Question

Cos²(90° -X).tan(180°-X).cos(180°+X)/sin(180°-X)

Answer 1

Recall some identities:

tan(x) = sin(x) / cos(x)

cos(x + y) = cos(x) cos(y) - sin(x) sin(y)

cos(x - y) = cos(x) cos(y) + sin(x) sin(y)

sin(x - y) = sin(x) cos(y) - cos(x) sin(y)

This means we have

• cos²(90° - x) = [cos(90°) cos(x) + sin(90°) sin(x)]²

… = sin²(x)

• tan(180° - x) = sin(180° - x) / cos(180° - x)

… = [sin(180°) cos(x) - cos(180°) sin(x)] / [cos(180°) cos(x) + sin(180°) sin(x)]

… = sin(x) / (-cos(x))

… = -tan(x)

(and we also get sin(180° - x) = sin(x))

• cos(180° + x) = cos(180°) cos(x) - sin(180°) sin(x)

… = -cos(x)

So, the given expression reduces to

sin²(x) (-tan(x)) (-cos(x)) / sin(x) = sin²(x)

since tan(x) and cos(x)/sin(x) = 1/tan(x) will cancel.