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,
$x=n\pi +\alpha$
$x=n\pi +\frac{5\pi }{6}(\because n\in Z)$