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..)

The mistake is in the third line of your derivation: the denominators should be $(1-\sin\beta)(1+\sin\beta)$, or $1-\sin^2\beta$, instead of $\sin^2\beta$.
