If y=left(log _{cot x} tan xright)left(log _{ | vedclass

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(B) Given $y = (\log_{\cot x} 	an x)(\log_{	an x} \cot x) + 	an^{-1}\left(\frac{4x}{4-x^2}ight)$.Since $\log_{\cot x} 	an x = \frac{1}{\log_{	a

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$\frac{4}{4+x^2}$

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(B) Given $y = (\log_{\cot x} 	an x)(\log_{	an x} \cot x) + 	an^{-1}\left(\frac{4x}{4-x^2}ight)$.Since $\log_{\cot x} 	an x = \frac{1}{\log_{	an x} \cot x}$,the product is $1$.So,$y = 1 + 	an^{-1}\left(\frac{4x}{4-x^2}ight)$.Let $x = 2 	an 	heta$,then $\frac{4x}{4-x^2} = \frac{8 	an 	heta}{4 - 4 	an^2 	heta} = 2 \left(\frac{2 	an 	heta}{1 - 	an^2 	heta}ight) = 2 	an(2	heta)$.Alternatively,differentiate directly: $\frac{dy}{dx} = \frac{d}{dx} 	an^{-1}\left(\frac{4x}{4-x^2}ight)$.Using the chain rule: $\frac{dy}{dx} = \frac{1}{1 + \left(\frac{4x}{4-x^2}ight)^2} \cdot \frac{d}{dx} \left(\frac{4x}{4-x^2}ight)$.$= \frac{(4-x^2)^2}{(4-x^2)^2 + 16x^2} \cdot \frac{4(4-x^2) - 4x(-2x)}{(4-x^2)^2}$.$= \frac{16 - 8x^2 + x^4 + 16x^2}{1} 	ext{ (denominator)} = 16 + 8x^2 + x^4 = (4+x^2)^2$.$= \frac{16 - 4x^2 + 8x^2}{(4+x^2)^2} = \frac{16 + 4x^2}{(4+x^2)^2} = \frac{4(4+x^2)}{(4+x^2)^2} = \frac{4}{4+x^2}$.

If $y=\left(\log _{\cot x} \tan x\right)\left(\log _{\tan x} \cot x\right)+\tan ^{-1}\left(\frac{4 x}{4-x^2}\right)$,then $\frac{d y}{d x}=$

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