Simplify $\frac{\cos(2x)}{\cot(x)-1}-\frac{\sin(2x)}{2}$ I am given this expression to simplify:
$\frac{\cos(2x)}{\cot(x)-1}-\frac{\sin(2x)}{2}$
and I know the correct answer is $\sin^2(x)$
I was able to reduce the second fraction to a bit nicer
$\frac{\sin(2x)}{2}=\frac{2\sin(x)\cos(x)}{2\tan(x)\cot(x)}=\frac{2\sin(x)\cos(x)}{\frac{2\cos(x)\sin(x)}{\cos(x)\sin(x)}}=\frac{2\sin(x)\cos(x)}{1}\cdot\frac{\cos(x)\sin(x)}{2\cos(x)\sin(x)}=\cos(x)\sin(x)$
which changes the original into this
$\frac{\cos(2x)}{\cot(x)-1}-\cos(x)\sin(x)$
which is the prettiest I've got. No matter what I do with the first fraction I get something too complicated. Does anyone have any idea? Thx

$$\frac { \cos  2x }{ \cot  x-1 } -\frac { \sin  2x }{ 2 } =\frac { \cos ^{ 2 }{ x } -\sin ^{ 2 }{ x }  }{ \frac { \cos { x } -\sin { x }  }{ \sin { x }  }  } -\sin { x } \cos { x } =$$
$$=\frac { \sin { x } \left( \cos { x } +\sin { x }  \right) \left( \cos { x } -\sin { x }  \right)  }{ \cos { x } -\sin { x }  } -\sin { x\cos { x } = } $$
$$=\sin { x } \cos { x } +\sin ^{ 2 }{ x } -\cos { x } \sin { x } =\sin ^{ 2 }{ x } $$
