frac{d}{dx}left( frac{cot^2 x - 1}{cot^2 x + | vedclass

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Question

(D) We are given the expression $y = \frac{\cot^2 x - 1}{\cot^2 x + 1}$.First,we simplify the expression using trigonometric identities:$\frac{\cot^2

Vedclass · Accepted Answer

$-2\sin 2x$

Vedclass · Answer

(D) We are given the expression $y = \frac{\cot^2 x - 1}{\cot^2 x + 1}$.First,we simplify the expression using trigonometric identities:$\frac{\cot^2 x - 1}{\cot^2 x + 1} = \frac{\frac{\cos^2 x}{\sin^2 x} - 1}{\frac{\cos^2 x}{\sin^2 x} + 1} = \frac{\cos^2 x - \sin^2 x}{\cos^2 x + \sin^2 x}$.Since $\cos^2 x + \sin^2 x = 1$ and $\cos^2 x - \sin^2 x = \cos 2x$,the expression simplifies to $\cos 2x$.Now,we differentiate with respect to $x$:$\frac{d}{dx}(\cos 2x) = -\sin 2x \cdot \frac{d}{dx}(2x) = -2\sin 2x$.Thus,the correct option is $D$.

$\frac{d}{dx}\left( \frac{\cot^2 x - 1}{\cot^2 x + 1} \right) = $

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