If y = sqrt{frac{1 + tan x}{1 - tan x}},then | vedclass

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(A) Given $y = \sqrt{\frac{1 + 	an x}{1 - 	an x}}$.Using the formula $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,we know that $	an\le

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$\frac{1}{2}\sqrt{\frac{1 - 	an x}{1 + 	an x}} \cdot \sec^2\left(\frac{\pi}{4} + xight)$

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(A) Given $y = \sqrt{\frac{1 + 	an x}{1 - 	an x}}$.Using the formula $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,we know that $	an\left(\frac{\pi}{4} + xight) = \frac{	an(\pi/4) + 	an x}{1 - 	an(\pi/4)	an x} = \frac{1 + 	an x}{1 - 	an x}$.So,$y = \sqrt{	an\left(\frac{\pi}{4} + xight)}$.Differentiating with respect to $x$ using the chain rule:$\frac{dy}{dx} = \frac{1}{2\sqrt{	an\left(\frac{\pi}{4} + xight)}} \cdot \frac{d}{dx}\left(	an\left(\frac{\pi}{4} + xight)ight)$.$\frac{dy}{dx} = \frac{1}{2\sqrt{	an\left(\frac{\pi}{4} + xight)}} \cdot \sec^2\left(\frac{\pi}{4} + xight)$.Since $\frac{1}{\sqrt{	an(\pi/4 + x)}} = \sqrt{\frac{1 - 	an x}{1 + 	an x}}$,we get:$\frac{dy}{dx} = \frac{1}{2}\sqrt{\frac{1 - 	an x}{1 + 	an x}} \cdot \sec^2\left(\frac{\pi}{4} + xight)$.

If $y = \sqrt{\frac{1 + \tan x}{1 - \tan x}}$,then $\frac{dy}{dx} = $

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