If secα + tanα = 2, then the value of sinα is (assume that 0 < α < 90°)

$$\eqalign{ & sec\alpha + \tan \alpha = 2\,.....(i) \cr & sec\alpha - \tan \alpha = \frac{1}{2}\,.....(ii) \cr & {\text{Adding equation (i) and (ii)}} \cr & {\text{2sec}}\alpha = 2 + \frac{1}{2} \cr & \sec \alpha = \frac{{5 \to {\text{H}}}}{{4 \to {\text{B}}}} \cr} $$

$$\sin \alpha = \frac{3}{5} = 0.6$$