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Question

Prove that (tanA+sinA)(tanA-sinA)=(secA+1)(secA-1).


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Solution

Solving the left hand side (LHS) of the equation:

Taking the LHS and solving we get,

tanA+sinAtanA-sinA

=sinAcosA+sinAsinAcosA-sinA tanA=sinAcosA

=sinA1cosA+1sinA1cosA-1 secA=1cosA

=secA+1secA-1=RHS

i.e. (tanA+sinA)(tanA-sinA)=(secA+1)(secA-1)

Hence, proved.


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