Proving of this trigonometric identity $$\frac{\cot \beta}{\csc \beta - 1} + \frac{\cot \beta}{\csc \beta + 1} = 2 \sec \beta$$
What I've done:
$$\frac{\frac{\cos \beta}{\sin \beta}} {\frac{1}{\sin \beta} +1} +
\frac{\frac{\cos \beta}{\sin \beta}} {\frac{1}{\sin \beta} -1}=\\
=\frac{\frac{\cos \beta}{\sin \beta}} {\frac{1-\sin\beta}{\sin \beta}} +
\frac{\frac{\cos \beta}{\sin \beta}} {\frac{1+\sin\beta}{\sin \beta}}=\\
=\cos \beta\frac{(1 - \sin\beta)}{\sin^2\beta} +  \cos \beta\frac{1 + \sin\beta}{\sin^2\beta}=\\
=\frac{ \cos \beta - \cos\beta \sin\beta + \cos\beta + \cos\beta \sin\beta}{\sin^2 \beta}=\\
=\frac{2\cos\beta}{\sin^2\beta}=\\
=\frac{2\cos\beta}{1-\cos^2\beta}$$
Right side:
$$2 \sec\beta=\frac{1}{2\cos\beta}$$
Could you tell me where I went wrong? I've tried using a proof program online (symbolab) though those steps are a bit hard for me to follow.
Note: I only want to use what I have on the left side to solve the left side.
Thank you very much.
(Might be some errors first time using math jax..)

You first mistake was listening to your teacher. This has turned you into a $\sin$ / $\cos$ robot, causing you go against your better intuition and find a common denominator. So let's do that now
$$\begin{array}{lll}
\frac{\cot\beta}{\csc\beta-1}+\frac{\cot\beta}{\csc\beta+1}&=&\frac{\cot\beta}{\csc\beta-1}\cdot\frac{\csc\beta+1}{\csc\beta+1}+\frac{\cot\beta}{\csc\beta+1}\cdot\frac{\csc\beta-1}{\csc\beta-1}\\
&=&\cot\beta\bigg(\frac{\csc\beta+1}{\csc^2\beta-1}+\frac{\csc\beta-1}{\csc^2\beta-1}\bigg)\\
&=&\cot\beta\bigg(\frac{\csc\beta+1+\csc\beta-1}{\csc^2\beta-1}\bigg)\\
&=&\cot\beta\bigg(\frac{2\csc\beta}{\csc^2\beta-1}\bigg)\\
&=&\cot\beta\bigg(\frac{2\csc\beta}{\cot^2\beta}\bigg)\\
&=&2\frac{\csc\beta}{\cot\beta}\\
&=&2\csc\beta\tan\beta\\
\end{array}$$
At this point feel free to do the $\sin$/$\cos$ thingy, use the following identity
$$\tan\beta=\frac{\sin\beta}{\cos\beta}=\frac{\sin\beta}{\cos\beta}\cdot \frac{\frac{1}{\sin \beta \cos \beta}}{\frac{1}{\sin \beta \cos \beta}}=\frac{\frac{1}{\cos\beta}}{\frac{1}{\sin\beta}}=\frac{\sec\beta}{\csc\beta}$$
Continuing we have
$$2\csc\beta\tan\beta=2\csc\beta\cdot\frac{\sec\beta}{\csc\beta}=2\sec\beta$$
without the mess.
