Aus Online Mathematik Brückenkurs 1

We rewrite \displaystyle \text{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}

If we then use the Pythagorean identity

\displaystyle \cos ^{2}v+\sin ^{2}v=1

and rewrite \displaystyle \text{cos}^{\text{2}}v in the denominator as \displaystyle \text{1}-\text{sin}^{\text{2}}v\text{ }, 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}