The general solution of $\tan \theta+\tan 2 \theta=\tan 3 \theta$ is
The general solution of $\tan \theta+\tan 2 \theta=\tan 3 \theta$ is
- $\theta=(2 n+1) \frac{\pi}{2}, n \in Z$
- $\theta=n \pi, n \in Z$ or $\theta=\frac{p \pi}{3}, p \in Z$
- $\theta=\frac{n \pi}{5}, n \in Z$
- $\theta=(2 n-1) \frac{\pi}{3}, n \in Z$
Solution
$\tan 3 \theta=\tan (2 \theta+\theta)=\tan \theta+\tan 2 \theta$
$\therefore \frac{\tan 2 \theta+\tan \theta}{1-\tan 2 \theta \tan \theta}=\tan \theta+\tan 2 \theta$
$\therefore 1-\tan 2 \theta \tan \theta=1 \Rightarrow \tan 2 \theta \tan \theta=0 \Rightarrow \tan \theta=0$
$\therefore \theta=\mathrm{n} \pi, \mathrm{n} \in \mathrm{Z}$
Asked in: MHT CET 2020 (13 Oct Shift 2)
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