# Find the value of $$tan^{-1} \sqrt 3 - cot^{-1} (-\sqrt 3)$$

$\begin{array} (A) \pi \quad & (B) -\frac{\pi}{2} \quad & (C) 0 \quad & (D) 2\sqrt3 \end{array}$

Toolbox:
• $$tan^{-1}\sqrt{3}=\large\frac{\pi}{3}$$
• $$cot^{-1}-\sqrt{3}=\pi-\large\frac{\pi}{6}$$
Ans (B)
since the principal interval of tan is ($$-\large\frac{\pi}{2},\large\frac{\pi}{2})$$ and
that of cot is (0,$$\pi$$), $$tan^{-1}\sqrt{3}=\large\frac{\pi}{3}\:$$ and $$\:cot^{-1}-\sqrt{3}=\pi-\large\frac{\pi}{6}$$
The given expression becomes $$\large\frac{\pi}{3}- \bigg(\pi- \large\frac{\pi}{6} \bigg) =- \large\frac{\pi}{2}$$
edited Jul 9, 2014