Recall the double angle identities:
cos(2θ) = 1 - 2 sin²(θ)
sin(2θ) = 2 sin(θ) cos(θ)
With some rewriting, we can get
cos(4θ) - cos(2θ) = 2 cos²(2θ) - 1 - cos(2θ) = (2 cos(2θ) + 1) (cos(2θ) - 1)
and
sin(4θ) + sin(2θ) = 2 sin(2θ) cos(2θ) + sin(2θ) = sin(2θ) (2 cos(2θ) + 1)
The numerator and denominator both share a factor of 2 cos(2θ) + 1; canceling them leaves us with
(cos(2θ) - 1) / sin(2θ)
Replace cos(2θ) = 1 - 2 sin²(θ) and sin(2θ) = 2 sin(θ) cos(θ) :
((1 - 2 sin²(θ)) - 1) / (2 sin(θ) cos(θ))
-2 sin²(θ) / (2 sin(θ) cos(θ))
- sin(θ) / cos(θ)
- tan(θ)
so the answer is C.
