Question

Consider the following statements
Statement I The principal value of ${\cos }^{-1}\left(-\frac{1}{2}\right)$ is $\frac{-\pi }{3}$
Statement II The principal value of ${\tan }^{-1}(-1)$ is $\frac{-\pi }{4}$
Choose the correct option.

• A. Only I is true
• B. Only II is true
• C. Both I and II are true
• D. Neither I nor II is true

Answer 1

Answer: (B)

I. Let ${\cos }^{-1}\left(-\frac{1}{2}\right)=\theta \Rightarrow \cos \theta =-\frac{1}{2}$
We know that, the range of principal value branch of ${\cos }^{-1}\theta$ is $[0,\pi ]$.
$\because \cos \theta =-\frac{1}{2}=-\cos \frac{\pi }{3}=\cos \left(\pi -\frac{\pi }{3}\right)[\because \cos (\pi -\theta )=-\cos \theta ]$
$=\cos \frac{2\pi }{3}\Rightarrow \theta =\frac{2\pi }{3}$ where ,$\theta \in [0,\pi ]$
$\Rightarrow {\cos }^{-1}\left(-\frac{1}{2}\right)=\frac{2\pi }{3}$
Hence, principal value of ${\cos }^{-1}\left(-\frac{1}{2}\right)$ is $\frac{2\pi }{3}$.
Note ${\cos }^{-1}(-\theta )\ne -{\cos }^{-1}\theta$
II. Let ${\tan }^{-1}(-1)=\theta \Rightarrow \tan \theta =-1$
We know that, the range of principal value branch of ${\tan }^{-1}\theta$ is $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$.
$\because \tan \theta =-1=-\tan \frac{\pi }{4}=\tan \left(-\frac{\pi }{4}\right)$
$(\because \tan (-\theta )=-\tan \theta )$
$\Rightarrow \theta =-\frac{\pi }{4}$ where, $\theta \in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)$
$\therefore {\tan }^{-1}(-1)=-\frac{\pi }{4}$
Hence, the principal value of ${\tan }^{-1}(-1)$ is $-\frac{\pi }{4}$.