Alternative area of a triangle formula The problem is as follows:
There is a triangle $ABC$ and I need to show that it's area is: $$\frac{1}{2} c^2 \frac{\sin A \sin B}{\sin (A+B)}$$
Since there is a half in front I decided that base*height is equivalent to $c^2 \frac{\sin A \sin B}{\sin (A+B)}$. So I made an assumption that base is $c$ and went on to prove that height is $c \frac{\sin A \sin B}{\sin (A+B)}$. But I end up expressing height in terms of itself.. i.e.  $h \equiv \frac{ch}{a\cos B + b \cos A}$. How do I prove this alternative area of triangle formula?

Let's denote by $[ABC]$ the area of $\triangle ABC$, its known that $$[ABC]=\frac{1}{2}ab\sin C$$
From Sine Law we have  $a\sin C=c\sin A$ and $b\sin C=c\sin B$, also $\sin (A+B)=\sin(\pi-C)=\sin C$, then
\begin{align*}
[ABC]&=\frac{1}{2}\frac{(a\sin C)(b\sin C)}{\sin C}\\
&=\frac{1}{2}\frac{(c\sin A)(c\sin B)}{\sin (A+B)}\\
&=\frac{1}{2}c^2\frac{\sin A\sin B}{\sin (A+B)}
\end{align*}
