Answer
The identity is verified.
See the graphs of $f(θ)=\frac{tan^2θ}{sec~θ}$ (Black) and $g(θ)=sin~θ~tan~θ$ (Red)
Work Step by Step
$\frac{tan^2θ}{sec~θ}=tan^2θ·\frac{1}{sec~θ}=\frac{sin^2θ}{cos^2θ}·cos~θ=\frac{sin^2θ}{cos~θ}=sin~θ·\frac{sin~θ}{cos^2θ}=sin~θ~tan~θ$
