If $$\tan \theta + 3 \cot \theta - 2\sqrt{3} = 0, 0^\circ < \theta < 90^\circ$$, then what is the value of $$(\cosec^2 \theta + \cos^2 \theta)?$$

$$\tan\theta+3\cot\theta-2\sqrt{3}=0$$

$$\tan\theta+\frac{3}{\tan\theta\ }-2\sqrt{3}=0$$

$$tan^2\theta-2\sqrt{3}\tan\theta\ +3=0$$

$$\left(\tan\theta\ -\sqrt{3}\right)^2=0$$

$$\tan\theta\ -\sqrt{3}=0$$

$$\tan\theta\ =\sqrt{3}$$

$$0^\circ < \theta < 90^\circ$$

$$\Rightarrow$$  $$\theta\ =60^{\circ\ }$$

$$\operatorname{cosec}^2\theta+\cos^2\theta=\operatorname{cosec}^260^{\circ\ }+\cos^260^{\circ\ }$$

= $$\left(\frac{2}{\sqrt{3}}\right)^2\ +\left(\frac{1}{2}\right)^2$$

= $$\frac{4}{3}\ +\frac{1}{4}$$

= $$\frac{16+3}{12}$$

= $$\frac{19}{12}$$

Hence, the correct answer is Option B