If y=tan ^{-1}left(frac{2-3 sin x}{3-2 sin x} | vedclass

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(B) Given,$y=	an ^{-1}\left(\frac{2-3 \sin x}{3-2 \sin x}ight)$.Using the chain rule,$\frac{d y}{d x}=\frac{1}{1+\left(\frac{2-3 \sin x}{3-2 \sin x

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$\frac{-5 \cos x}{13 \sin ^2 x-24 \sin x+13}$

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(B) Given,$y=	an ^{-1}\left(\frac{2-3 \sin x}{3-2 \sin x}ight)$.Using the chain rule,$\frac{d y}{d x}=\frac{1}{1+\left(\frac{2-3 \sin x}{3-2 \sin x}ight)^2} 	imes \frac{d}{d x}\left(\frac{2-3 \sin x}{3-2 \sin x}ight)$.Simplify the denominator: $1+\left(\frac{2-3 \sin x}{3-2 \sin x}ight)^2 = \frac{(3-2 \sin x)^2 + (2-3 \sin x)^2}{(3-2 \sin x)^2} = \frac{9+4 \sin^2 x - 12 \sin x + 4 + 9 \sin^2 x - 12 \sin x}{(3-2 \sin x)^2} = \frac{13 \sin^2 x - 24 \sin x + 13}{(3-2 \sin x)^2}$.Now,differentiate the inner function using the quotient rule: $\frac{d}{d x}\left(\frac{2-3 \sin x}{3-2 \sin x}ight) = \frac{(3-2 \sin x)(-3 \cos x) - (2-3 \sin x)(-2 \cos x)}{(3-2 \sin x)^2}$.$= \frac{-9 \cos x + 6 \sin x \cos x + 4 \cos x - 6 \sin x \cos x}{(3-2 \sin x)^2} = \frac{-5 \cos x}{(3-2 \sin x)^2}$.Combining these,$\frac{d y}{d x} = \frac{(3-2 \sin x)^2}{13 \sin^2 x - 24 \sin x + 13} 	imes \frac{-5 \cos x}{(3-2 \sin x)^2} = \frac{-5 \cos x}{13 \sin^2 x - 24 \sin x + 13}$.

If $y=\tan ^{-1}\left(\frac{2-3 \sin x}{3-2 \sin x}\right)$,then $\frac{d y}{d x}=$

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