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# Three by One

Here is a hint to help on you on the way :

Let $\alpha + \beta = \gamma$ and $\tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}$

where $\alpha = \tan^{-1}{1\over3}$, $\beta = \tan^{-1}{1\over2}$, $\gamma = \tan^{-1}1$.

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Age 16 to 18

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Here is a hint to help on you on the way :

Let $\alpha + \beta = \gamma$ and $\tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}$

where $\alpha = \tan^{-1}{1\over3}$, $\beta = \tan^{-1}{1\over2}$, $\gamma = \tan^{-1}1$.