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- Jan 26, 2012

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Remember to read the POTW submission guidelines to find out how to submit your answers!

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- Thread starter
- Admin
- #1

- Jan 26, 2012

- 4,093

--------------------

Remember to read the POTW submission guidelines to find out how to submit your answers!

- Thread starter
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- #2

- Jan 26, 2012

- 4,093

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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