Question

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

Answer 1

L.H.S. $(\text{cosec}\theta -\cot \theta {)}^2$

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

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

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

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

$=\frac{1-\cos \theta }{1+\cos \theta }=R.H.S.$