If secθ + tanθ = 2 + 5 , then the value of sinθ + cosθ is?

Answer: Option A

$$\begin{array}{rl}& sec\theta +\text{tan}\theta =2+\sqrt{5}.....(i)\\ & \text{se}{\text{c}}^2\theta -\text{ta}{\text{n}}^2\theta =1\\ & \left(sec\theta -\text{tan}\theta \right)\left(sec\theta +\text{tan}\theta \right)=1\\ & \left(sec\theta -\text{tan}\theta \right)=\frac{1}{2+\sqrt{5}}=\frac{1}{\sqrt{5}+2}=\sqrt{5}-2.....(ii)\\ & \text{Adding equation }\text{ (i) and (ii)}\\ & \text{2}sec\theta =2+\sqrt{5}+\sqrt{5}-2\\ & \Rightarrow 2sec\theta =2\sqrt{5}\\ & \Rightarrow sec\theta =\sqrt{5}\\ & \Rightarrow \text{cos}\theta =\frac{1}{\sqrt{5}}\\ & \Rightarrow {\sin }^2\theta +\text{co}{\text{s}}^2\theta =1\\ & \Rightarrow {\sin }^2\theta =1-{\left(\frac{1}{\sqrt{5}}\right)}^2\\ & \Rightarrow {\sin }^2\theta =\frac{4}{5}\\ & \Rightarrow \sin \theta =\frac{2}{\sqrt{5}}\\ & \therefore \sin \theta +\text{cos}\theta =\frac{2}{\sqrt{5}}+\frac{1}{\sqrt{5}}\\ & =\frac{3}{\sqrt{5}}\end{array}$$