If sqrt 3 tan 2theta + sqrt 3 tan 3theta + ta | vedclass

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(b) $\sqrt 3 	an 2	heta + \sqrt 3 	an 3	heta + 	an 2	heta 	an 3	heta = 1$

$ \Rightarrow $ $\frac{{	an 2	heta + 	an 3	heta }}{{1 - 	an

Vedclass · Accepted Answer

$\left( {n + \frac{1}{6}} \right)\frac{\pi }{5}$

Vedclass · Answer

(b) $\sqrt 3 	an 2	heta + \sqrt 3 	an 3	heta + 	an 2	heta 	an 3	heta = 1$

$ \Rightarrow $ $\frac{{	an 2	heta + 	an 3	heta }}{{1 - 	an 2	heta 	an 3	heta }} = \frac{1}{{\sqrt 3 }}$ 

==> $	an 5	heta = 	an \frac{\pi }{6}$

$ \Rightarrow $ $5	heta = n\pi + \frac{\pi }{6} $

$\Rightarrow 	heta = \left( {n + \frac{1}{6}} ight)\frac{\pi }{5}$.

If $\sqrt 3 \tan 2\theta + \sqrt 3 \tan 3\theta + \tan 2\theta \tan 3\theta = 1$, then the general value of $\theta $ is

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