Lösung 4.3:8a

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We rewrite
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<math>\text{tan }v</math>
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on the left-hand side as
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<math>\frac{\sin v}{\cos v}</math>, so that
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<math>\tan ^{2}v=\frac{\sin ^{2}v}{\cos ^{2}v}</math>
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If we then use the Pythagorean identity
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<math>\cos ^{2}v+\sin ^{2}v=1</math>
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and rewrite
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<math>\text{cos}^{\text{2}}v</math>
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in the denominator as
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<math>\text{1}-\text{sin}^{\text{2}}v\text{ }</math>, we get what we are looking for on the right-hand side. The whole calculation is
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<math>\tan ^{2}v=\frac{\sin ^{2}v}{\cos ^{2}v}=\frac{\sin ^{2}v}{1-\sin ^{2}v}</math>

Version vom 10:37, 30. Sep. 2008


We rewrite \displaystyle \text{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}


If we then use the Pythagorean identity


\displaystyle \cos ^{2}v+\sin ^{2}v=1


and rewrite \displaystyle \text{cos}^{\text{2}}v in the denominator as \displaystyle \text{1}-\text{sin}^{\text{2}}v\text{ }, 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}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_4.3:8a“