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