Question

For $0<\theta <\frac{\pi }{2},\tan \theta +\tan 2\theta +\tan 3\theta =0$ if

• A. $\tan \theta =0$
• B. $\tan 2\theta =0$
• C. $\tan 3\theta =0$
• D. $\tan \theta \tan 2\theta =2$

Answer 1

Answer: (C), (D)

$\tan \theta \tan 2\theta =2$
$\tan 3\theta =0$

Clearly $\tan \theta \ne 0$ and $\tan 2\theta \ne 0$ for $0<\theta <x/2$

We have $\tan 3\theta =\tan (2\theta +\theta )$

$\Rightarrow \tan 3\theta =\frac{\tan 2\theta +\tan \theta }{1-\tan 2\theta \tan \theta }\therefore 0=\tan \theta +\tan 2\theta +\tan 3\theta =\tan 3\theta (1-\tan 2\theta \tan 3\theta )+\tan 3\theta =\tan 3\theta [2-\tan 2\theta \tan \theta ]\Rightarrow \tan 3\theta =0or\tan 2\theta \tan \theta =2$

Therefore, Correct Answer is $\tan 3\theta =0or\tan 2\theta \tan \theta =2$