Question

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

Answer 1

$LHS=(\text{cosec}\theta -\cot \theta {)}^2=(\frac{1}{\sin \theta }-\frac{\cos \theta }{\sin \theta }{)}^2=\frac{(1-\cos \theta {)}^2}{{\sin }^2\theta }$

$=\frac{(1-\cos \theta {)}^2}{1-{\cos }^2\theta }=\frac{(1-\cos \theta {)}^2}{(1-\cos \theta )(1+\cos \theta )}=\frac{1-\cos \theta }{1+\cos \theta }=RHS$

$LHS=RHS$

Hence proved