Prove the following identity,where the angles involved are acute angles for which the expressions are defined:
$(\operatorname{cosec} \theta - \cot \theta)^2 = \frac{1 - \cos \theta}{1 + \cos \theta}$

(A) To prove: $(\operatorname{cosec} \theta - \cot \theta)^2 = \frac{1 - \cos \theta}{1 + \cos \theta}$
Consider the $L.H.S. = (\operatorname{cosec} \theta - \cot \theta)^2$
Using the trigonometric identities $\operatorname{cosec} \theta = \frac{1}{\sin \theta}$ and $\cot \theta = \frac{\cos \theta}{\sin \theta}$,we get:
$L.H.S. = \left( \frac{1}{\sin \theta} - \frac{\cos \theta}{\sin \theta} \right)^2$
$= \left( \frac{1 - \cos \theta}{\sin \theta} \right)^2 = \frac{(1 - \cos \theta)^2}{\sin^2 \theta}$
Using the identity $\sin^2 \theta = 1 - \cos^2 \theta$,we get:
$= \frac{(1 - \cos \theta)^2}{1 - \cos^2 \theta}$
Since $a^2 - b^2 = (a - b)(a + b)$,we can write $1 - \cos^2 \theta = (1 - \cos \theta)(1 + \cos \theta)$:
$= \frac{(1 - \cos \theta)^2}{(1 - \cos \theta)(1 + \cos \theta)}$
$= \frac{1 - \cos \theta}{1 + \cos \theta} = R.H.S.$
Hence,$L.H.S. = R.H.S.$