Question

Sin theta - 2Sin^2theta / 2Cos^3 theta - Cos theta =Tan theta

Answer fast please​

Answer 1

Here I am using A instead of theta.

It may like this :

LHS = (sinA-2sin³A)/(2cos³A-cosA)

=[sinA(1-2sin²A)]/[cosA(2cos²A-1]

= tanA[(1-2sin²A)]/[2(1-sin²A)-1]

= tanA[(1-2sin²A)]/(1-2sin²A)

= tanA

= RHS

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