Question

Find the principal value $ta{n}^{-1}\left(-\sqrt{3}\right)+se{c}^{-1}\left(-2\right)-cose{c}^{-1}\left(\frac{2}{\sqrt{3}}\right)$

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

Answer 1

Answer: (D)

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