Question

please answer ............

Answer 1

$given \: \: \sec A - \tan A = \frac{1}{2} - - - (1) \\ we \: \: know \\ { \sec }^{2} A - { \tan }^{2} A = 1 \\ i.e. \: ( \sec A + \tan A)( \sec A - \tan A) = 1 \\ \sec A + \tan A = \frac{1}{ \sec A - \tan A} \\ \sec A + \tan A = 2 - - - (2) \\ (1) + (2) \: \: gives \\ 2 \sec A = \frac{1}{2} + 2 = \frac{5}{2} \\ so \: \: \sec A = \frac{5}{4}$