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Question

Prove that:

cosA1+sinA+1+sinAcosA=2SecA


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Solution

To prove:

cosA1+sinA+1+sinAcosA=2SecA

Taking LHS

cosA1+sinA+1+sinAcosA

=(cosA)(cosA)+(1+sinA)(1+sinA)(1+sinA)(cosA) ( Taking LCM of denominator)

=cos2A+(1+sinA)2(1+sinA)cosA=cos2A+1+sin2A+2sinA(1+sinA)cosA

=2+2sinA(1+sinA)cosA (sin2A+cos2A=1)

=2(1+sinA)(1+sinA)cosA=2cosA=2SecA

=RHS

Hence, proved cosA1+sinA+1+sinAcosA=2SecA


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