Lösung 4.3:8b

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Because <math>\tan v = \frac{\sin v}{\cos v}</math>, the left-hand side can be written using <math>\cos v</math> as the common denominator,
Because <math>\tan v = \frac{\sin v}{\cos v}</math>, the left-hand side can be written using <math>\cos v</math> as the common denominator,
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{{Displayed math||<math>\frac{1}{\cos v} - \tan v = \frac{1}{\cos v} - \frac{\sin v}{\cos v} = \frac{1-\sin v}{\cos v}\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>\frac{1}{\cos v} - \tan v = \frac{1}{\cos v} - \frac{\sin v}{\cos v} = \frac{1-\sin v}{\cos v}\,\textrm{.}</math>}}
Now, we observe that if we multiply top and bottom with <math>1+\sin v</math>, the denominator will contain the denominator of the right-hand side as a factor and, in addition, the numerator can be simplified to give <math>1-\sin^2\!v = \cos ^2\!v\,</math>, using the difference of two squares,
Now, we observe that if we multiply top and bottom with <math>1+\sin v</math>, the denominator will contain the denominator of the right-hand side as a factor and, in addition, the numerator can be simplified to give <math>1-\sin^2\!v = \cos ^2\!v\,</math>, using the difference of two squares,
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{{Displayed math||<math>\begin{align}
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{{Abgesetzte Formel||<math>\begin{align}
\frac{1-\sin v}{\cos v}
\frac{1-\sin v}{\cos v}
&= \frac{1-\sin v}{\cos v}\cdot\frac{1+\sin v}{1+\sin v}\\[5pt]
&= \frac{1-\sin v}{\cos v}\cdot\frac{1+\sin v}{1+\sin v}\\[5pt]
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Eliminating <math>\cos v</math> then gives the answer,
Eliminating <math>\cos v</math> then gives the answer,
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{{Displayed math||<math>\frac{\cos^2\!v}{\cos v\,(1+\sin v)} = \frac{\cos v}{1+\sin v}\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>\frac{\cos^2\!v}{\cos v\,(1+\sin v)} = \frac{\cos v}{1+\sin v}\,\textrm{.}</math>}}

Version vom 08:57, 22. Okt. 2008

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{1-\sin v}{\cos v}\,\textrm{.}

Now, we observe that if we multiply top and bottom with \displaystyle 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 1-\sin^2\!v = \cos ^2\!v\,, using the difference of two squares,

\displaystyle \begin{align}

\frac{1-\sin v}{\cos v} &= \frac{1-\sin v}{\cos v}\cdot\frac{1+\sin v}{1+\sin v}\\[5pt] &= \frac{1-\sin^2\!v}{\cos v\,(1+\sin v)}\\[5pt] &= \frac{\cos^2\!v}{\cos v\,(1+\sin v)}\,\textrm{.} \end{align}

Eliminating \displaystyle \cos v then gives the answer,

\displaystyle \frac{\cos^2\!v}{\cos v\,(1+\sin v)} = \frac{\cos v}{1+\sin v}\,\textrm{.}
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_4.3:8b“