Lösung 4.3:8a
Aus Online Mathematik Brückenkurs 1
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Version vom 15:08, 22. Okt. 2008
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{.} |
