Question

General solution of the equation $\tan \theta \tan 2 \theta=1$ is given by

• A. $(2 n+1) \frac{\pi}{4}, n \in I$
• B. $n \pi+\frac{\pi}{6}, n \in I$
• C. $n \pi-\frac{\pi}{6}, n \in I$
• D. $n \pi \pm \frac{\pi}{6}, n \in I$

Answer 1

Answer: (D)

$\tan \,\theta \,\tan \,2 \,\theta=1$
$ \therefore \tan \,\theta \frac{2 \,\tan\, \theta}{1-\tan ^{2} \theta}=1$
$\therefore 2 \,\tan ^{2} \theta=1-\tan ^{2} \theta $
$\therefore 3 \,\tan ^{2} \theta=1$
$ \therefore \theta= n \pi \pm \frac{\pi}{6}$