જો cos ^{-1}left(frac{12}{13}right)+sin ^{-1} | vedclass

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(B) આપણને સમીકરણ $\sin ^{-1}\left(\frac{3}{5}\right)+\cos ^{-1}\left(\frac{12}{13}\right)=\sin ^{-1} P$ આપેલ છે. પ્રથમ,$\cos ^{-1}\left(\frac{12}{13}\

Vedclass · Accepted Answer

$\frac{56}{65}$

Vedclass · Answer

(B) આપણને સમીકરણ $\sin ^{-1}\left(\frac{3}{5}ight)+\cos ^{-1}\left(\frac{12}{13}ight)=\sin ^{-1} P$ આપેલ છે. પ્રથમ,$\cos ^{-1}\left(\frac{12}{13}ight)$ ને $\sin ^{-1}$ સ્વરૂપમાં ફેરવો. ધારો કે $\cos ^{-1}\left(\frac{12}{13}ight) = 	heta$,તો $\cos 	heta = \frac{12}{13}$. $\sin 	heta = \sqrt{1-\cos^2 	heta} = \sqrt{1-\left(\frac{12}{13}ight)^2} = \sqrt{1-\frac{144}{169}} = \sqrt{\frac{25}{169}} = \frac{5}{13}$ નો ઉપયોગ કરતા. તેથી,$\cos ^{-1}\left(\frac{12}{13}ight) = \sin ^{-1}\left(\frac{5}{13}ight)$. હવે,સમીકરણ $\sin ^{-1}\left(\frac{3}{5}ight)+\sin ^{-1}\left(\frac{5}{13}ight) = \sin ^{-1} P$ બને છે. સૂત્ર $\sin ^{-1} x + \sin ^{-1} y = \sin ^{-1}\left(x \sqrt{1-y^2} + y \sqrt{1-x^2}ight)$ નો ઉપયોગ કરતા: $\sin ^{-1} P = \sin ^{-1}\left[\frac{3}{5} \sqrt{1-\left(\frac{5}{13}ight)^2} + \frac{5}{13} \sqrt{1-\left(\frac{3}{5}ight)^2}ight]$ $= \sin ^{-1}\left[\frac{3}{5} 	imes \frac{12}{13} + \frac{5}{13} 	imes \frac{4}{5}ight]$ $= \sin ^{-1}\left(\frac{36}{65} + \frac{20}{65}ight) = \sin ^{-1}\left(\frac{56}{65}ight)$. તેથી,$P = \frac{56}{65}$.

જો $\cos ^{-1}\left(\frac{12}{13}\right)+\sin ^{-1}\left(\frac{3}{5}\right)=\sin ^{-1} P$ હોય,તો $P$ નું મૂલ્ય શોધો.

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