Solution 4.3:8b

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Current revision (08:27, 10 October 2008) (edit) (undo)
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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,
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Because
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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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<math>\tan v=\frac{\sin v}{\cos v}</math>, the left-hand side can be written using
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<math>\cos v</math>
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as the common denominator:
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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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<math>\frac{1}{\cos v}-\tan v=\frac{1}{\cos v}-\frac{\sin v}{\cos v}=\frac{\text{1-}\sin v}{\cos v}</math>
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{{Displayed math||<math>\begin{align}
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\frac{1-\sin v}{\cos v}
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&= \frac{1-\sin v}{\cos v}\cdot\frac{1+\sin v}{1+\sin v}\\[5pt]
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&= \frac{1-\sin^2\!v}{\cos v\,(1+\sin v)}\\[5pt]
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&= \frac{\cos^2\!v}{\cos v\,(1+\sin v)}\,\textrm{.}
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\end{align}</math>}}
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Eliminating <math>\cos v</math> then gives the answer,
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Now, we observe that if we multiply top and bottom by with
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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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<math>\text{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
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<math>\text{1}-\sin ^{2}v\text{ }=\cos ^{2}v</math>, using the conjugate rule:
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<math>\begin{align}
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& \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)} \\
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& =\frac{\cos ^{2}v}{\cos v\left( 1+\sin v \right)}. \\
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\end{align}</math>
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Eliminating
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<math>\cos v</math>
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then gives the answer:
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<math>\frac{\cos ^{2}v}{\cos v\left( 1+\sin v \right)}=\frac{\cos v}{1+\sin v}</math>
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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{.}
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