Solution 4.3:8b
From Förberedande kurs i matematik 1
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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, |
| - | < | + | |
| - | {{ | + | {{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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| + | 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} | ||
| + | \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}</math>}} | ||
| + | |||
| + | Eliminating <math>\cos v</math> then gives the answer, | ||
| + | |||
| + | {{Displayed math||<math>\frac{\cos^2\!v}{\cos v\,(1+\sin v)} = \frac{\cos v}{1+\sin v}\,\textrm{.}</math>}} | ||
Current revision
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{.} |
