# How do you prove tan^-1(1/3)+tan^-1(1/5)=tan^-1(4/7)?

Oct 20, 2016

See explanation...

#### Explanation:

For any $\alpha$ and $\beta$ we have:

$\tan \left(\alpha + \beta\right) = \frac{\tan \left(\alpha\right) + \tan \left(\beta\right)}{1 - \tan \left(\alpha\right) \tan \left(\beta\right)}$

Let $\alpha = {\tan}^{- 1} \left(\frac{1}{3}\right)$ and $\beta = {\tan}^{- 1} \left(\frac{1}{5}\right)$

Note that $\alpha \in \left[0 , 1\right)$ and $\beta \in \left[0 , 1\right)$, so $\alpha < {\tan}^{- 1} = \frac{\pi}{4}$ and $\beta < \frac{\pi}{4}$. Hence $\alpha + \beta < \frac{\pi}{2}$ is in the range of ${\tan}^{- 1}$

Then we find:

$\tan \left(\alpha + \beta\right) = \frac{\tan \left(\alpha\right) + \tan \left(\beta\right)}{1 - \tan \left(\alpha\right) \tan \left(\beta\right)}$

$\textcolor{w h i t e}{\tan \left(\alpha + \beta\right)} = \frac{\frac{1}{3} + \frac{1}{5}}{1 - \frac{1}{3} \cdot \frac{1}{5}}$

$\textcolor{w h i t e}{\tan \left(\alpha + \beta\right)} = \frac{\frac{8}{15}}{\frac{14}{15}}$

$\textcolor{w h i t e}{\tan \left(\alpha + \beta\right)} = \frac{8}{14}$

$\textcolor{w h i t e}{\tan \left(\alpha + \beta\right)} = \frac{4}{7}$

So:

${\tan}^{- 1} \left(\frac{1}{3}\right) + {\tan}^{- 1} \left(\frac{1}{5}\right) = \alpha + \beta = {\tan}^{- 1} \left(\frac{4}{7}\right)$

Oct 20, 2016

See the proof below
${\tan}^{-} 1 \left(\frac{4}{7}\right) = {\tan}^{-} 1 \left(\frac{1}{3}\right) + {\tan}^{-} 1 \left(\frac{1}{5}\right)$

#### Explanation:

Let us have two angles a and b
Then $\tan \left(a + b\right) = \sin \frac{a + b}{\cos} \left(a + b\right) = \frac{\sin a \cos b + \sin b \cos a}{\cos a \cos b - \sin a \sin b}$
Divide throughout by $\cos a \cos b$
tan(a+b)=(((sinacosb)/(cosacosb))+((sinbcosa)/(cosacosb)))/((cosacosb)/(cosacosb)-(sinasinb)/(cosacosb)
Simplifyng we get
$\tan \left(a + b\right) = \frac{\tan a + \tan b}{1 - \tan a \tan b}$
Here we have $\tan a = \frac{1}{3}$ and $\tan b = \frac{1}{5}$
so $\tan \left(a + b\right) = \frac{\frac{1}{3} + \frac{1}{5}}{1 - \left(\frac{1}{3} \cdot \frac{1}{5}\right)} = \frac{\frac{8}{15}}{\frac{14}{15}} = \frac{8}{14} = \frac{4}{7}$
${\tan}^{-} 1 \left(\frac{4}{7}\right) = {\tan}^{-} 1 \left(\frac{1}{3}\right) + {\tan}^{-} 1 \left(\frac{1}{5}\right)$