Question

The solution of $\tan 2\theta\: \tan \theta = 1$ is

• A. $\frac{\pi}{3}$
• B. $(6n\pm1)\frac{\pi}{6}$
• C. $(4n\pm1)\frac{\pi}{6}$
• D. $(2n\pm1)\frac{\pi}{6}$

Answer 1

Answer: (B)

$(6n\pm1)\frac{\pi}{6}$

Answer 2

$\tan\,2 \theta \,\tan\,\theta=1$
$\tan3\theta=\frac{\tan\theta+\tan2\theta}{1-\tan\theta.\tan2\theta}$

$\tan3\theta=$∞

i,e, $3\theta=\frac{2n+1}{2}\pi$

$\theta=\frac{{(2n+1)}}{6}\pi$