If ^2 ^2 - ^2 = 3, 0° < θ < 90°, then the value of cotθ + cosecθ is:

Answer: Option A

$$\begin{array}{rl}& \frac{{\cos }^2\theta }{{\cot }^2\theta -{\cos }^2\theta }=3\\ & \frac{{\cos }^2\theta }{{\cos }^2\theta \left(\frac{1-{\sin }^2\theta }{{\sin }^2\theta }\right)}=3\\ & \frac{{\sin }^2\theta }{{\cos }^2\theta }=3\\ & \tan \theta =\sqrt{3}\\ & \theta =60^{\circ }\\ & \cot \theta +\text{cosec}\theta \\ & =\cot 60^{\circ }+\text{cosec}60^{\circ }\\ & =\frac{1}{\sqrt{3}}+\frac{2}{\sqrt{3}}\\ & =\frac{3}{\sqrt{3}}\\ & =\sqrt{3}\end{array}$$