tan ^{-1}left[frac{sin x}{1+cos x}right] का t | vedclass

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(B) माना $u = 	an ^{-1}\left(\frac{\sin x}{1+\cos x}ight)$ और $v = 	an ^{-1}\left(\frac{\cos x}{1+\sin x}ight)$.$u$ के लिए:$u = 	an ^{-1}\left(

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$-1$

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(B) माना $u = 	an ^{-1}\left(\frac{\sin x}{1+\cos x}ight)$ और $v = 	an ^{-1}\left(\frac{\cos x}{1+\sin x}ight)$.$u$ के लिए:$u = 	an ^{-1}\left(\frac{2 \sin(x/2) \cos(x/2)}{2 \cos^2(x/2)}ight) = 	an ^{-1}(	an(x/2)) = x/2$.अतः,$\frac{du}{dx} = \frac{1}{2}$.$v$ के लिए:$v = 	an ^{-1}\left(\frac{\cos^2(x/2) - \sin^2(x/2)}{(\cos(x/2) + \sin(x/2))^2}ight) = 	an ^{-1}\left(\frac{(\cos(x/2) - \sin(x/2))(\cos(x/2) + \sin(x/2))}{(\cos(x/2) + \sin(x/2))^2}ight)$.$v = 	an ^{-1}\left(\frac{\cos(x/2) - \sin(x/2)}{\cos(x/2) + \sin(x/2)}ight) = 	an ^{-1}\left(\frac{1 - 	an(x/2)}{1 + 	an(x/2)}ight) = 	an ^{-1}\left(	an\left(\frac{\pi}{4} - \frac{x}{2}ight)ight) = \frac{\pi}{4} - \frac{x}{2}$.अतः,$\frac{dv}{dx} = -\frac{1}{2}$.अब,$\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{1/2}{-1/2} = -1$.

$\tan ^{-1}\left[\frac{\sin x}{1+\cos x}\right]$ का $\tan ^{-1}\left[\frac{\cos x}{1+\sin x}\right]$ के सापेक्ष अवकलज क्या है?

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