Question

In (0, π), the number of solutions of the equation $\tan \mathrm{\theta }+\tan 2\mathrm{\theta }+\tan 3\mathrm{\theta }=\mathrm{tan\theta }\tan 2\mathrm{\theta }\tan 3\mathrm{\theta }$ is
(a) 7
(b) 5
(c) 4
(d) 2.

Answer 1

(d) 2
Given equation:

$$\begin{gathered} \tan \theta +\tan 2\theta +\tan 3\theta =\tan \theta \tan 2\theta \tan 3\theta \\ \Rightarrow \tan \theta +\tan 2\theta =-\tan 3\theta +\tan \theta \tan 2\theta \tan 3\theta \\ \Rightarrow \tan \theta +\tan 2\theta =-\tan 3\theta (1-\tan \theta \tan 2\theta ) \\ \Rightarrow \frac{\tan \theta +\tan 2\theta }{1-\tan \theta \tan 2\theta }=-\tan 3\theta \\ \Rightarrow \tan (\theta +2\theta )=-\tan 3\theta \\ \Rightarrow \tan 3\theta =-\tan 3\theta \\ \Rightarrow 2\tan 3\theta =0 \\ \Rightarrow \tan 3\theta =0 \\ \Rightarrow 3\theta =n\mathrm{\pi } \\ \Rightarrow \theta =\frac{n\mathrm{\pi }}{3} \end{gathered}$$

Now,
$\theta =\frac{\mathrm{\pi }}{3},n=1$
$\theta =\frac{2\mathrm{\pi }}{3},n=2$
$\theta =\frac{3\mathrm{\pi }}{3}=180°$, which is not possible, as it is not in the interval $(0,2\mathrm{\pi })$.

Hence, the number of solutions of the given equation is 2.