Question

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

Answer 1

$(\cot \theta -\cos ec\theta {)}^2=\frac{1-\cos \theta }{1+\cos \theta },$

Let LHS

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