Trigonometric identities: $ \frac{1+\cos(a)}{1-\cos(a)} + \frac{1-\cos(a)}{1+\cos(a)} = 2+4\cot^2(a)$ I don't really know how to begin, so if I'm missing some information please let me know what it is and I'll fill you guys in :).
This is the question I can't solve:
$$
\frac{1+\cos(a)}{1-\cos(a)} + \frac{1-\cos(a)}{1+\cos(a)} = 2+4\cot^2(a)       
$$
I need to prove their trigonometric identities.
I have the $5$ basic set of rules, I could write them all here but I suppose it's not needed, if it is please let me know since it's not gonna be simple to type.
I have over $40$ questions like these and I just couldn't seem to understand how to prove them equal, my best was $4 \cot^2(a) = 2 + 4 \cot^2(a)$
Thanks for everything!

Simplify: 
$$\frac{1 + \cos a}{1 - \cos a} + \frac{1-\cos a}{1 + \cos a} \equiv \frac{(1+ \cos a)^2 + (1-\cos a)^2}{(1-\cos a)(1+\cos a)}$$
The denominator is the difference of two squares, expand the numerator: 
$$\begin{align} \frac{2 + 2\cos^2 a}{1 - \cos^2 a} &\equiv \frac{2(1 + \cos^2 a)}{\sin^2 a} \\ & \equiv \frac{2}{\sin^2 x} + 2\frac{\cos^2 x}{\sin^2 x} \\ & \equiv 2\csc^2 x + 2\cot^2 x \ \\ & \equiv 2(1 + \cot^2 x) + 2\cot^2 x \\ & \equiv 2 + 4\cot^2 x\end{align}$$

Some explanations: 


*

*The first line follows from $\frac{a+b}{c} \equiv \frac{a}{c} +
    \frac{b}{c}$

*Since $\frac{1}{\sin x} = \csc x \Rightarrow \frac{1}{\sin^2 x}
    \equiv \csc^2 x$. We used this in the second line. 

*Since $\cot x \equiv \frac{1}{ \tan x} \equiv \frac{\cos x}{\sin x}$
then $\cot^2 x = \frac{\cos^2 x}{\sin^2 x}$ we used this in the
third line. 

*Note that since $\sin^2 x + \cos^2 x \equiv 1$ then we can divide
both sides of this identity by $\sin^2 x$ to get $1 + \cot^2 x
\equiv \csc^2 x$, we used this in the penultimate line.
