frac{d}{dx} left[ left( frac{tan^2 2x - tan^2 | vedclass

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(C) Let $y = \frac{	an^2 2x - 	an^2 x}{1 - 	an^2 2x 	an^2 x}$.Using the identity $a^2 - b^2 = (a-b)(a+b)$,we have:$y = \frac{(	an 2x - 	an x)(

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$\sec^2 x$

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(C) Let $y = \frac{	an^2 2x - 	an^2 x}{1 - 	an^2 2x 	an^2 x}$.Using the identity $a^2 - b^2 = (a-b)(a+b)$,we have:$y = \frac{(	an 2x - 	an x)(	an 2x + 	an x)}{(1 + 	an 2x 	an x)(1 - 	an 2x 	an x)}$.Using the tangent addition and subtraction formulas $	an(A-B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$ and $	an(A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,we get:$y = 	an(2x - x) 	an(2x + x) = 	an x 	an 3x$.Now,we need to find $\frac{d}{dx} [y \cdot \cot 3x]$.Substituting $y$,we get $\frac{d}{dx} [	an x 	an 3x \cdot \cot 3x]$.Since $	an 3x \cdot \cot 3x = 1$,the expression simplifies to $\frac{d}{dx} [	an x]$.Therefore,$\frac{d}{dx} [	an x] = \sec^2 x$.

$\frac{d}{dx} \left[ \left( \frac{\tan^2 2x - \tan^2 x}{1 - \tan^2 2x \tan^2 x} \right) \cot 3x \right] =$

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