Question

Principal value of ${\tan }^{-1}1+{\cos }^{-1}\left(\frac{-1}{2}\right)+{\sin }^{-1}\left(\frac{-1}{2}\right)$ is equal to

• A. $\frac{2\pi }{3}$
• B. $\frac{3\pi }{4}$
• C. $\frac{\pi }{2}$
• D. $6\pi$

Answer 1

Answer: (B)

Iet ${\tan }^{-1}(1)=\theta$
$\Rightarrow \tan \theta =1=\tan \frac{\pi }{4}$
$\Rightarrow \theta =\frac{\pi }{4}\in \left(\frac{-\pi }{2},\frac{\pi }{2}\right)$
$\therefore$ Principal value of ${\tan }^{-1}(1)$ is $\frac{\pi }{4}$
Let ${\cos }^{-1}\left(\frac{-1}{2}\right)=\varphi$
$\Rightarrow \cos \varphi =\frac{-1}{2}=-\cos \frac{\pi }{3}$
$=\cos \left(\pi -\frac{\pi }{3}\right)=\cos \frac{2\pi }{3}$
$\Rightarrow \varphi =\frac{2\pi }{3}\in [0,\pi ]$
$\therefore$ Principal value of ${\cos }^{-1}\left(\frac{-1}{2}\right)$ is $\frac{2\pi }{3}$
Also, principal value of ${\sin }^{-1}\left(\frac{-1}{2}\right)$ is $\left(\frac{-\pi }{6}\right)$
Principal value of ${\tan }^{-1}(1)+{\cos }^{-1}\left(\frac{-1}{2}\right)+{\sin }^{-1}\left(\frac{-1}{2}\right)$
$=\frac{\pi }{4}+\frac{2\pi }{3}-\frac{\pi }{6}=\frac{3\pi }{4}$