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!
 A: 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.
A: A general approach is to try to convert the left side to the right side. In practice, this often ends up being done by simplifying BOTH sides, but making sure that each step in the simplification is invertible. So if you have something like
$$
\sin^2 x + \cos^2 x + \ldots = \ldots
$$
it's OK to turn that into 
$$
1 + \ldots = \ldots.
$$
On the other hand, if you have something like
$$
\sin^2 x = \cos^2 y
$$
you cannot turn this into 
$$
\sin x = \cos y
$$
because it's possible that 
$$
\sin x = -\cos y;
$$
both simpler equations imply the one with the squares, but the one with the squares doesn't imply either of the simpler ones. 
My own preference is to start by changing everything to sines and cosines, getting rid of all tangents, secants, cotangents, etc. In the case of your problem, that means changing
$$
\frac{1+\cos(a)}{1-\cos(a)} + \frac{1-\cos(a)}{1+\cos(a)} = 2+4\cot^2(a)
$$
into
$$
\frac{1+\cos(a)}{1-\cos(a)} + \frac{1-\cos(a)}{1+\cos(a)} = 2+4\frac{\cos^2(a)}{\sin^2 (a)}.
$$
The next step, often ignored, is to note for which values of $a$ the equality makes no sense; in this case, it's values of $a$ for which $\cos a = \pm 1$, or 
$\sin a = 0$. In either of these cases, you've got a division by zero. So at this point, I write down
"From here on, assume that $a$ is not a multiple of $\pi$, so that $\cos a \ne \pm 1$, and $\sin a \ne 0.$"
Then I multiply through by the denominators to get rid of fractions. That means multiplying by $1-\cos a$, then by $1 + \cos a$, adn then by $\sin^2 a$. The first two can be combined: $(1 - \cos a)(1+\cos a) = 1 - \cos^2 a= \sin^2 a$, so it turns out that this also combines with the third. So: multiply everything on the left by $(1-\cos a)(1+\cos a)$ and everything on the right by $\sin^2 a$ (which is the same thing!). 
This is allowable because mutliplying by a nonzero number can be inverted by multiplying by its multiplicative inverse. 
So the two original sides are equal if and only if they're equal after multiplying by the stuff in the previous paragraph. 
Expand out, apply the $\sin^2 x + \cos^2 x = 1$ rule as often as you can, and you'll probably find they're equal. 
A: $\dfrac{1+\cos x}{1-\cos x}=\dfrac{(1+\cos x)^2}{1-\cos^2x}$
$=\left(\dfrac{1+\cos x}{\sin x}\right)^2=(\csc x+\cot x)^2$
As $\dfrac{1+\cos x}{1-\cos x}\cdot\dfrac{1-\cos x}{1+\cos x}=1$ for $\cos x\ne\pm1$
$\dfrac{1-\cos x}{1+\cos x}=(\csc x-\cot x)^2$ as $(\csc x+\cot x)(\csc x-\cot x)=1$
Can you take it from here?
A: Put $t=\tan (\frac a2)$.You have $$\frac{1+\cos(a)}{1-\cos(a)}=\frac{1+\frac{1-t^2}{1+t^2}}{1-\frac{1-t^2}{1+t^2}}=\frac{1}{t^2}$$
Besides $\cot(a)=\frac{1-t^2}{2t}$ so we have to prove
$$\frac{1}{t^2}+t^2=2+4\left(\frac{1-t^2}{2t}\right)^2$$
Immediate calculation gives $$\frac{1+t^4}{t^2}=\frac{1+t^4}{t^2}$$ which is obviously true.
