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If $\cot \theta+\tan \theta=2 \operatorname{cosec} \theta$, then find the general value of $\theta$.
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Verified Answer
We have, $\cot \theta+\tan \theta=2 \operatorname{cosec} \theta$
$$
\begin{aligned}
&\Rightarrow \frac{\cos \theta}{\sin \theta}+\frac{\sin \theta}{\cos \theta}=\frac{2}{\sin \theta} \\
&\Rightarrow \frac{\cos ^2+\sin ^2 \theta}{\sin \theta \cdot \cos \theta}=\frac{2}{\sin \theta} \\
&\Rightarrow \frac{1}{\cos \theta}=2\left[\because \sin ^2 \theta+\cos ^2 \theta=1\right] \\
&\Rightarrow \cos \theta=\frac{1}{2} \Rightarrow \cos \theta=\cos \frac{\pi}{3} \\
&\therefore \theta=2 n \pi \pm \frac{\pi}{3}
\end{aligned}
$$
$$
\begin{aligned}
&\Rightarrow \frac{\cos \theta}{\sin \theta}+\frac{\sin \theta}{\cos \theta}=\frac{2}{\sin \theta} \\
&\Rightarrow \frac{\cos ^2+\sin ^2 \theta}{\sin \theta \cdot \cos \theta}=\frac{2}{\sin \theta} \\
&\Rightarrow \frac{1}{\cos \theta}=2\left[\because \sin ^2 \theta+\cos ^2 \theta=1\right] \\
&\Rightarrow \cos \theta=\frac{1}{2} \Rightarrow \cos \theta=\cos \frac{\pi}{3} \\
&\therefore \theta=2 n \pi \pm \frac{\pi}{3}
\end{aligned}
$$
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