If $A$ and $B$ are solutions to $7\cos\theta+4\sin\theta+5=0$, then $\cot\frac{A}{2}+\cot\frac{B}{2}=-\frac{2}{3}$ If $A$ and $B$ are the solutions to ${\displaystyle 7\cos\theta+4\sin\theta+5=0\mbox{ where }A>0,0<B<2\pi;}$
Without finding the solutions to the trig equation, show that;
$$\cot\left(\frac{A}{2}\right)+\cot\left(\frac{B}{2}\right)=-\frac{2}{3}$$
This is my effort so far:
Since A and B are solutions then:
\begin{align*}
7\cos A+4\sin A+5 & =0\\
7\left( 2\cos^{2}\frac{A}{2}-1\right) +4\times2\sin\frac{A}{2}\cos\frac{A}{2}+5 & =0\\
14\cos^{2}\frac{A}{2}+8\sin\frac{A}{2}\cos\frac{A}{2}-2 & =0\\
\mbox{Now divide by }\sin\frac{A}{2}\cos\frac{A}{2} & \mbox{ gives:}\\
14\cot\frac{A}{2}+8-\frac{2}{\sin\frac{A}{2}\cos\frac{A}{2}} & =0\\
\mbox{similarly for B;}\\
14\cot\frac{B}{2}+8-\frac{2}{\sin\frac{B}{2}\cos\frac{B}{2}} & =0
\end{align*}
When I add these two equations I have $\cot\left(\frac{A}{2}\right)+\cot\left(\frac{B}{2}\right)$,
but am unsure how to progress.

We use following relations:
$sin(\theta)=\frac {2 tan\frac{(\theta)}2}{1+tan^2(\frac{\theta}2}$
$cos(\theta)=\frac {1-tan^2(\frac{\theta}2}{1+tan^2(\frac{\theta}2}$
plugging these in equation we get:
$6cot^2(\frac{(\theta)}2)+4cot(\frac{(\theta)}2)-1=0$
Comparing with $ax^2+bx+c=0$ we have:
$x_1+x_2=-\frac ba=-\frac 23$
which gives what is required.
