Prove the identity... $$\frac{\cos{2x}-\sin{4x}-\cos{6x}}{\cos{2x}+\sin{4x}-\cos{6x}}=\tan{(x-15^{\circ})}cot{(x+15^{\circ})}$$
So, here's what I've done so far, but don't know what do do next:
$$\frac{\cos{2x}-2\sin{2x}\cos{2x}-\cos{6x}}{\cos{2x}+2\sin{2x}\cos{2x}-\cos{6x}}=$$
$$\frac{\cos{2x}-4\sin{x}\cos{x}\cos{2x}-\cos{(4x+2x)}}{\cos{2x}+4\sin{x}\cos{x}\cos{2x}-\cos{(4x+2x)}}=$$
$$\frac{\cos{2x}-4\sin{x}\cos{x}\cos{2x}-\cos{4x}\cos{2x}+\sin{4x}\sin{2x}}{\cos{2x}+4\sin{x}\cos{x}\cos{2x}-\cos{4x}\cos{2x}+\sin{4x}\sin{2x}}=...$$
I have no idea what to do next. I have a big feeling that I'm going in a totally wrong direction. Is there anything I can do with the expression in the beginning?

Using Prosthaphaeresis Formula, $$\frac{\cos2x-\cos6x}{\sin4x}=\frac{2\sin4x\sin2x}{2\sin4x}=\frac{\sin2x}{\dfrac12}$$
$$\implies \frac{\cos2x-\cos6x}{\sin4x}=\frac{\sin2x}{\sin30^\circ}$$
Now apply Componendo and dividendo and again apply Prosthaphaeresis formulae $$\sin C\pm\sin D$$
