Answer
$\csc \theta \sec \theta$
Work Step by Step
We know that $1+\tan^2 x=\sec^2 x $
$\dfrac{\cos \theta}{\sin \theta}+\dfrac{\sin \theta}{\cos \theta}=\dfrac{\cos^2 \theta+\sin^2 \theta}{\sin \theta \cos \theta}$
or, $=\dfrac{1}{\sin \theta \cos \theta}$
or, $=\csc \theta \sec \theta$
