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

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#### Jameson

##### Administrator
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Prove that $$\displaystyle \tan \left( \alpha + \beta \right)=\frac{\tan(\alpha)+\tan(\beta)}{1-\tan(\alpha)\tan(\beta)}$$
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#### Jameson

##### Administrator
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Congratulations to the following members for their correct solutions:

1) BAdhi
2) Sudharaka

Solution (from BAdhi):

\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*}

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