pls help me solve part(a). thank you.
This should have been posted in the Trigonometry subforum.
$\displaystyle \begin{aligned}
\tan(45^{\circ} + \alpha) &= p \\
\frac{\tan 45^{\circ} + \tan \alpha}{1 - \tan 45^{\circ}\tan \alpha} &= p \\
\frac{1 + \tan \alpha}{1 - \tan \alpha} &= p \\
1 + \tan \alpha &= p(1 - \tan \alpha) \\
1 + \tan \alpha &= p - (\tan \alpha)p \\
\end{aligned}$
$\displaystyle \begin{aligned}
(\tan \alpha)p + \tan \alpha&= p - 1 \\
(\tan \alpha)(p + 1) &= p - 1 \\
\tan \alpha &= \frac{p - 1}{p + 1}
\end{aligned}$
What you have scribbled on the top is actually correct, just not simplified:
$\displaystyle \frac{1 - p}{-1 - p} = \frac{-(p - 1)}{-(p + 1)} = \frac{p - 1}{p + 1}$
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