Question

prove that

sin A - 2 sin^3 A/
2 sin^3A- cos A
=tan A​

Answer 1

Answer:

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Answer 2

Answer:

LHS =  $\frac{sinA - 2sin^3A}{2cos^3A - cosA}$ =  $\frac{sinA(1-2sin^2A)}{cosA(2cos^2A-1)}$

= $\frac{sinA (sin^2A+cos^2A- 2sin^2A)}{cosA (2cos^2A - sin^2A- cos^2A)}$

=$\frac{sinA (cos^2A- sin^2A)}{cosA(cos^2A-sin^2A)}$

= $\frac{sinA}{cosA}$ = tanA

hence proved.

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