If $$\sin^2 \theta - \cos^2 \theta - 3 \sin \theta + 2 = 0, 0^\circ < \theta < 90^\circ$$, then what is the value of $$\frac{1}{\sqrt{\sec \theta - \tan \theta}}$$ is:

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

$$\sin^2\theta-\left(1-\sin^2\theta\ \right)-3\sin\theta+2=0$$

$$2\sin^2\theta-3\sin\theta+1=0$$

$$2\sin^2\theta-2\sin\theta-\sin\theta\ +1=0$$

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

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

$$\sin\theta\ -1=0$$  or  $$2\sin\theta\ -1=0$$

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

$$\theta\ =90^{\circ\ }$$  or  $$\theta\ =30^{\circ\ }$$

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

$$\Rightarrow$$  $$\theta\ =30^{\circ\ }$$

$$\frac{1}{\sqrt{\sec\theta-\tan\theta}}=\frac{1}{\sqrt{\sec30^{\circ\ }-\tan30^{\circ\ }}}$$

$$=\frac{1}{\sqrt{\frac{2}{\sqrt{3}}\ -\frac{1}{\sqrt{3}}}}$$

$$=\frac{1}{\sqrt{\frac{1}{\sqrt{3}}}}$$

$$=\sqrt[\ 4]{3}$$

Hence, the correct answer is Option A