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# Find the value of $\; \tan^{-1}\bigg(\tan{\frac{5\pi}{6}}\bigg)+\cos^{-1}\bigg(\cos{\frac{13\pi}{6}}\bigg)$

$\begin{array}{1 1} 0 \\ 3 \pi \\ \pi \\ \frac{\pi}{3} \end{array}$

Toolbox:
• Principal interval of tan is $(-\large\frac{\pi}{2},\large\frac{\pi}{2})$
• Principal interval of cos is $[0,\pi]$
• $-tan\theta=tan(-\theta)$
• $tan(\pi-\theta)=-tan\theta$
• $cos(2\pi+\theta)=cos\theta$
Since $\large\frac{5\pi}{6}$is not within the principal interval  reduce it within the principal interval of tan.
$\large\frac{5\pi}{6}=\pi-\large\frac{\pi}{6}$

Similarly reduce $\large\frac{13\pi}{6}$ within the principal interval of cos
$\large\frac{13\pi}{6}=2\pi+\large\frac{\pi}{6}$

By taking $\theta=\large\frac{\pi}{6},\:tan(\pi-\large\frac{\pi}{6})=-\tan\large\frac{\pi}{6}=tan(-\large\frac{\pi}{6})\:and$
$cos(2\pi+\large\frac{\pi}{6})=cos\large\frac{\pi}{6}$

$\Rightarrow\:tan\large\frac{5\pi}{6}=tan(-\large\frac{\pi}{6})$ and
$cos\large\frac{13\pi}{6}=cos\large\frac{\pi}{6}$
$tan^{-1}tan(\large\frac{5\pi}{6})+cos^{-1}cos(\large\frac{13\pi}{6})=$
$tan^{-1} \bigg[ tan \bigg( \pi - \large\frac{\pi}{6} \bigg)\bigg] + cos^{-1} \bigg( cos (2\pi+\large\frac{\pi}{6} )\bigg)$

$=tan^{-1}tan(-\large\frac{\pi}{6})+cos^{-1}cos\large\frac{\pi}{6}$
$= \large\frac{-\pi}{6}+\large\frac{\pi}{6} = 0$

edited Mar 15, 2013