Question

Solve the following question in easiest way with explanation and I will mark him/her as the brainliest:

Answer 1

RHS=1/(seca-tana)
LHS=(sina-cosa+1)/(sina+cosa-1)
=(sina-cosa+1)/(sina+cosa-1)×(sina-cosa+1)/(sina-cosa+1)
=(sin^2a-sinacosa+sina-cosasina+cos^2a-cosa+sina-cosa+1)/(sin^2a-sinacosa+sina+cosasina-cos^2a+cosa-sina+cosa-1)
=(2-2cosasina+2sina-2cosa)/(sin^2a-cos^2a+2cosa-1)
=2(1-cosasina+sina-cosa)/(sin^2a-cos^2a+2cosa-sin^2a-cos^2a)
=2{1-cosa+sina(1-cosa)}/(-2cos^2a+2cosa)
=2{(1-cosa)(1+sina)}/2{cosa-cos^2a}
=(1-cosa)(1+sina)/cosa(1-cosa)
=(1+sina)/cosa
=1/cosa+sina/cosa
=seca+tana
=(seca+tana)×(seca-tana)/(seca-tana)
=(sec^2a-tan^2a)/(seca-tana)
=1/(seca-tana)
=RHS
Hence proves that,
(sina-cosa+1)/(sina+cosa-1)=1/(seca-tana)