If sec²θ + tan²θ = $$3\frac{1}{2},$$  0°

If sec²θ + tan²θ = $$3\frac{1}{2},$$  0° Correct Answer $$\frac{{2 + \sqrt 5 }}{3}$$

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