# Solve the following $$\tan^{-1} \sqrt{3} -sec^{-1} (-2)$$:

$(A) \quad \pi \qquad (B) \quad -\frac{\pi}{3} \qquad (C) \quad \frac{\pi}{3} \qquad (D) \quad \frac{2\pi}{3} \qquad$

Ans : B
$$\frac{\pi}{3} - \bigg( \pi-\frac{\pi}{3} \bigg) = \frac{-\pi}{3}$$

Toolbox:
• The range of the principal value of $\tan^{-1}x$ is $\left [ -\large\frac{\pi}{2},\large \frac{\pi}{2} \right ]$
• The range of the principal value of $\sec^{-1}x$ is $\left [0, \pi \right ] - (\large\frac{\pi}{2})$
• $sec (\pi - \theta) = - sec \theta$
Let $\tan^{-1} \sqrt3 = x$
$\Rightarrow \tan x = \sqrt 3= tan\large\frac{\pi}{3}$

$$\therefore$$, $x = \large\frac{\pi}{3} \; \epsilon \left [ -\large\frac{\pi}{2}, \large\frac{\pi}{2} \right ]$
Let $\sec^{-1} (-2) = y$
$\Rightarrow \sec y = -2$

$$\therefore$$, $y = -sec \large\frac{\pi}{2}$
$\Rightarrow sec y = sec (\pi -\large \frac{\pi}{2}) = sec(\large\frac{2\pi}{3})$

$$\therefore$$, $y = \large\frac{2\pi}{3} \;\; \epsilon \;\left [0, \pi \right ] - (\large\frac{\pi}{2})$
$\tan^{-1} \sqrt3 - \sec^{-1} (-2) = x- y =\large \frac{\pi}{3} - \large\frac{2\pi}{3} = -\large\frac{\pi}{3}$

The correct answer is $(B): -\large\frac{\pi}{3}$

edited Mar 15, 2013