^{-1}(\sqrt{1 - \frac{144}{169}}) = \cos^{-1}(\sqrt{\frac{25}{169}}) = \cos^{-1}(\frac{5}{13})$. So,$\sin^{-1}(\frac{x}{13}) + \cos^{-1}(\frac{5}{13}) = \frac{\pi}{2}$. Comparing this with $\sin^{-1}(	heta) + \cos^{-1}(	heta) = \frac{\pi}{2}$,we must have $\frac{x}{13} = \frac{5}{13}$. Thus,$x = 5$.

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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