Question

Find the principal value $tan^{-1} + cos^{-1} \left(\frac{-1}{2}\right) + sin^{-1}\left(\frac{-1}{2}\right)$

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

Answer 1

Answer: (B)

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