Question
Find the principal solutions of $\cot \theta=-\sqrt{3}$
We know that $\tan \frac{\pi}{6}=\frac{1}{\sqrt{3}}$
Using identities, $\tan (\pi-\theta)=-\tan \theta$ and $\tan (2 \pi-\theta)=-\tan \theta$, we get
$\tan \frac{5 \pi}{6}=-\frac{1}{\sqrt{3}}$ and $\tan \frac{11 \pi}{6}=-\frac{1}{\sqrt{3}}$
An $0 \leq \frac{5 \pi}{6}<2 \pi$ and $0 \leq \frac{11 \pi}{6}<2 \pi$
$\therefore \frac{5 \pi}{6}$ and $\frac{11 \pi}{6}$ are required principal solutions.
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| X: | 0.5 | 1 | 1.5 | 2 |
| P(X): | $k$ | $k^2$ | $2k^2$ | $k$ |