Solution 4.3:8a
From Förberedande kurs i matematik 1
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- | {{ | + | We rewrite <math>\tan v</math> on the left-hand side as <math>\frac{\sin v}{\cos v}</math>, so that |
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- | {{ | + | {{Displayed math||<math>\tan^2\!v = \frac{\sin^2\!v}{\cos^2\!v}\,\textrm{.}</math>}} |
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+ | If we then use the Pythagorean identity | ||
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+ | {{Displayed math||<math>\cos^2\!v + \sin^2\!v = 1</math>}} | ||
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+ | and rewrite <math>\cos^2\!v</math> in the denominator as <math>1 - \sin^2\!v</math>, we get what we are looking for on the right-hand side. The whole calculation is | ||
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+ | {{Displayed math||<math>\tan^2\!v = \frac{\sin^2\!v}{\cos^2\!v} = \frac{\sin^2\!v}{1-\sin^2\!v}\,\textrm{.}</math>}} |
Current revision
We rewrite \displaystyle \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}\,\textrm{.} |
If we then use the Pythagorean identity
\displaystyle \cos^2\!v + \sin^2\!v = 1 |
and rewrite \displaystyle \cos^2\!v in the denominator as \displaystyle 1 - \sin^2\!v, 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}\,\textrm{.} |