If $$\sin \theta \sec^2 \theta = \frac{2}{3}, 0^\circ < \theta < 90^\circ$$, then the value of $$(\tan^2 \theta + \cos^2 \theta)$$ is:

Given,

$$\sin \theta \sec^2 \theta = \frac{2}{3}$$,  $$0^\circ < \theta < 90^\circ$$

$$=$$>  $$\frac{\sin\theta}{\cos^2\theta\ }=\frac{2}{3}$$

$$=$$>  $$\frac{\sin\theta}{1-\sin^2\theta\ \ }=\frac{2}{3}$$

$$=$$>  $$3\sin\theta\ \ =2-2\sin^2\theta\ $$

$$=$$>  $$2\sin^2\theta+3\sin\theta\ -2=0\ $$

$$=$$>  $$2\sin^2\theta+4\sin\theta-\sin\theta -2=0\ $$

$$=$$>  $$2\sin\theta\left(\sin\theta\ +2\right)-1\left(\sin\theta+2\right)=0\ $$

$$=$$>  $$\left(\sin\theta\ +2\right)\left(2\sin\theta\ -1\right)=0\ $$

$$=$$>  $$\sin\theta\ =-2$$  or $$\sin\theta\ =\frac{1}{2}$$

$$=$$>  $$\sin\theta\ =\frac{1}{2}$$

$$=$$>  $$\cos\theta\ =\frac{\sqrt{3}}{2}$$  and  $$\tan\theta\ =\frac{1}{\sqrt{3}}$$

$$\therefore\ \tan^2\theta\ +\cos^2\theta\ =\left(\frac{1}{\sqrt{3}}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2=\frac{1}{3}+\frac{3}{4}=\frac{13}{12}$$

Hence, the correct answer is Option A