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If $\tan\theta+\tan\big(\theta+\large\frac{\pi}{3}\big)$$+\tan\big(\theta+\large\frac{2\pi}{3}\big)$$=3$ then which of the following is equal to 1.

$\begin{array}{1 1}(a)\;\tan 2\theta&(b)\;\tan 3\theta\\(c)\;\tan^2\theta&(d)\;\tan^3\theta\end{array}$

1 Answer

Given :
$\tan\theta+\tan\big(\theta+\large\frac{\pi}{3}\big)$$+\tan\big(\theta+\large\frac{2\pi}{3}\big)$$=3$
$\Rightarrow \tan\theta+\large\frac{\tan\theta+3}{1-\sqrt 3\tan\theta}+\frac{\tan\theta-\sqrt 3}{1-\sqrt 3\tan\theta}$$=3$
$\Rightarrow \tan\theta+\large\frac{8\tan\theta}{1-3\tan^2\theta}$$=3$
$\Rightarrow \large\frac{9\tan\theta-2\tan^3\theta}{1-3\tan^2\theta}$$=3$
$\Rightarrow 3\tan 3\theta=3$
$\tan 3\theta=1$
Hence (b) is the correct option.
answered Oct 16, 2013 by sreemathi.v
 

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