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If $\tan \theta+\tan 2 \theta+\sqrt{3} \tan \theta \tan 2 \theta=\sqrt{3}$, then the general values of $\theta$ are
Options:
Solution:
2809 Upvotes
Verified Answer
The correct answer is:
$(3 n+1) \frac{\pi}{9}, n \in Z$
It is given that
$$
\begin{aligned}
& \tan \theta+\tan 2 \theta+\sqrt{3} \tan \theta \tan 2 \theta=\sqrt{3} \\
& \Rightarrow \frac{\tan \theta+\tan 2 \theta}{1-\tan \theta \tan 2 \theta}=\sqrt{3}=\tan \frac{\pi}{3} \\
& \Rightarrow \tan (3 \theta)=\tan \frac{\pi}{3} \Rightarrow 3 \theta=n \pi+\frac{\pi}{3}, n \in Z \\
& \Rightarrow \theta=(3 n+1) \frac{\pi}{9}, n \in Z
\end{aligned}
$$
Hence, option (b) is correct.
$$
\begin{aligned}
& \tan \theta+\tan 2 \theta+\sqrt{3} \tan \theta \tan 2 \theta=\sqrt{3} \\
& \Rightarrow \frac{\tan \theta+\tan 2 \theta}{1-\tan \theta \tan 2 \theta}=\sqrt{3}=\tan \frac{\pi}{3} \\
& \Rightarrow \tan (3 \theta)=\tan \frac{\pi}{3} \Rightarrow 3 \theta=n \pi+\frac{\pi}{3}, n \in Z \\
& \Rightarrow \theta=(3 n+1) \frac{\pi}{9}, n \in Z
\end{aligned}
$$
Hence, option (b) is correct.
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