Answer
$\dfrac{26+6 \sqrt {13}}{39}$
Work Step by Step
It can be seen from the given figure that
$\tan \theta=\dfrac{2}{3}$
Thus, $\sin \theta= \dfrac{2}{\sqrt{(2)^2+(3)^2}}=\dfrac{2 \sqrt {13}}{13}$
Now, $\tan \theta+\sin \theta=\dfrac{2}{3}+\dfrac{2 \sqrt {13}}{13}$
This gives:
$\tan \theta+\sin \theta=\dfrac{26+6 \sqrt {13}}{39}$
