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 $\cos ^{-1}(-\theta) \neq-\cos ^{-1} \theta$
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 $-\frac{\pi}{4}$.