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Question #56c6f

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Feb 21, 2017

$tan 2 x - cot 2 x$ $tan2x-cot2x$

$= \frac{sin 2 x}{cos 2 x} - \frac{cos 2 x}{sin 2 x}$ $=(sin2x)/(cos2x)-(cos2x)/(sin2x)$

$= \frac{{sin}^{2} 2 x - {cos}^{2} 2 x}{sin 2 x cos 2 x}$ $=(sin^2 2x-cos^2 2x)/(sin2xcos2x)$

$= - \frac{2 ({cos}^{2} 2 x - {sin}^{2} 2 x)}{2 \cdot sin 2 x cos 2 x}$ $=-(2(cos^2 2x-sin^2 2x))/(2*sin2xcos2x)$

$= - \frac{2 cos 4 x}{sin 4 x}$ $=-(2cos4x)/(sin4x)$

$= - cot 4 x$ $=-cot4x$

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