Prove that $\frac{{-\cos(x-y)-\cos(x+y)}}{-\cos(x-y)+\cos(x+y)} = \cot x \cot y$ I solved this from my implicit differentiation, and i end up with this answer, they say it's right but not simplified, I tried to simply it but I get $\cot(x)\cot(y)-\tan(x)\tan(y)$
$$\frac{{-\cos(x-y)-\cos(x+y)}}{-\cos(x-y)+\cos(x+y)} =\cot x \cot y$$ 

we know that
$$\cos(A+B)=\cos A\cos B-\sin A\sin B$$
$$\sin(-A)=-\sin A$$
$$\cos(-A)=\cos A$$
Now just use these facts to solve the problem
$$\begin{array}{lll}
\frac{-\cos(x-y)-\cos(x+y)}{-\cos(x-y)+\cos(x+y)}&=&\frac{\cos(x-y)+\cos(x+y)}{\cos(x-y)-\cos(x+y)}\\
&=&\frac{(\cos x\cos(-y)-\sin x\sin(-y))+(\cos x\cos y-\sin x\sin y)}{(\cos x\cos(-y)-\sin x\sin(-y))-(\cos x\cos y-\sin x\sin y)}\\
&=&\frac{(\cos x\cos y+\sin x\sin y)+(\cos x\cos y-\sin x\sin y)}{(\cos x\cos y+\sin x\sin y)-(\cos x\cos y-\sin x\sin y)}\\
&=&\frac{(\cos x\cos y+\sin x\sin y)+(\cos x\cos y-\sin x\sin y)}{(\cos x\cos y+\sin x\sin y)+(-\cos x\cos y+\sin x\sin y)}\\
&=&\frac{2\cos x\cos y}{2\sin x\sin y}\\
&=&\cot x\cot y
\end{array}$$
