Question

sinx-sin3x/sin²x-cos²x​

Answer 1

Answer:

2 sin x

Step-by-step explanation:

$\frac{sinx - sin3x}{sin^{2}x-cos^{2}x }$

=$\frac{2 cos(\frac{x + 3x}{2} ) . sin(\frac{x - 3x}{2})}{-cos2x}$

=$\frac{- 2 cos(2x) . sin(x)}{-cos2x}$  [cos2x = cos²x - sin²x]

=2  sin x