Prove that $\displaystyle \frac{1+\sec \mathrm{A}}{\sec \mathrm{A}}=\frac{\sin ^{2} \mathrm{~A}}{1-\cos \mathrm{A}}$.

\[\begin{aligned} \mathrm{LHS}=\frac{1+\sec \mathrm{A}}{\sec A} & =\frac{1+\dfrac{1}{\cos \mathrm{~A}}}{\dfrac{1}{\cos A}} \\ & =1+\cos \mathrm{A} \\ & =\frac{(1-\cos \mathrm{A})(1+\cos \mathrm{A})}{(1-\cos \mathrm{A})} \\ & =\frac{1-\cos ^{2} A}{1-\cos A} \\ & =\frac{\sin ^{2} A}{1-\cos A}=\mathrm{RHS} \end{aligned} \]