From the given figure From the given figure, prove that 

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


My work. 

I have got this solution from one of my friends. But I didn't understand all the process. I understood the application of ceva theorem in the first step but the steps after that I can’t understand. Also how has be $\prod $ used here, why? How it works here?  Please make me understand 
 A: The $\Pi$ (capital Pi) notation is just a shorthand for the products. Your friend is taking the product over cyclic permutations of $A,B,C$.  So he (I assume it's a he!) didn't really compute anything different when he wrote the $\Pi$ in the second step.
On the last line of the products, he's using the difference angle formula:
$$\sin(A - \theta) = \sin A \cos \theta - \cos A \sin \theta.$$
Can you fill in the rest?
A: *

*Well, what I can say at most is that $\prod$ sign is used to express a "product". For example, $\prod_\limits{i=1}^3 a_i=a_1a_2a_3$. Here, the product of the $3$ trigonometric quantities has been expressed using the $\prod$ sign.

*And on the second line, it should be $$\frac{\sin^2 A}{\sin B \sin C}\cdot \frac{\sin^2 B}{\sin A \sin C}\cdot \frac{\sin^2 C}{\sin B \sin A}$$


Now the last part of the work is simple solution of a quadratic equation. And I think you will understand it if you make a careful study of it.
So I am restating the middle part of the solution, where he uses the short-hand sign for product.
$$\frac{\sin (A-\theta)}{\sin \theta}\cdot \frac{\sin (B-\theta)}{\sin \theta}\cdot \frac{\sin (C-\theta)}{\sin \theta}=\frac{\sin^2 A}{\sin B \sin C}\cdot \frac{\sin^2 B}{\sin A \sin C}\cdot \frac{\sin^2 C}{\sin B \sin A}$$
$$\implies \frac{\sin (A-\theta)}{\sin \theta\cdot \sin A}\cdot \frac{\sin (B-\theta)}{\sin \theta\cdot \sin B}\cdot \frac{\sin (C-\theta)}{\sin \theta\cdot \sin C}=\frac{\sin A}{\sin B \sin C}\cdot \frac{\sin B}{\sin A \sin C}\cdot \frac{\sin C}{\sin B \sin A}$$
$$\implies (\cot \theta-\cot A)\cdot (\cot \theta-\cot B)\cdot (\cot \theta-\cot C)=\frac{\sin (B+C)}{\sin B \sin C}\cdot \frac{\sin (C+A)}{\sin A \sin C}\cdot \frac{\sin (B+A)}{\sin B \sin A}$$
$$\implies (\cot \theta-\cot A)\cdot (\cot \theta-\cot B)\cdot (\cot \theta-\cot C)=(\cot B+\cot C)\cdot (\cot C+\cot A)\cdot (\cot A+\cot B)$$
I hope you can carry on the rest.
