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

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(A) આપેલ છે, $\sin ^{-1}\left(\frac{x}{13}\right)+\operatorname{cosec}^{-1}\left(\frac{13}{12}\right)=\frac{\pi}{2} \quad ...(i)$ આપણે જાણીએ છીએ કે $\

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) આપેલ છે, $\sin ^{-1}\left(\frac{x}{13}ight)+\operatorname{cosec}^{-1}\left(\frac{13}{12}ight)=\frac{\pi}{2} \quad ...(i)$ આપણે જાણીએ છીએ કે $\operatorname{cosec}^{-1}(z) = \sin^{-1}(\frac{1}{z})$. તેથી, $\operatorname{cosec}^{-1}\left(\frac{13}{12}ight) = \sin^{-1}\left(\frac{12}{13}ight)$. આ કિંમત સમીકરણ $(i)$ માં મૂકતા: $\sin ^{-1}\left(\frac{x}{13}ight)+\sin ^{-1}\left(\frac{12}{13}ight)=\frac{\pi}{2}$. આપણે નિત્યસમ $\sin^{-1}(	heta) + \cos^{-1}(	heta) = \frac{\pi}{2}$ જાણીએ છીએ. વળી, $\sin^{-1}(\frac{12}{13}) = \cos^{-1}(\sqrt{1 - (\frac{12}{13})^2}) = \cos^{-1}(\sqrt{1 - \frac{144}{169}}) = \cos^{-1}(\sqrt{\frac{25}{169}}) = \cos^{-1}(\frac{5}{13})$. તેથી, $\sin^{-1}(\frac{x}{13}) + \cos^{-1}(\frac{5}{13}) = \frac{\pi}{2}$. આને $\sin^{-1}(	heta) + \cos^{-1}(	heta) = \frac{\pi}{2}$ સાથે સરખાવતા, આપણને $\frac{x}{13} = \frac{5}{13}$ મળે. આમ, $x = 5$.

જો $\sin ^{-1}\left(\frac{x}{13}\right)+\operatorname{cosec}^{-1}\left(\frac{13}{12}\right)=\frac{\pi}{2}$ હોય, તો $x$ ની કિંમત શોધો.

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