Question

Prove that:
$(\csc \theta -\cot \theta {)}^2=\frac{1-\cos \theta }{1+\cos \theta }$

Answer 1

$\begin{array}{c}\frac{\left(1-\cos \theta \right)}{\left(1+\cos \theta \right)}={\left(\cos ec\theta -\cot \theta \right)}^2\\ \Rightarrow {\left(\frac{1}{\sin \theta }-\frac{\cos \theta }{\sin \theta }\right)}^2\\ \Rightarrow \frac{{\left(1-\cos \theta \right)}^2}{{\sin }^2\theta }\\ \Rightarrow \frac{\left(1-\cos \theta \right)\left(1-\cos \theta \right)}{1-{\cos }^2\theta }\\ \Rightarrow \frac{\left(1-\cos \theta \right)\left(1-\cos \theta \right)}{\left(1-\cos \theta \right)\left(1+\cos \theta \right)}\\ \Rightarrow \frac{1-\cos \theta }{1+\cos \theta }\end{array}$