sinA/1+cosA+tanA/1-cosA
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here is your answer.
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sinA/1+cosA + tanA/1-cosA=[sinA(1-cosA)+tanA(1+cosA)]/sin^2A
=[sinA.2.sin^2(A/2) + tanA.2cos^2(A/2)]/sin^2A
=[2.sin^2(A/2)+2.cos^2(A/2)/cosA]/sinA
=tan(A/2)+[{1+cosA}/cosA]/sinA
=tan(A/2)+{secA+1}/sinA
further factorising tanA & secA in terms of sinA & cosA...we can get the complete solution of the given question.
=[sinA.2.sin^2(A/2) + tanA.2cos^2(A/2)]/sin^2A
=[2.sin^2(A/2)+2.cos^2(A/2)/cosA]/sinA
=tan(A/2)+[{1+cosA}/cosA]/sinA
=tan(A/2)+{secA+1}/sinA
further factorising tanA & secA in terms of sinA & cosA...we can get the complete solution of the given question.
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