Answer
$$ y'= - \frac{-\csc \theta \cot \theta}{2\sqrt{1+\csc \theta }} \sec^2\sqrt{1+\csc \theta }.$$
Work Step by Step
Recall that $(\tan x)'=\sec^2 x$.
Recall that $(\csc x)'=-\csc x \cot x$.
Since $ y=\tan\sqrt{1+\csc \theta }$, then by the chain rule, the derivative $ y'$ is given by
$$ y'= \sec^2(\sqrt{1+\csc \theta } ) \left(\frac{-\csc \theta \cot \theta}{2\sqrt{1+\csc \theta }}\right)\\=- \frac{-\csc \theta \cot \theta}{2\sqrt{1+\csc \theta }} \sec^2\sqrt{1+\csc \theta }.$$
