Lines $OA, OB$ and $OC$ are drawn in a triangle so that the angles $OAB$, $OBC$ and $OCA$ are each equal to $\theta$ In the triangle $ABC$, lines $OA, OB$ and $OC$ are drawn so that the angles $OAB$, $OBC$ and $OCA$ are each equal to $\theta$, prove that:
$1.$ $\cot \theta= \cot A +\cot B +\cot C$ 
$2.$ $\csc(2 \theta)=\csc(2 A)+\csc(2 B) +\csc(2 C)$
I verified the result for equilateral triangle but how to prove it in general? 
Could someone give me slight hint?
 A: The problem can be solved by repeatedly applying the sine rule. One strategy can be taking ratios of sides, and observing patterns. As in this case:
$$\frac{AO}{BO}=\frac{\sin(B-\theta)}{\sin\theta}=\sin B\cot\theta-\cos B$$
Note that you can also refer to the ratios $\frac{BO}{CO}$ or $\frac{CO}{AO}$, but the method remains the same.
But this alone will not tell you the whole story. However, on further inspection, note that $\angle AOB=\pi-B$,  and $\angle AOC=\pi-A$. Above, we found out the ratio $\frac{AO}{BO}$. So, lets again find that ratio, but now in other terms:
$$\frac{AO}{AC}=\frac{\sin x}{\sin A}\\
\frac{BO}{AB}=\frac{\sin x}{\sin B}
$$
Using the above two ratios, compute $\frac{AO}{BO}$, to get:
$$\frac{AO}{BO}=\frac{\sin B\times AC}{\sin A\times AB}$$
Apply sine rule in the biggest triangle:
$$\frac{AO}{BO}=\frac{\sin^2 B}{\sin A\sin C}$$
Now, equate the RHS we obtained:
$$\sin B\cot\theta-\cos B=\frac{\sin^2 B}{\sin A\sin C}$$
And from there the claim. I leave the trigonometric simplification to you.  
As for the second equation, square the LHS and RHS of the first equation.
