Question #ac9e3

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Answer 1 · 2016-12-18T07:21:52

See the proof below

We use

#tanx=sinx/cosx#

#sin^2x+cos^2x=1#

LHS, #tan^2x/(secx+1)=(sin^2x/cos^2x)/(1/cosx+1)#

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

#=(1-cos^2x)/(cosx(1+cosx))#

#=(cancel(1+cosx)(1-cosx))/(cosxcancel(1+cosx))#

#=(1-cosx)/cosx#

#=RHS#

#Q.E.D#

Answer 2 · 2016-12-18T07:26:25

#LHS=tan^2x/(secx+1)#

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

#=(1-cos^2x)/(cosx*cosx(secx+1))#

#=((1-cosx)(1+cosx))/(cosx(cosx*secx+cosx))#

#=((1-cosx)cancel((1+cosx)))/(cosxcancel((1+cosx)))#

#=(1-cosx)/cosx=RHS#

Proved