Show that (\cosec \theta - \cot \theta)^2 = \frac{1 - \cos \theta}{1 + \cos \theta}

Step 1: Note down the given equation
\left(\operatorname{cosec}\theta-\cot\theta\right)^2=\frac{1-\cos\theta}{1+\cos\theta}
Step 2: Manipulate the L.H.S by the equivalent trigonometric ratios
L.H.S: = \left(\operatorname{cosec}\theta-\cot\theta\right)^2
= (\frac{1}{\sin \theta} - \frac{\cos \theta}{\sin \theta})^2
Do the L.C.M
= (\frac{1 - \cos \theta}{\sin \theta})^2
= (\frac{(1 - \cos \theta)^2}{\sin^2 \theta}
= (\frac{(1 - \cos \theta)^2}{1 - \cos^2 \theta}
[since, \sin^2 \theta + \cos^2 \theta = 1
Step 3: Use the a^2 - b^2 = (a - b) (a - b) formula to simplify
EXAMPLE: = \frac{(1-\cos\theta)^2}{\left(1^2-\cos^2\theta\right)}
= \frac{(1-\cos\theta)^2}{(1+\cos\theta)(1-\cos\theta)}
Cancel the equal terms
= \frac{(1-\cos\theta)}{(1+\cos\theta)}
Hence, proved R.H.S