If $$0^\circ < \theta < 90^\circ, \sqrt{\frac{\sec^2 \theta + \cosec^2 \theta}{\tan^2 \theta - \sin^2 \theta}}$$ is equal to:

$$\sqrt{\frac{\sec^2\theta+\operatorname{cosec}^2\theta}{\tan^2\theta-\sin^2\theta}}=\sqrt{\frac{\frac{1}{\cos^2\theta}+\frac{1}{\sin^2\theta}}{\frac{\sin^2\theta\ }{\cos^2\theta}-\sin^2\theta}}$$

$$=\sqrt{\frac{\frac{\sin^2\theta+\cos^2\theta\ }{\cos^2\theta\sin^2\theta}}{\frac{\sin^2\theta-\sin^2\theta\cos^2\theta\ }{\cos^2\theta}}}$$

$$=\sqrt{\frac{1\ }{\cos^2\theta\sin^2\theta}\times\frac{\cos^2\theta\ }{\sin^2\theta\left(1-\cos^2\theta\right)\ }}$$

$$=\sqrt{\frac{1\ }{\sin^2\theta}\times\frac{1}{\sin^2\theta\left(\sin^2\theta\right)\ }}$$

$$=\sqrt{\frac{1\ }{\sin^6\theta}\ }$$

$$=\sqrt{\operatorname{cosec}^6\theta\ }$$

$$=\operatorname{cosec}^3\theta\ $$

Hence, the correct answer is Option B