Thread: help with angle and figure
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Old June 27th, 2009, 10:37 PM
yeongil yeongil is offline
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This should have been posted in the Trigonometry subforum.

\begin{aligned}\tan(45^{\circ} + \alpha) &= p \\\frac{\tan 45^{\circ} + \tan \alpha}{1 - \tan 45^{\circ}\tan \alpha} &= p \\\frac{1 + \tan \alpha}{1 - \tan \alpha} &= p \\1 + \tan \alpha &= p(1 - \tan \alpha) \\1 + \tan \alpha &= p - (\tan \alpha)p \\\end{aligned}
\begin{aligned}(\tan \alpha)p + \tan \alpha&= p - 1 \\(\tan \alpha)(p + 1) &= p - 1 \\\tan \alpha &= \frac{p - 1}{p + 1}\end{aligned}

What you have scribbled on the top is actually correct, just not simplified:
\frac{1 - p}{-1 - p} = \frac{-(p - 1)}{-(p + 1)} = \frac{p - 1}{p + 1}

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