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यदि $\sqrt 3 \tan 2\theta + \sqrt 3 \tan 3\theta + \tan 2\theta \tan 3\theta = 1$, तो $\theta $ का व्यापक मान है
A
$n\pi + \frac{\pi }{5}$
B
$\left( {n + \frac{1}{6}} \right)\frac{\pi }{5}$
C
$\left( {2n \pm \frac{1}{6}} \right)\frac{\pi }{5}$
D
$\left( {n + \frac{1}{3}} \right)\frac{\pi }{5}$
Solution
(b) $\sqrt 3 \tan 2\theta + \sqrt 3 \tan 3\theta + \tan 2\theta \tan 3\theta = 1$
$ \Rightarrow $ $\frac{{\tan 2\theta + \tan 3\theta }}{{1 – \tan 2\theta \tan 3\theta }} = \frac{1}{{\sqrt 3 }}$
==> $\tan 5\theta = \tan \frac{\pi }{6}$
$ \Rightarrow $ $5\theta = n\pi + \frac{\pi }{6} $
$\Rightarrow \theta = \left( {n + \frac{1}{6}} \right)\frac{\pi }{5}$.
Standard 11
Mathematics
