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

Answer: Option B

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