(1+cosA+sinA)/(1+cosA-sinA)=(1+sinA)/cosA

Asked by  | 29th Aug, 2012, 09:33: PM

Expert Answer:

LHS = (1+cosA+sinA)/(1+cosA-sinA)
= (1+cosA+sinA)(1+cosA+sinA)/(1+cosA-sinA)(1+cosA+sinA)
= (1+cosA+sinA)2/(1+cosA)2 -sin2A
= 1+cos2A+sin2A+2cosA+2sinA+2sinAcosA/(1+cos2A+2cosA)-sin2A
= 2+2cosA+2sinA+2sinAcosA/2cos2A+2cosA
= (1+cosA)+sinA(1+cosA)/cosA(cosA+1)
= (1+cosA)(1+sinA)/cosA(cosA+1)
= 1+sinA/cosA
= RHS

Answered by  | 29th Aug, 2012, 10:13: PM

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