Question

Find the principal value $tan^{-1}\left(-\sqrt{3}\right) + sec^{-1}\left(-2\right)-cosec^{-1}\left(\frac{2}{\sqrt{3}}\right)$

• A. $\frac{5\pi}{6}$
• B. $\frac{2\pi}{3}$
• C. $\frac{\pi}{3}$
• D. $0$

Answer 1

Answer: (D)

Let $tan^{-1}\left(-\sqrt{3}\right) = \alpha$
$\Rightarrow tan \,\alpha = -\sqrt{3} = -tan \frac{\pi}{3}$
$= tan\left(-\frac{\pi}{3}\right)$
$\Rightarrow \, \alpha = \frac{-\pi }{3} \in \left( \frac{-\pi }{2}, \frac{\pi }{2}\right)$
$\therefore $ Principal value of $tan^{-1} \left(-\sqrt{3}\right)$ is $\left(\frac{-\pi }{3}\right)$
Let $sec^{-1} \left(-2\right) = \beta$
$\Rightarrow sec\,\beta = -2$
$= -sec \frac{\pi}{3}$
$= sec\left(\pi-\frac{\pi }{3}\right)$
$= sec \left(\frac{2\pi }{3}\right)$
$\Rightarrow \beta = \frac{2\pi }{3} \in \left[0, \,\pi\right]-\left\{\frac{\pi}{2}\right\}$
$\therefore $ Principal value of $sec^{-1} \left(-2\right)$ is $\frac{2\pi }{3}$
Let $cosec^{-1} \left(\frac{2}{\sqrt{3}}\right) = \gamma$
$\Rightarrow cosec\,\gamma = \frac{2}{\sqrt{3}}$
$= cosec \frac{\pi}{3}$
$\Rightarrow \gamma = \frac{\pi }{3}\in\left[\frac{-\pi }{2}, \frac{\pi }{2}\right] - \left\{0\right\}$
$\therefore $ Principal value of $cosec^{-1} \left(\frac{2}{\sqrt{3}}\right)$ is $\frac{\pi }{3}$
So, the principal value of
$tan^{-1}\left(-\sqrt{3}\right) + sec^{-1}\left(-2\right)-cosec^{-1}\left(\frac{2}{\sqrt{3}}\right)$
$= \frac{-\pi }{3}+\frac{2\pi }{3} - \frac{\pi }{3} $
$= 0$