If sec2θ + tan2θ = 31/2, 0° < θ < 90°, then (cosθ + sinθ) is equal to

Answer: Option B

$$\begin{array}{rl}& {\sec }^2\theta +{\tan }^2\theta =3\frac{1}{2}\\ & \Rightarrow 1+{\tan }^2\theta +{\tan }^2\theta =\frac{7}{2}\\ & \Rightarrow 2{\tan }^2\theta =\frac{7}{2}-1\\ & \Rightarrow 2{\tan }^2\theta =\frac{5}{2}\\ & \Rightarrow 3{\tan }^2\theta =\frac{5}{4}\\ & \Rightarrow \tan \theta =\frac{\sqrt{5}\to P}{2\to B}\\ & H=\sqrt{5+4}=3\\ & \therefore \cos \theta +\sin \theta \\ & =\frac{2}{3}+\frac{\sqrt{5}}{3}\\ & =\frac{2+\sqrt{5}}{3}\end{array}$$