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

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(A) Given,$y = \sqrt{\frac{1+	an x}{1-	an x}} = \sqrt{	an(\frac{\pi}{4}+x)}$.Using the chain rule,$\frac{dy}{dx} = \frac{1}{2\sqrt{	an(\frac{\pi}{

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

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(A) Given,$y = \sqrt{\frac{1+	an x}{1-	an x}} = \sqrt{	an(\frac{\pi}{4}+x)}$.Using the chain rule,$\frac{dy}{dx} = \frac{1}{2\sqrt{	an(\frac{\pi}{4}+x)}} \cdot \sec^2(\frac{\pi}{4}+x) \cdot \frac{d}{dx}(\frac{\pi}{4}+x)$.Since $\frac{d}{dx}(\frac{\pi}{4}+x) = 1$,we have:$\frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{\sqrt{	an(\frac{\pi}{4}+x)}} \cdot \sec^2(\frac{\pi}{4}+x)$.Substituting $\sqrt{	an(\frac{\pi}{4}+x)} = \sqrt{\frac{1+	an x}{1-	an x}}$,we get:$\frac{dy}{dx} = \frac{1}{2} \sqrt{\frac{1-	an x}{1+	an x}} \sec^2(\frac{\pi}{4}+x)$.Thus,option $A$ is correct.

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

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