Geometry with Trigonometry From the given figure, prove that 

$$\cot\ \theta=\cot\ A+\cot\ B+\cot\ C$$


I could not even start to the first point. Please help me to solve this. 
Thanks in advance 

If $O$ is the intersection of the inner lines,
$\angle ABO=B-\theta\implies\angle AOB=\pi-B,$
$\angle BCO=C-\theta\implies\angle COB=\pi-C$
Writing $AB=c$, $BC=a$, $CA=b$ (conventionally)
Using Sine Law for $\triangle AOB,$ $$\dfrac{AB}{\sin(\pi-B)}=\dfrac{OB}{\sin\theta}\implies\dfrac c{\sin B}=\dfrac{OB}{\sin\theta}\  \ \ \  (1)$$
Similarly from $\triangle COB,$ $$\dfrac{OB}{\sin(C-\theta)}=\dfrac a{\sin C}\  \ \ \  (2)$$
Equating the values of $OB,$
$$\sin C\sin\theta=\dfrac ac\sin B\sin(C-\theta)\ \ \ \ (3)$$
Now using Sine Law for $\triangle ABC,$ $$\dfrac a{\sin A}=\dfrac c{\sin C}\iff\dfrac ac=\dfrac{\sin A}{\sin C}\ \ \ \  (4)$$
By $(3),(4)$
$$\sin\theta\cdot\sin^2C=\sin A\sin B\sin(C-\theta)=\sin A\sin B\{\sin C\cos\theta-\cos C\sin\theta\}$$
$$\implies\dfrac{\cos\theta}{\sin\theta}=\dfrac{\sin^2C+\sin A\sin B\cos C}{\sin A\sin B\sin C}\ \ \ \ (5)$$
Again, $$\cot A+\cot B+\cot C=\dfrac{\cos A\sin B\sin C+\sin A\cos B\sin C+\sin A\sin B\cos C}{\sin A\sin B\sin C}$$
$$=\dfrac{(\cos A\sin B+\sin A\cos B)\sin C+\sin A\sin B\cos C}{\sin A\sin B\sin C}\ \ \ \ (6)$$
Finally use $(\cos A\sin B+\sin A\cos B)\sin C$
$=\sin(A+B)\sin C=\sin(\pi-C)\cdot\sin C=\sin^2C$ on $(6)$
and compare $(5),(6)$.
