<s>[INST] 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?
 [/INST] Here is what I did:
$$\frac{1}{2}bh \equiv \frac{1}{2}c^2\frac{\sin A \sin B}{\sin (A+B)}$$
Assume that $b$ = $c$.
Then, $c\frac{\sin A \sin B}{\sin (A+B)} \equiv \frac{c}{\cot A + \cot B}$. But $\cot A \equiv \frac{b \cos A}{h}$ and $\cot B \equiv \frac{a\cos B}{h}$. Therefore $\frac{c}{\cot A + \cot B} \equiv \frac{ch}{a\cos B + b\cos A}$. Since $a\cos B + b\cos A \equiv c$, we have $h \equiv \frac{ch}{c}$. $h\equiv h$.
I realised this after I posted the question...
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