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..)
 A: 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.
A: 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 $\sin^2\beta$.
