Answer
$\dfrac{4-3\sqrt 2}{4}$
Work Step by Step
Apply the trigonometric identities:
$\sin^2 \theta+\cos^2 \theta=1 \implies \sin \theta=-\sqrt{1-\cos^2\theta}$
$\sin \theta=-\sqrt{1-({\dfrac{1}{3})^2}}=-\sqrt{1-{\dfrac{1}{9}}}=-\dfrac{2\sqrt 2}{3}$
Now, $\tan \theta\cot \theta+\csc \theta=(\dfrac{1}{\cot \theta})\times \cot \theta+(\dfrac{1}{\sin\theta})$
This gives:
$\tan \theta\cot \theta+\csc \theta=(\dfrac{1}{\cot \theta})\times \cot \theta+(-\dfrac{1}{\sin\theta})=1-\dfrac{3\sqrt 2}{4}$
or, $\tan \theta\cot \theta+\csc \theta=(\dfrac{1}{\cot \theta})\times \cot \theta+(-\dfrac{1}{\sin\theta})=\dfrac{4-3\sqrt 2}{4}$
