<s>[INST] How do I show $( \csc\theta- \cot\theta)^2 \equiv \frac{1-\cos\theta}{1+\cos\theta} $? How do I show that the left hand side of the below equation is equal to the right hand side?
$$( \csc\theta- \cot\theta)^2 \equiv  \frac{1-\cos\theta}{1+\cos\theta} $$
 [/INST] 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
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