If secA + tanA = a, then the value of cosA is?

$$\eqalign{ & {\text{secA}} + {\text{tanA}} = a \cr & {\text{we know that}} \cr & \Rightarrow {\text{se}}{{\text{c}}^2}{\text{A}} - {\text{ta}}{{\text{n}}^2}{\text{A}} = 1 \cr & \Rightarrow \left( {{\text{secA}} - {\text{tanA}}} \right)\left( {{\text{secA}} + {\text{tanA}}} \right) = 1 \cr & \Rightarrow {\text{secA}} - {\text{tanA}} = \frac{1}{a} \cr & \Rightarrow {\text{secA}} + {\text{tanA}} = a \cr & \Rightarrow {\text{2secA}} = a + \frac{1}{a} \cr & \Rightarrow {\text{2secA}} = \frac{{{a^2} + 1}}{a} \cr & \Rightarrow \sec \theta = \frac{{{a^2} + 1}}{{2a}} \cr & {\text{So, }}\cos \theta = \frac{{2a}}{{{a^2} + 1}} \cr} $$