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

Question

1418 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-05T19:02:42

see below

#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