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Trigonometric Functions — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSTrigonometric Functions1 Mark
Question
If $\tan\theta+\tan2\theta+\sqrt{3}\tan\theta\tan2\theta=\sqrt{3},$ then $\theta=\frac{\text{n}\pi}{3}+\frac{\pi}{9}$
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Answer

True.
Solution:
Given that, $\tan\theta+\tan2\theta=-\sqrt3\tan\theta\tan2\theta+\sqrt3$
$\Rightarrow\tan\theta+\tan2\theta=\sqrt3-\sqrt3\tan\theta\tan2\theta$
$\Rightarrow\tan\theta+\tan2\theta=\sqrt3(1-\tan\theta\tan2\theta)$
$\Rightarrow\frac{\tan\theta+\tan2\theta}{1-\tan\theta\tan2\theta}=\sqrt3$
$\Rightarrow\tan(\theta+2\theta)=\sqrt3$
$\Rightarrow\tan3\theta=\tan\frac{\pi}{3}\therefore3\theta=\text{n}\pi+\frac{\pi}{3}$
so $\theta=\frac{\text{n}\pi}{3}+\frac{\pi}{9}$
Hence, the given statement is 'true'.

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