Proving a trigonometric expression is identical to $2(\operatorname{cosec}^{2}{B}-1).$ I came across this trigonometric identity:
$$\frac{\operatorname{cosec}{B} - \cot{B}}{\operatorname{cosec}{B} + \cot{B}} + \frac{\operatorname{cosec}{B} + \cot{B}}{\operatorname{cosec}{B} - \cot{B}} = 2(\operatorname{cosec}^{2}{B} - 1) = 2(\frac{1+\cos^{2}{B}}{1 - \cos^{2}{B}})$$
And as I solved it, the equation came down to:
$2\operatorname{cosec}^{2}{B}+2\cot^{2}{B}.$
This can be written as $2(\operatorname{cosec}^{2}{B} + \cot^{2}{B}),$ which can further be written as $2\frac{1+\cos^{2}{B}}{1 - \cos^{2}{B}}.$
But I can't seem to get my mind around the middle part of the question, that is, $2(\operatorname{cosec}^{2}{B} - 1)$
Is it possible to write it like this, or is this an error in the question paper itself?

$$(\csc B-\cot B)(\csc B+\cot B)=1\implies\dfrac{\csc B+\cot B}{\csc B-\cot B}=(\csc B+\cot B)^2$$
$$(\csc B+\cot B)^2+(\csc B-\cot B)^2=2(\csc^2B+\cot^2B)$$
Now $\cot^2B=\csc^2B-1$
Finally,  $$\dfrac{1+\cos^2B}{1-\cos^2B}=\dfrac{1+\cos^2B}{\sin^2B}=\csc^2B+\cot^2B$$
