Can you use both sides of an equation to prove equality? For example:

$\color{red}{\text{Show that}}$$$\color{red}{\frac{4\cos(2x)}{1+\cos(2x)}=4-2\sec^2(x)}$$

In high school my maths teacher told me

To prove equality of an equation; you start on one side and manipulate it algebraically until it is equal to the other side.

So starting from the LHS: $$\frac{4\cos(2x)}{1+\cos(2x)}=\frac{4(2\cos^2(x)-1)}{2\cos^2(x)}=\frac{2(2\cos^2(x)-1)}{\cos^2(x)}=\frac{4\cos^2(x)-2}{\cos^2(x)}=4-2\sec^2(x)$$ $\large\fbox{}$
At University, my Maths Analysis teacher tells me

To prove a statement is true, you must not use what you are trying to prove.

So using the same example as before:
LHS = $$\frac{4\cos(2x)}{1+\cos(2x)}=\frac{4(2\cos^2(x)-1)}{2\cos^2(x)}=\frac{2(2\cos^2(x)-1)}{\cos^2(x)}=\frac{2\Big(2\cos^2(x)-\left[\sin^2(x)+\cos^2(x)\right]\Big)}{\cos^2(x)}=\frac{2(\cos^2(x)-\sin^2(x))}{\cos^2(x)}=\bbox[yellow]{2-2\tan^2(x)}$$
RHS =$$4-2\sec^2(x)=4-2(1+\tan^2(x))=\bbox[yellow]{2-2\tan^2(x)}$$
So I have shown that the two sides of the equality in $\color{red}{\rm{red}}$ are equal to the same highlighted expression. But is this a sufficient proof?
Since I used both sides of the equality (which is effectively; using what I was trying to prove) to show that $$\color{red}{\frac{4\cos(2x)}{1+\cos(2x)}=4-2\sec^2(x)}$$
One of the reasons why I am asking this question is because I have a bounty question which is suffering from the exact same issue that this post is about.

EDIT:
Comments and answers below seem to indicate that you can use both sides to prove equality. So does this mean that my high school maths teacher was wrong?
$$\bbox[#AFF]{\text{Suppose we have an identity instead of an equality:}}$$ $$\bbox[#AFF]{\text{Is it possible to manipulate both sides of an identity to prove that the identity holds?}}$$
Thank you.

It is enough.. Consider this example:
To prove:
$a=b$
Proof:
$$a=c$$
$$b=c$$
Since $a$ and $b$ are equal to the same thing, $a=b$.
That is the exact technique you are using and it sure can be used.
