\textbf{To Prove:}
$ \left[1 + \frac{1}{\tan^2\theta} \right] + \left[1 + \frac{1}{\cot^2\theta} \right] = \frac{1}{\sin^2\theta - \sin^4\theta} $
\textbf{Proof:}
We start by simplifying each term inside the brackets:
Since $ \tan\theta = \frac{\sin\theta}{\cos\theta} $ and $ \cot\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.}