Problem of the week #33 - November 12th, 2012

[Jameson]

Prove that \(\displaystyle \tan \left( \alpha + \beta \right)=\frac{\tan(\alpha)+\tan(\beta)}{1-\tan(\alpha)\tan(\beta)}\)
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Remember to read the POTW submission guidelines to find out how to submit your answers!

 

[Jameson]

Congratulations to the following members for their correct solutions:

1) BAdhi
2) Sudharaka

Solution (from BAdhi):

Spoiler

$$
\begin{align*}
\tan(\alpha +\beta)&= \frac{\sin(\alpha+\beta)}{\cos(\alpha+\beta)}\\
&=\frac{\sin \alpha \cos \beta + \cos \alpha \sin \beta}{\cos \alpha \cos \beta- \sin \alpha \sin \beta}\\
\end{align*}$$

by dividing the denominator and divisor by $\cos\alpha \cos \beta$

$$\begin{align*}
\tan (\alpha +\beta) &= \frac{\frac{\sin \alpha}{\cos \alpha}+\frac{\sin \beta}{\cos \beta}}{1- \frac{\sin \alpha \sin \beta}{\cos \alpha \cos \beta}}\\
&= \frac{\tan \alpha + \tan \beta}{1- \tan \alpha \tan \beta}
\end{align*}$$