27. question please solve it step by step
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In the numerator take sinA common and in the denominator take cosA as common
Therefore,LHS= sinA(1-2sin²A)/cosA(2cos²A-1)
sinA(1-2sin²A)= sinA(1-sin²A-sin²A)
= sinA(cos²A-1+cos²A)
= sinA(2cos²A-1)
Therefore, LHS = sinA(2cos²A-1)/cosA(2cos²A-1)
=sinA/cosA [(2cos²A-1) gets cancelled}
= tanA = RHS
Hence, LHS= RHS
Hence, proved
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