MCQ
The principal solution of the equation $\cot x=-\sqrt{3}$ is
- A$\frac{\pi}{6}$
- B$\frac{\pi}{3}$
- C$\frac{5 \pi}{6}$
- D$-\frac{5 \pi}{6}$
$\frac{5 \pi}{6}$
$\cot x=-\sqrt{3}$
$\cot x=-\cot x$
we know $\cot x$;is negative in 2nd and 4th quadrant so, we convert this 2nd or 4th
now
$\cot x=-\cot x=\cot \left(\pi-\frac{5 \pi}{6}\right)$
$x=\frac{5 \pi}{6}$
also
$\cot x=\cot \left(2 \pi-\frac{11 \pi}{6}\right)$
$x=\frac{11 \pi}{6}$
only option given is $\frac{5 \pi}{6}$
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