How do you prove $\frac{1 - 2 {cos}^{2} θ}{sin θ cos θ} = tan θ - cot θ$ $(1-2cos^2theta)/(sinthetacostheta)=tantheta-cottheta$ ?

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Answer 1 · 2016-10-07T15:57:30

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$\frac{1 - 2 {cos}^{2} θ}{sin θ cos θ} = tan θ - cot θ$ $(1-2cos^2theta)/(sintheta cos theta) = tan theta-cot theta$

Right Side: $= tan θ - cot θ$ $=tan theta-cot theta$

$= \frac{sin θ}{cos θ} - \frac{cos θ}{sin θ}$ $=sin theta/cos theta - cos theta/sin theta$

$= \frac{{sin}^{2} θ - {cos}^{2} θ}{sin θ cos θ}$ $=(sin^2 theta-cos^2 theta)/(sin theta cos theta)$

$= \frac{1 - {cos}^{2} θ - {cos}^{2} θ}{sin θ cos θ}$ $=(1-cos^2 theta -cos^2 theta)/(sin theta cos theta)$

$= \frac{1 - 2 {cos}^{2} θ}{sin θ cos θ}$ $=(1-2cos^2 theta)/(sin theta cos theta)$

$:.=$ Left Side

How do you prove (1-2cos^2theta)/(sinthetacostheta)=tantheta-cottheta1−2cos2θsinθcosθ=tanθ−cotθ(1-2cos^2theta)/(sinthetacostheta)=tantheta-cottheta?