In a ΔABC, right-angled at B, AB = 5 cm and (AC - BC) = 1 cm.  The value \(\sec {C} + \cot {A}\) of is?

\(\frac{3}{2}\) \(\frac{2}{3}\) \(\frac{3}{4}\) \(\frac{4}{5}\)

\(\frac{3}{2}\)

AC - BC = 1 triplets = 5, 12, 13 Sec C + Cot A = \(\frac{13}{12}\) + \(\frac{5}{12}\) = \(\frac{18}{12}\) = \(\frac{3}{2}\)