Aus Online Mathematik Brückenkurs 1

Because \displaystyle \tan v=\frac{\sin v}{\cos v}, the left-hand side can be written using \displaystyle \cos v as the common denominator:

\displaystyle \frac{1}{\cos v}-\tan v=\frac{1}{\cos v}-\frac{\sin v}{\cos v}=\frac{\text{1-}\sin v}{\cos v}

Now, we observe that if we multiply top and bottom by with \displaystyle \text{1}+\sin v, the denominator will contain the denominator of the right-hand side as a factor and, in addition, the numerator can be simplified to give \displaystyle \text{1}-\sin ^{2}v\text{ }=\cos ^{2}v, using the conjugate rule:

\displaystyle \begin{align} & \frac{\text{1-}\sin v}{\cos v}=\frac{\text{1-}\sin v}{\cos v}\centerdot \frac{1+\sin v}{1+\sin v}=\frac{1-\sin ^{2}v}{\cos v\left( 1+\sin v \right)} \\ & =\frac{\cos ^{2}v}{\cos v\left( 1+\sin v \right)}. \\ \end{align}

Eliminating \displaystyle \cos v then gives the answer:

\displaystyle \frac{\cos ^{2}v}{\cos v\left( 1+\sin v \right)}=\frac{\cos v}{1+\sin v}