Solution 4.3:8a
From Förberedande kurs i matematik 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{.} |