Simplify: $$\sin(x-2\pi)\cos\left(\frac{3\pi}{2}-x\right)+\tan(\pi-x)\tan\left(\frac{3\pi}{2}+x\right)$$
What I did was convert each term to its simplest form.
$$\sin(x-2\pi)\cos\left(\frac{3\pi}{2}-x\right) = \sin(x)\sin(x) = \sin^2(x)$$
$$\tan(\pi-x)\tan\left(\frac{3\pi}{2}+x\right) = (-\tan(x))(-\cot(x))=1 $$
So than I added those two together and got:
$$\sin^2(x)+1$$
However this is wrong somehow. Any ideas?
Hint: $$\begin{split} &\sin(x-2\pi)\cos(3\pi/2-x)+\tan(\pi-x)\tan(3\pi/2+x)\\ = &-\sin(x)\cos(\pi/2-x) + \left(-\tan(x)\right)\left(-\tan(x-\pi/2)\right)\\ =& -\sin^2(x) + \tan(x)\tan(x-\pi/2) \end{split}$$