Aus Online Mathematik Brückenkurs 1

We rewrite \displaystyle \tan v on the left-hand side as \displaystyle \frac{\sin v}{\cos v}, so that

\displaystyle \tan^2\!v = \frac{\sin^2\!v}{\cos^2\!v}\,\textrm{.}

If we then use the Pythagorean identity

\displaystyle \cos^2\!v + \sin^2\!v = 1

and rewrite \displaystyle \cos^2\!v in the denominator as \displaystyle 1 - \sin^2\!v, we get what we are looking for on the right-hand side. The whole calculation is

\displaystyle \tan^2\!v = \frac{\sin^2\!v}{\cos^2\!v} = \frac{\sin^2\!v}{1-\sin^2\!v}\,\textrm{.}