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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Aicharya · Accepted Answer

Given, $\sin ^{-1}\left(\frac{x}{13}ight)+\operatorname{cosec}^{-1}\left(\frac{13}{12}ight)=\frac{\pi}{2} ..(i)$ Let $\operatorname{cosec}^{-1} \frac{13}{12}=y$ Then. $\operatorname{cosec} y=\frac{13}{12} \Rightarrow \sin y=\frac{12}{13}$ $	herefore \cos y=\sqrt{1-\sin ^{2} y}$ $\begin{array}{l} =\sqrt{1-\left(\frac{12}{13}ight)^{2}}=\sqrt{1-\frac{144}{169}} \ =\sqrt{\frac{25}{169}}=\frac{5}{13} \Rightarrow y=\cos ^{-1} \frac{5}{13} \end{array}$ Eq. (i) becomes, $\sin ^{-1}\left(\frac{x}{13}ight)+\cos ^{-1}\left(\frac{5}{13}ight)=\frac{\pi}{2}$ We know that $\sin ^{-1} 	heta+\cos ^{-1} 	heta=\frac{\pi}{2}$ $	herefore$ Both angles of Eq. (i) should be same. $	herefore x=5$

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

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