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cosA/1+sinA. + 1+sinA/cosA. =2sec A
LHS: cos2A+(1+sin2A)/(1+sinA)cosA. [(a+b2=a2+b2+2ac]
=cos2A+1+2sinA+sin2A/(1+sinA)cosA
=cos2A+sin2A+1+2sinA/(1+sinA)cosA
=1+1+2sinA/(1+sinA)cosA. [cos2A+sin2A=1]
=2+2sinA/(1+sinA)cosA
=2/cosA=2secA =RHS
Hence Proved
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