જો sin ^{-1}left(frac{x}{5}right)+operatornam | vedclass

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(D) આપેલ સમીકરણ: $\sin ^{-1}\left(\frac{x}{5}\right)+\operatorname{cosec}^{-1}\left(\frac{5}{4}\right)=\frac{\pi}{2}$ આપણે જાણીએ છીએ કે $\operatorname

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$3$

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(D) આપેલ સમીકરણ: $\sin ^{-1}\left(\frac{x}{5}\right)+\operatorname{cosec}^{-1}\left(\frac{5}{4}\right)=\frac{\pi}{2}$ આપણે જાણીએ છીએ કે $\operatorname{cosec}^{-1}(y) = \sin ^{-1}\left(\frac{1}{y}\right)$. તેથી,$\operatorname{cosec}^{-1}\left(\frac{5}{4}\right) = \sin ^{-1}\left(\frac{4}{5}\right)$. આ કિંમત સમીકરણમાં મૂકતા: $\sin ^{-1}\left(\frac{x}{5}\right)+\sin ^{-1}\left(\frac{4}{5}\right)=\frac{\pi}{2}$ $\Rightarrow \sin ^{-1}\left(\frac{x}{5}\right)=\frac{\pi}{2}-\sin ^{-1}\left(\frac{4}{5}\right)$ નિત્યસમ $\sin ^{-1}(z)+\cos ^{-1}(z)=\frac{\pi}{2}$ નો ઉપયોગ કરતા,આપણને મળે $\frac{\pi}{2}-\sin ^{-1}\left(\frac{4}{5}\right) = \cos ^{-1}\left(\frac{4}{5}\right)$. $\Rightarrow \sin ^{-1}\left(\frac{x}{5}\right)=\cos ^{-1}\left(\frac{4}{5}\right)$ હવે,$\cos ^{-1}\left(\frac{4}{5}\right)$ ને $\sin ^{-1}$ સ્વરૂપમાં ફેરવતા: $\cos ^{-1}\left(\frac{4}{5}\right) = \sin ^{-1}\left(\sqrt{1-\left(\frac{4}{5}\right)^2}\right) = \sin ^{-1}\left(\sqrt{1-\frac{16}{25}}\right) = \sin ^{-1}\left(\sqrt{\frac{9}{25}}\right) = \sin ^{-1}\left(\frac{3}{5}\right)$. $\Rightarrow \sin ^{-1}\left(\frac{x}{5}\right)=\sin ^{-1}\left(\frac{3}{5}\right)$ બંને બાજુ સરખાવતા,આપણને મળે $\frac{x}{5} = \frac{3}{5}$,જેનો અર્થ છે કે $x = 3$.

જો $\sin ^{-1}\left(\frac{x}{5}\right)+\operatorname{cosec}^{-1}\left(\frac{5}{4}\right)=\frac{\pi}{2}$ હોય,તો $x$ ની કિંમત શોધો.

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