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?
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.