How do you prove that #costheta/(1-sintheta) + costheta/(1+sintheta) = 2 /costheta#?

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Answer 1 · 2018-03-08T18:42:14

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#LHS : cos theta/(1-sin theta)+costheta/(1+sintheta)#

#=(costheta(1+sintheta)+costheta(1-sintheta))/((1-sintheta)(1+sin theta))#->common denominator

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

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

#=(2costheta)/cos^2theta#->use the property #sin^2theta+cos^2theta=1#

#=(2cancelcostheta)/cos^cancel2theta#

#=2/costheta#

#=RHS#