Prove that $\displaystyle [1 + \frac{1}{tan^2\theta}] + [1 + \frac{1}{cot^2\theta}] = \frac{1}{sin^2\theta - sin^4\theta}$

$= \frac{\cos\theta}{\sin\theta} $, we have:

$ \frac{1}{\tan^2\theta} = \frac{\cos^2\theta}{\sin^2\theta}, \quad \frac{1}{\cot^2\theta} $

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

Substituting these values into the given expression:

$ \left[ 1 + \frac{\cos^2\theta}{\sin^2\theta} \right] + \left[ 1 + \frac{\sin^2\theta}{\cos^2\theta} \right] $

Rewriting each term:

$ \left[ \frac{\sin^2\theta + \cos^2\theta}{\sin^2\theta} \right] + \left[ \frac{\sin^2\theta + \cos^2\theta}{\cos^2\theta} \right] $

Since $ \sin^2\theta + \cos^2\theta = 1 $, we simplify:

$ \left[ \frac{1}{\sin^2\theta} \right] + \left[ \frac{1}{\cos^2\theta} \right] $

Taking the LCM:

$ \frac{\cos^2\theta + \sin^2\theta}{\sin^2\theta \cos^2\theta} $

Again, using $ \sin^2\theta + \cos^2\theta = 1 $:

$ \frac{1}{\sin^2\theta \cos^2\theta} $

Rewriting $ \cos^2\theta $ as $ 1 - \sin^2\theta $:

$ \frac{1}{\sin^2\theta (1 - \sin^2\theta)} $

Thus, we obtain:

$ \frac{1}{\sin^2\theta - \sin^4\theta} $

\textbf{Hence, proved.}