How do you prove #(cosx/(1 + sinx)) + (cosx/(1 - sinx)) = 2secx#?

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Answer 1 · 2018-03-06T17:27:01

See Below

#LHS#: #cosx /(1+sinx) + cos x/(1-sinx)#

#=(cosx(1-sinx)+cosx(1+sinx))/(1-sin^2x)#-->common denominator

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

#=(cosxcancel(-sin x cos x)+cosx+cancel(sinx cos x))/cos^2x#

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

#=(2cancelcosx)/cos^cancel2x#

#=2/cosx#

#=2*1/cosx#

#=sec x#

#=RHS#