Question

SinA-2sin³A/2cos²-cosA

Answer 1

SOLUTION:-

We have,

sinA -2sin³A/2cos² -cosA

Take L.H.S

$\frac{sinA - 2sin {}^{3} A}{2 {cos}^{3} - cosA} \\ \\ = > \frac{sina(1 - 2sin {}^{2} A )}{cosa(2cos {}^{2} A - 1) } \\ \\ = > \frac{sinA}{cosA} ( \frac{1 - 2(1 - {cos}^{2} A }{2 {cos}^{2} A - 1 } ) \\ \\ = > tanA( \frac{1 - 2 + 2 {cos}^{2} A }{2 {cos}^{2} A - 1} ) \\ \\ = > tan A( \frac{ - 1 + 2 {cos}^{2} A}{2 {cos}^{2} A - 1 } \\ \\ = > tanA( \frac{2cos {}^{2} A - 1 }{2cos {}^{2} A - 1 } ) \\ \\ = > tanA \times 1 \\ \\ = > tanA \: \: \: \: \: \: \: [R.H.S]$

Proved.

Answer 2

Step-by-step explanation:

here comes your answer