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?

$$\triangle =\dfrac12bc\sin A=\dfrac12(2R\sin B)c\sin A\cdot\dfrac c{2R\sin C}$$
Now $A+B=\pi-C\implies\sin(A+B)=\sin(\pi-C)=?$
