\(\sqrt{\frac{1\;+ \;sin\;A}{1\;- \;sin\;A}}\) = sec A + tan A

L.H.S = \(\sqrt{\frac{1\;+ \;sin\;A}{1\;- \;sin\;A}}\)

First divide the numerator and denominator of L.H.S. by cos A,

\(\sqrt{\frac{\frac{1}{cos A}+ \frac{sin A}{cos A}}{\frac{1}{cos A}- \frac{sin A}{cos A}}}\)

We know that 1/cos A = sec A and sin A/ cos A = tan A and it becomes,

= \(\sqrt{\frac{secA + tan A}{sec A - tan A}}\)

Now using rationalization, we get

= \(\sqrt{\frac{secA + tan A}{sec A - tan A}}\times \)\(\sqrt{\frac{secA + tan A}{sec A + tan A}}\)

= \(\sqrt{\frac{(secA + tan A)^2}{sec^2 A - tan^2 A}}\)

\(\frac{secA + tan A}{1}\)

= sec A + tan A = R.H.S

Hence proved