How do you prove ${tan}^{2} (\frac{1}{2} θ) = \frac{tan θ - sin θ}{tan θ + sin θ}$ $tan^2(1/2theta)=(tantheta-sintheta)/(tantheta+sintheta)$ ?

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How do you prove ${tan}^{2} (\frac{1}{2} θ) = \frac{tan θ - sin θ}{tan θ + sin θ}$ $tan^2(1/2theta)=(tantheta-sintheta)/(tantheta+sintheta)$ ?

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Answer 1 · 2016-10-05T19:02:42

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Explanation:

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

Right Side $= \frac{tan θ - sin θ}{tan θ + sin θ}$ $=(tan theta-sin theta)/(tan theta + sin theta)$

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

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

$= \frac{sin θ - sin θ cos θ}{cos θ} \cdot \frac{cos θ}{sin θ + sin θ cos θ}$ $=(sin theta-sin theta cos theta) / cos theta * cos theta/(sin theta+sin theta cos theta)$

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

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

$= \frac{1 - cos θ}{1 + cos θ}$ $=(1-cos theta)/(1+cos theta)$

$= {(tan (\frac{1}{2} θ))}^{2}$ $=(tan (1/2 theta))^2$

$= {tan}^{2} (\frac{1}{2} θ)$ $=tan^2(1/2 theta)$

$=$ $=$ Left Side

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How do you prove ${tan}^{2} (\frac{1}{2} θ) = \frac{tan θ - sin θ}{tan θ + sin θ}$ $tan^2(1/2theta)=(tantheta-sintheta)/(tantheta+sintheta)$ ?

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Explanation:

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How do you prove ${tan}^{2} (\frac{1}{2} θ) = \frac{tan θ - sin θ}{tan θ + sin θ}$ $tan^2(1/2theta)=(tantheta-sintheta)/(tantheta+sintheta)$ ?