Answer
The identity is verified.
$\frac{tan~β+cot~β}{tan~β}=csc^2β$
Work Step by Step
Use:
$\frac{1}{tan~β}=cot~β$
$1+cot^2β=csc^2β$
$\frac{tan~β+cot~β}{tan~β}=\frac{tan~β}{tan~β}+\frac{cot~β}{tan~β}=1+\frac{cot~β}{\frac{1}{cot~β}}=1+cot~β~\frac{cot~β}{1}=1+cot^2β=csc^2β$
