The first thing that comes to my mind is I want everything in terms of sin's and cos's if I can, so let's multiply through by cosx so that we can at least get rid of that secx=1cosx term.

\(\displaystyle{\left(\frac{{{\tan}^{{2}}{x}}}{{\sec{{x}}}}\right)}{\cos{{x}}}={1}-{{\cos}^{{2}}{x}}\)

Now that 1−cos^2x is popping out at me because I remember the Pythagorean identity \(\displaystyle{{\sin}^{{2}}{x}}+{{\cos}^{{2}}{x}}={1}\)

Using this we see that \(\displaystyle{1}−{{\cos}^{{2}}{x}}={{\sin}^{{2}}{x}}\) and so what we're trying to prove again simplifies to \(\displaystyle{\left(\frac{{{\tan}^{{2}}{x}}}{{\sec{{x}}}}\right)}{\cos{{x}}}={{\sin}^{{2}}{x}}\)

Now let's work on the LHS and see if we can show that it is equal to \(\displaystyle{{\sin}^{{2}}{x}}\). Firstly, let's unpack the definitions of tan and sec.

\((tan^2x/secx)cosx=(sin^2x/cos^2x/1/cosx)*cosx =sin^2x\)

which is exactly the modified RHS of the equation. Thus the identity is proven.