If sinα .sec(30° + α) = 1, (0 < α < 60°), then the value of sinα + cos2α is?

Answer: Option A

$$\begin{array}{rl}& \text{sin }\alpha .\sec \left({30}^{\circ }+\alpha \right)=1\\ & \mathbf{Shortcut}\mathbf{method}:\\ & \text{Put }\alpha \text{ value between }{\text{0}}^{\circ }\text{ to 6}{\text{0}}^{\circ }\\ & \text{If }\alpha =30^{\circ }\\ & \Rightarrow \text{sin 3}{0}^{\circ }.\sec \left({30}^{\circ }+{30}^{\circ }\right)=1\\ & \Rightarrow \text{sin 3}{0}^{\circ }.\sec 60^{\circ }=1\\ & \Rightarrow \frac{1}{2}\times 2\\ & \Rightarrow 1\\ & 1=1\left(\text{Satisfy}\right)\\ & \text{So},\alpha =30^{\circ }\\ & =\sin \alpha +\cos 2\alpha \\ & =\sin 30^{\circ }+\cos 2\times 30^{\circ }\\ & =\sin 30^{\circ }+\cos 60^{\circ }\\ & =\frac{1}{2}+\frac{1}{2}\\ & =1\end{array}$$