How do you prove $tan^2(1/2theta)=(tantheta-sintheta)/(tantheta+sintheta)$ ?

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Answer 1 · 2016-10-05T19:02:42

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$tan^2(1/2 theta)=(tan theta-sin theta)/(tan theta + sin theta)$

Right Side $=(tan theta-sin theta)/(tan theta + sin theta)$

$=(sin theta / cos theta -sin theta)/(sin theta/cos theta + sin theta)$

$=((sin theta-sin theta cos theta) / cos theta) / ((sin theta+sin theta cos theta)/cos theta)$

$=(sin theta-sin theta cos theta) / cos theta * cos theta/(sin theta+sin theta cos theta)$

$=(sin theta-sin theta cos theta)/(sin theta+sin theta cos theta)$

$=(sin theta(1-cos theta))/(sin theta(1+cos theta))$

$=(1-cos theta)/(1+cos theta)$

$=(tan (1/2 theta))^2$

$=tan^2(1/2 theta)$

$=$ Left Side

How do you prove tan^2(1/2theta)=(tantheta-sintheta)/(tantheta+sintheta)tan^2(1/2theta)=(tantheta-sintheta)/(tantheta+sintheta)?