How to prove #(1+sintheta-costheta)/(1+costheta +sintheta)=tan(theta/2)#?

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Answer 1 · 2017-07-10T10:41:23

#LHS=(1+sintheta-costheta)/(1+costheta +sintheta) #

#=(sintheta(1+sintheta-costheta))/(sintheta(1+costheta +sintheta))#

#=(sintheta+sin^2theta-sinthetacostheta)/(sintheta(1+costheta +sintheta))#

#=(sintheta-sinthetacostheta+(1-cos^2theta))/(sintheta(1+costheta +sintheta))#

#=(sintheta(1-costheta)+(1-cos^2theta))/(sintheta(1+costheta +sintheta))#

#=((1-costheta)(sintheta+1+costheta))/(sintheta(1+costheta +sintheta))#

#=(1-costheta)/sintheta#

#=(2sin^2(theta/2))/(2sin(theta/2)cos(theta/2))#

#=tan(theta/2)#