If 4 - 2sin2θ - 5cosθ = 0, 0° < θ < 90°, then the value of sinθ + tanθ is:

Answer: Option B

$$\begin{array}{rl}& 4-2{\sin }^2\theta -5\cos \theta =0\\ & \text{Let }\theta =60^{\circ }\\ & 4-2{\sin }^{2}60^{\circ }-5\cos 60^{\circ }=0\\ & 4-2\times \frac{3}{4}-5\times \frac{1}{2}=0\\ & 4-4=0\\ & \sin \theta +\tan \theta =\sin 60^{\circ }+\tan 60^{\circ }\\ & =\frac{\sqrt{3}}{2}+\sqrt{3}\\ & =\frac{3\sqrt{3}}{2}\end{array}$$