# Using the principal values, evaluate the following $tan^{-1}\frac{1}{\sqrt{3}}+sin^{-1} \bigg( -\frac{1}{2} \bigg)$

Ans:

$\frac{\pi}{6}-\frac{\pi}{6}=0$

We know that $$tan\frac{\pi}{6}=\frac{1}{\sqrt3}$$ and $$\frac{\pi}{6}\:\in\:(-\frac{\pi}{2},\frac{\pi}{2})$$

which is the principal interval of tan

Therefore,$$tan^{-1}\frac{1}{\sqrt3}=tan^{-1}tan\frac{\pi}{6}=\frac{\pi}{6}$$

similarly  we know that  $$sin(-\frac{\pi}{6})=-\frac{1}{2}$$  and

$$-\frac{\pi}{6}\in\:[-\frac{\pi}{2},\frac{\pi}{2}]$$ which is the principal interval of sin

Therefore, $$sin^{-1}(-\frac{1}{2})=sin^{-1}sin(-\frac{\pi}{6})=-\frac{\pi}{6}$$

$$\Rightarrow\:tan^{-1}\frac{1}{\sqrt3}+sin^{-1}-\frac{1}{2}=\frac{\pi}{6}-\frac{\pi}{6}=0$$
edited Mar 19, 2013