If $$\cos x = \frac{-1}{2}  and  \pi < x < \frac{3\pi}{2}$$, then the value of $$2 \tan^2 x + 3 \cosec^2 x$$ is:

Given,  $$\cos x = \frac{-1}{2}$$

$$=$$>  $$\sec x=-2$$

$$\therefore\ $$ $$2\tan^2x+3\operatorname{cosec}^2x=2\left(\sec^2x-1\right)+\frac{3}{\sin^2x}$$

$$=2\left(\sec^2x-1\right)+\frac{3}{1-\cos^2x}$$

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

$$=2\left(4-1\right)+\frac{3}{1-\frac{1}{4}}$$

$$=2\left(3\right)+\frac{3}{\frac{3}{4}}$$

$$=6+4$$

$$=10$$

Hence, the correct answer is Option A