Question

Find the principal value $ta{n}^{-1}\left(\sqrt{3}\right)$

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

Answer 1

Answer: (B)

Let $ta{n}^{-1}\left(\sqrt{3}\right)=\theta$
$\Rightarrow tan\theta =\sqrt{3}$
$=tan\frac{\pi }{3}$
$\Rightarrow \theta =\frac{\pi }{3}\in \left(\frac{-\pi }{2},\frac{\pi }{2}\right)$
$\therefore$ Principal value of $ta{n}^{-1}\left(\sqrt{3}\right)$ is $\frac{\pi }{3}$