If y = frac{tan x + cot x}{tan x - cot x},the | vedclass

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(D) Given $y = \frac{	an x + \cot x}{	an x - \cot x}$.We can simplify the expression by converting it to $\sin x$ and $\cos x$ or using trigonometri

Vedclass · Accepted Answer

$-2	an 2x \sec 2x$

Vedclass · Answer

(D) Given $y = \frac{	an x + \cot x}{	an x - \cot x}$.We can simplify the expression by converting it to $\sin x$ and $\cos x$ or using trigonometric identities.$y = \frac{\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x}}{\frac{\sin x}{\cos x} - \frac{\cos x}{\sin x}} = \frac{\frac{\sin^2 x + \cos^2 x}{\sin x \cos x}}{\frac{\sin^2 x - \cos^2 x}{\sin x \cos x}} = \frac{1}{-(\cos^2 x - \sin^2 x)} = -\frac{1}{\cos 2x} = -\sec 2x$.Now,differentiate $y = -\sec 2x$ with respect to $x$ using the chain rule:$\frac{dy}{dx} = -(\sec 2x 	an 2x) \cdot \frac{d}{dx}(2x) = -2 \sec 2x 	an 2x$.Thus,the correct option is $D$.

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

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