# How do you evaluate arctan((arctan(9/7) - arctan(7/6)) / (arctan(5/3) - arctan(3/2)))?

Apr 1, 2016

$\frac{\pi}{4} + K \cdot \pi , K \in \mathbb{N}$

#### Explanation:

Let's rewrite the expression as

$\epsilon = \arctan \left(\frac{\theta}{\rho}\right)$
with
$\theta = \arctan \left(\frac{9}{7}\right) - \arctan \left(\frac{7}{6}\right)$ => $\tan \theta = \tan \left(\arctan \left(\frac{9}{7}\right) - \arctan \left(\frac{7}{6}\right)\right)$
$\rho = \arctan \left(\frac{5}{3}\right) - \arctan \left(\frac{3}{2}\right)$ => $\tan \rho = \tan \left(\arctan \left(\frac{5}{3}\right) - \arctan \left(\frac{3}{2}\right)\right)$

Using the formula of the tangent of the difference between two angles
$\tan \left(x - y\right) = \frac{\tan x - \tan y}{1 + \tan x \cdot \tan y}$
we get

$\tan \theta = \frac{\frac{9}{7} - \frac{7}{6}}{1 + \frac{9}{\cancel{7}} \cdot \frac{\cancel{7}}{6}} = \frac{\frac{54 - 49}{42}}{1 + \frac{3}{2}} = \frac{2}{\cancel{5}} \cdot \frac{\cancel{5}}{42} = \frac{1}{21}$
$\to \theta = \arctan \left(\frac{1}{21}\right)$
and
$\tan \rho = \frac{\frac{5}{3} - \frac{3}{2}}{1 + \frac{5}{\cancel{3}} \cdot \frac{\cancel{3}}{2}} = \frac{\frac{10 - 9}{6}}{1 + \frac{5}{2}} = \frac{2}{7} \cdot \frac{1}{6} = \frac{1}{21}$
$\to \rho = \arctan \left(\frac{1}{21}\right)$

So
$\epsilon = \arctan \left(\frac{\theta}{\rho}\right) = \arctan \left(\frac{\cancel{\arctan \left(\frac{1}{21}\right)}}{\cancel{\arctan \left(\frac{1}{21}\right)}}\right) = \arctan \left(1\right)$
=> $\epsilon = \frac{\pi}{4} + K \cdot \pi , K \in \mathbb{N}$