# Find the principal value of $$tan^{-1} \sqrt 3 - sec^{-1}(-2)$$

This is Q.No.14 Sec1 of chapter 2

## 1 Answer

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 $\left [0, \pi \right ] - (\large\frac{\pi}{2})$
• $sec (\pi - \theta) = - sec \theta$
Ans :
$$\frac{\pi}{3} - \bigg( \pi-\frac{\pi}{3} \bigg) = \frac{-\pi}{3}$$
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 $-\large\frac{\pi}{3}$

answered Feb 13, 2013
edited Mar 19, 2013

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