Question
Find the general solution of : $\cot \theta=-\sqrt{3}$

Answer

$ \text { (iii) } \cot \theta=-\sqrt{3} \quad \therefore \tan \theta=-\frac{1}{\sqrt{3}}$
$\therefore \quad \tan \theta=\tan \frac{5 \pi}{6}\left(\text { As tan } \frac{\pi}{6}=\frac{1}{\sqrt{3}} \text { and } \tan (\pi-A)=-\tan A\right) $
The general solution of $\tan \theta=\tan \alpha$ is $\theta= n \pi+\alpha$, where $n \in Z$.
$\therefore$ The general solution of $\tan \theta=\tan \frac{5 \pi}{6}$ is $\theta= n \pi+\frac{5 \pi}{6}$, where $n \in Z$.
$\therefore$ The general solution of $\cot \theta=-\sqrt{3}$ is $\theta=n \pi+\frac{5 \pi}{6}$, where $n \in Z$.

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