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Question

Prove 1+sinAcosA+cosA1+sinA=2secA

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Solution

Simplifying the LHS of 1+sinAcosA+cosA1+sinA=2secA.


1+sinAcosA+cosA1+sinA=(1+sinA)2+(cosA)2(1+sinA)(cosA)

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

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

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

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

=2cosA

=2secA

=RHS


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