# How do you evaluate tan^-1(tan((11pi)/10))?

${\tan}^{-} 1 \left(\tan \left(\frac{11 \pi}{10}\right)\right) = {\tan}^{-} 1 \left(\tan \left(\frac{\pi}{10}\right)\right) = \frac{\pi}{10}$
The restriction for ${\tan}^{-} 1 x$ is $\left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$ but $\frac{11 \pi}{10}$ is in quadrant III. Since tangent is positive in quadrant III our answer will be in the positive quadrant from the restriction which is quadrant I. So let's find the reference angle which is $\frac{11 \pi}{10} - \pi = \frac{\pi}{10}$. Then since the reference angle equals the angle $\theta$ in quadrant one our argument in quadrant one is also $\frac{\pi}{10}$. Now apply the property ${f}^{-} 1 \left(f \left(x\right)\right) = x$
${\tan}^{-} 1 \left(\tan \left(\frac{11 \pi}{10}\right)\right) = {\tan}^{-} 1 \left(\tan \left(\frac{\pi}{10}\right)\right) = \frac{\pi}{10}$