How do I show $( \csc\theta- \cot\theta)^2 \equiv \frac{1-\cos\theta}{1+\cos\theta} $?

Question

How do I show that the left hand side of the below equation is equal to the right hand side?

$$( \csc\theta- \cot\theta)^2 \equiv \frac{1-\cos\theta}{1+\cos\theta} $$

Answer 1

There is

$$( \csc\theta- \cot\theta)^2 =\left( \frac{1}{\sin \theta} - \frac {\cos \theta}{\sin \theta}\right)^2 =\left( \frac{1-\cos \theta}{\sin \theta} \right)^2$$

Since $$ \sin^2 \theta = 1- \cos^2 \theta$$ the expression becomes:

$$( \csc\theta- \cot\theta)^2 \equiv \frac{1-\cos\theta}{1+\cos\theta} $$

and the identity is verified

Answer 2

$\csc2t-\cot2t=\dfrac{1-\cos2t}{\sin2t}=\tan t$ for $\sin t\ne0$

$\dfrac{1-\cos2t}{1+\cos2t}=\dfrac{2\sin^2t}{2\cos^2t}=?$