Question

Show that: $\frac{1+\cos A}{1-\cos A}=\frac{{\tan }^{2}A}{{(\sec A-1)}^2}$

Answer 1

$RHS=\frac{{\tan }^{2}A}{({\text{sec}}^{2}A-1{)}^2}$

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

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

$=\frac{\text{sec}A+1}{\text{sec}A-1}$

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

Hence Proved.