# How do you find the exact value of arctan(cot(11π / 6))?

Aug 25, 2016

$- \frac{2}{3} \pi$

#### Explanation:

$\cot \left(\frac{11}{6} \pi\right) = \tan \left(\frac{\pi}{2} - \frac{11}{6} \pi\right) = \tan \left(- \frac{4}{3} \pi\right)$

So, the exact value of

arc tan(cot(11/6pi)

=arc tan (tan (-4/3pi)

= the operand $- \frac{4}{3} \pi$,

against the given operand $\frac{11}{6} \pi$ in $\cot \left(\frac{11}{6} \pi\right)$,

If the solution is sought as a principal value in $\left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$.

use the general solution $n \pi + {\left(- 1\right)}^{n} \left(- \frac{4}{3} \pi\right) , n = 0 , \pm 1 , \pm 2 , \pm 3 , . .$

For, $n = - 1$, it is $- \pi + \left(- 1\right) \left(- \frac{4}{3} \pi\right) = - \frac{\pi}{3}$.

Your calculator might gives the answer as $- {60}^{o}$.

All these are attributed to the convention that inverse trigonometric

functions give principal values only. The algorithms in the

software for computer approximations of inverse trigonometric

functions are based on this convention..For example,

$\tan \left(- {240}^{o}\right)$ is displayed as

$- 1. .732030808$, but its inverse is displayed as$- {60}^{o}$.

For applications, it has to be remembered that trigonometric

functions are periodic, and are bijective only in a part of one period,

in which the principal value is defined,. For example, sin x is

bijective in $\left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$, within $\left[- \pi , \pi\right]$..