If 3sinθ = 2cos2θ, 0° < θ < 90°, then the value of (tan2θ + sec2θ - cosec2θ) is:

Answer: Option B

$$\begin{array}{rl}& 3\sin \theta =2{\cos }^2\theta \\ & \text{Let }\theta =30^{\circ }\\ & 3\times \frac{1}{2}=2\times \frac{{\left(\sqrt{3}\right)}^2}{4}\\ & \frac{3}{2}=\frac{3}{2}\\ & {\tan }^2\theta +{\sec }^2\theta -\text{cose}{\text{c}}^2\theta \\ & ={\tan }^{2}30^{\circ }+{\sec }^{2}30^{\circ }-\text{cose}{\text{c}}^{2}30^{\circ }\\ & =\frac{1}{3}+\frac{4}{3}-4\\ & =\frac{5}{3}-4\\ & =-\frac{7}{3}\end{array}$$