How do you prove #(1-cos2theta)/(1+cos2theta) = tan^2theta#?

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Answer 1 · 2018-04-13T12:28:22

See below.

#LHS=(1-cos2theta)/(1+cos2theta)#

#=(1-(cos^2theta-sin^2theta))/(1+cos^2theta-sin^2theta)# since #cos2theta=cos^2theta-sin^2theta#

#=(1-cos^2theta+sin^2theta)/(1-sin^2theta+cos^2theta)# expanding brackets

#=(sin^2x+sin^2x)/(cos^2x+cos^2x)# since #sin^2x+cos^x=1#

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

#=sin^2x/cos^2x#

#=tan^2x# since #sinx/cosx=tanx#

#=RHS#

Answer 2 · 2018-04-13T12:29:16

Please see the proof below

We need

#cos2theta=1-2sin^2theta#

#cos2theta=2cos^2theta-1#

#tan theta=sintheta/costheta#

Therefore,

#LHS=(1-cos2theta)/(1+cos2theta)#

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

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

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

#=tan^2theta#

#=RHS#

#QED#