Question

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

Answer 1

R.H.S.

= tanA

L.H.S.

$= \frac{sinA - 2 {sin}^{3}A }{2 {cos}^{3}A - cosA } \\ = \frac{sinA}{cosA} ( \frac{1 - 2 {sin }^{2}A }{ 2{cos}^{2}A - 1 } ) \\ = tanA( \frac{1 - 2(1 - {cos}^{2}A) }{ 2{cos}^{2}A - 1} ) \: \: \: \: \: \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ... \: ( {sin}^{2}A + {cos}^{2}A = 1) \\ = \: tanA (\frac{1 - 2 + 2 {cos}^{2} A}{2 {cos}^{2} A - 1} ) \\ = tanA( \frac{2 {cos}^{2}A - 1 }{2 {cos}^{2}A - 1} ) \\ = tanA$

= R.H.S.

•.• L.H.S. = R.H.S.

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

... Hence Proved!