# How do you find the exact value of arc tan (1/2)+arc tan (1/3)?

May 22, 2015

In this way:

$\arctan \left(\frac{1}{2}\right) + \arctan \left(\frac{1}{3}\right) = \alpha$.

If we search the tangent of both members:

$\tan \left(\arctan \left(\frac{1}{2}\right) + \arctan \left(\frac{1}{3}\right)\right) = \tan \alpha$

And now, using the sum angle formula of the tangent:

$\frac{\tan \arctan \left(\frac{1}{2}\right) + \tan \arctan \left(\frac{1}{3}\right)}{1 - \tan \arctan \left(\frac{1}{2}\right) \tan \arctan \left(\frac{1}{3}\right)} = \tan \alpha \Rightarrow$

$\frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2} \cdot \frac{1}{3}} = \tan \alpha \Rightarrow \tan \alpha = \frac{\frac{5}{6}}{\frac{5}{6}} = 1 \Rightarrow$

alpha=45°+k180°, or, in radians: $\alpha = \frac{\pi}{4} + k \pi$.