Question

prove that sinA -2sin3A / 2cos3A-cosA = tanA

Answer 1

I hope it should be helpful for you

Answer 2

This question is incorrect. It would be "2sin^3A" in place of "2sin3A" and "2cos^3A" in place of "2cos3A".

Hence the question will be -

$\frac{sinA - 2sin^3A}{2cos^3A - cosA} }{}  = tanA$

L.H.S.

$\frac{sinA - 2sin^3A}{2cos^3A - cosA} }{}  </p><p>[tex]\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$

∴ L.H.S. = R.H.S [PROVED]