Question

If $\cot x=-\sqrt{3}$, then

• A. Its principal value is $\frac{5 \pi}{6}$
• B. Its general solution is $n \pi+\frac{5 \pi}{6}, n \in Z$
• C. Only (a) is true
• D. Both (a) and (b) are true

Answer 1

Answer: (D)

$\cot x=-\sqrt{3}$
$\Rightarrow \tan x=-\frac{1}{\sqrt{3}}$ $\Rightarrow \tan x=-\tan \frac{\pi}{6}$
$\Rightarrow \tan x=\tan \left(\pi-\frac{\pi}{6}\right)$ or $\tan \left(2 \pi-\frac{\pi}{6}\right)$
$[\because \tan x$ is negative in 2nd and 4th quadrant and $\tan (\pi-\theta)=-\tan \theta$ and $\tan (2 \pi-\theta)=-\tan \theta]$
$\tan x =\tan \frac{5 \pi}{6} \text { or } \tan \frac{11 \pi}{6}$
$\Rightarrow x =\frac{5 \pi}{6} \text { or } \frac{11 \pi}{6}$
Here, principal value is $x=\frac{5 \pi}{6}$
We know that, if $\tan x=\tan \alpha$, then
General solution,
$ x=n \pi+\alpha $
$ x=n \pi+\frac{5 \pi}{6} (\because n \in Z)$