Correct Answer: $\tan A$

Solution : Given: $\frac{\left(1+\tan ^2 A\right)}{\operatorname{cosec}^2 A \cdot \tan A}$
$=\frac{\left(\sec^2 A\right)}{\operatorname{\frac{1}{\sin^2A}} \cdot \frac{\sin A}{\cos A}}$
$=\frac{\left(\frac{1}{\cos^2 A}\right)}{\operatorname{\frac{1}{\sin A \cdot \cos A }}}$
$=\frac{\sin A \cdot \cos A }{\cos^2 A}$
$=\frac{\sin A}{\cos A}$
$=\tan A$
Hence, the correct answer is $\tan A$.