If secθ + tanθ = 2 + $$\sqrt 5 {\text{,}}$$  then the value of sinθ + cosθ is?

Solution (By Examveda Team)

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