જો sin ^{-1} frac{1}{3}+sin ^{-1} frac{3}{5}+ | vedclass

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(B) આપેલ સમીકરણ: $\sin ^{-1} \frac{1}{3}+\sin ^{-1} \frac{3}{5}+\sin ^{-1} x=\frac{\pi}{2}$.આપણે જાણીએ છીએ કે $\sin ^{-1} x = \cos ^{-1} \sqrt{1-x^2}$

Vedclass · Accepted Answer

$\frac{8 \sqrt{2}-3}{15}$

Vedclass · Answer

(B) આપેલ સમીકરણ: $\sin ^{-1} \frac{1}{3}+\sin ^{-1} \frac{3}{5}+\sin ^{-1} x=\frac{\pi}{2}$.આપણે જાણીએ છીએ કે $\sin ^{-1} x = \cos ^{-1} \sqrt{1-x^2}$.તેથી,$\sin ^{-1} \frac{1}{3} = \cos ^{-1} \frac{2\sqrt{2}}{3}$ અને $\sin ^{-1} \frac{3}{5} = \cos ^{-1} \frac{4}{5}$.સમીકરણમાં કિંમતો મૂકતા: $\sin ^{-1} x = \frac{\pi}{2} - (\sin ^{-1} \frac{1}{3} + \sin ^{-1} \frac{3}{5})$.ધારો કે $A = \sin ^{-1} \frac{1}{3}$ અને $B = \sin ^{-1} \frac{3}{5}$.તો $\sin A = \frac{1}{3}, \cos A = \frac{2\sqrt{2}}{3}$ અને $\sin B = \frac{3}{5}, \cos B = \frac{4}{5}$.$x = \sin(\frac{\pi}{2} - (A+B)) = \cos(A+B) = \cos A \cos B - \sin A \sin B$.$x = (\frac{2\sqrt{2}}{3} 	imes \frac{4}{5}) - (\frac{1}{3} 	imes \frac{3}{5}) = \frac{8\sqrt{2}-3}{15}$.આમ,સાચો વિકલ્પ $B$ છે.

જો $\sin ^{-1} \frac{1}{3}+\sin ^{-1} \frac{3}{5}+\sin ^{-1} x=\frac{\pi}{2}$ હોય,તો $x=$

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