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Let $\cot^{-1} {7}$ , $\cot^{-1} {18}$ and $\cot^{-1} {8}$ be $\alpha$ , $\beta$ and $\gamma$ , then we have:

$\theta = \alpha + \beta + \gamma$ , $\quad$ and $\quad \tan {\alpha} = \frac {1}{7}$ , $\quad \tan {\beta} = \frac {1}{18}\quad$ and $\quad \tan {\gamma} = \frac {1}{8}$ .

And, $\quad \tan {\theta} = \tan {\alpha+\beta + \gamma} = \dfrac {\tan {(\alpha + \beta) } + \tan {\gamma } } {1 - \tan {(\alpha + \beta) } \tan {\gamma } }$

$\quad \quad = \dfrac {\frac {\tan{\alpha} + \tan {\beta} } {1- \tan {\alpha} \tan { \beta} } + \tan {\gamma} } {1 - \frac {\tan{\alpha} + \tan {\beta} } {1- \tan {\alpha} \tan { \beta} } \times \tan {\gamma } } = \dfrac {\frac {\frac {1} {7} + \frac {1}{18} } {1- \frac {1} {7} \times \frac {1} {18} } + \frac {1}{8} } {1 - \frac {\frac {1} {7} + \frac {1}{18} } {1- \frac {1} {7} \times \frac {1} {8} } \times \frac {1}{8} }$

$\quad \quad = \dfrac {\frac {25} {125} + \frac {1}{8} } {1 - \frac {25}{125} \times \frac {1}{8} } = \dfrac {\frac {1} {5} + \frac {1}{8} } {1 - \frac {1}{5} \times \frac {1}{8} } = \dfrac {13} {39} = \boxed {3}$