Question

Prove that $(\cot A-co\sec A{)}^2=\frac{1-\cos A}{1+\cos A}$

Answer 1

LHS$=(\cot A-cosecA{)}^2$

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

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

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

$=\frac{1-\cos A}{1+\cos A}$

$=RHS$

Hence proved.