The value of $2\sin^{2}\theta \cdot \cos^{2}\theta (\sec^{2}\theta + \csc^{2}\theta)$ is …

$\sec^{2}\theta = 1 + \tan^{2}\theta$ and $\csc^{2}\theta = 1 + \cot^{2}\theta$. So, $\sec^{2}\theta + \csc^{2}\theta = 2 + \tan^{2}\theta + \cot^{2}\theta$. Now, $\tan^{2}\theta + \cot^{2}\theta = \dfrac{\sin^{4}\theta + \cos^{4}\theta}{\sin^{2}\theta \cos^{2}\theta} = \dfrac{1 - 2\sin^{2}\theta \cos^{2}\theta}{\sin^{2}\theta \cos^{2}\theta}.$ Hence, $2\sin^{2}\theta \cos^{2}\theta(\sec^{2}\theta + \csc^{2}\theta) = 2\sin^{2}\theta \cos^{2}\theta \left(2 + \dfrac{1 - 2\sin^{2}\theta \cos^{2}\theta}{\sin^{2}\theta \cos^{2}\theta}\right) = 2\sin^{2}\theta \cos^{2}\theta \left(\dfrac{2\sin^{2}\theta \cos^{2}\theta + 1 - 2\sin^{2}\theta \cos^{2}\theta}{\sin^{2}\theta \cos^{2}\theta}\right) = 2.$