Tangent

Geometry Level 2

$\large \tan^{-1} \frac {1}{3} - \tan^{-1} \frac{1}{5} + \tan^{-1} \frac{1}{7} = \ ?$

\text{tan}^{-1} \dfrac{5}{7}

\text{tan}^{-1} \dfrac{3}{11}

\text{tan}^{-1} \dfrac{2}{4}

\text{tan}^{-1} \dfrac{4}{9}

1 solution

Chew-Seong Cheong
Sep 9, 2017

$\large \begin{aligned} \theta & = \tan^{-1} \frac 13 - \tan^{-1} \frac 15 + \tan^{-1} \frac 17 \\ \tan \theta & = \tan \left(\tan^{-1} \frac 13 - \tan^{-1} \frac 15 + \tan^{-1} \frac 17 \right) \\ & = \frac {\frac {\frac 13 - \frac 15}{1+\frac 13 \times \frac 15} + \frac 17}{1-\frac {\frac 13 - \frac 15}{1+\frac 13 \times \frac 15} \times \frac 17} \\ & = \frac {\frac 18 + \frac 17}{1-\frac 18 \times \frac 17} = \frac {15}{55} = \frac 3{11} \\ \implies \theta & = \boxed{\tan^{-1} \dfrac 3{11}} \end{aligned}$

Tangent

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