Question

sin²x + 4 sin x +3. by cos²x. = 3+sinx / 1- sin x

Answer 1

sin²x + 4 sin x +3 / cos²x = 3+sinx / 1- sin x
(sin²x + 4 sin x +3) * (1- sin x) = (3+sinx) * cos²x
sin²x + 4 sin x + 3 - sin3x - 4 sin²x - 3sin x = 3cos²x + sinx * cos²x
- sin3x - 3sin²x + sin x +3 = 3cos²x + sinx * cos²x
- sin3x - 3sin²x - 3cos²x + sin x +3 = sinx * cos²x
- sin3x - 3 (sin²x+cos²x ) + sin x +3 = sinx * cos²x
- sin3x - 3 * 1 + sin x +3 = sinx * cos²x
- sin3x + sin x = sinx * cos²x
- sin3x - sinx * cos²x + sin x = 0
- sinx ( sin²x+cos²x ) + sin x = 0
- sinx + sin x = 0
0 = 0

Answer 2

Last few steps: >-3sin^2x+3sin^2x+sinxcos^2x+3cos^2x.
>3cos^2x + sinxcos^2x.=RHS..
Hence Proved.
Hope it helps.