Question

Solve : $\tan \theta +\tan [\theta +\frac{\pi }{3}]+\tan [\theta +\frac{2\pi }{3}]=3$.

• A. $\theta =(4n+1)\frac{\pi }{12};n\in I$
• B. $\theta =(4n+1)\frac{\pi }{3};n\in I$
• C. $\theta =(4n+1)\frac{\pi }{6};n\in I$
• D. None of these

Answer 1

The correct option is A $\theta =(4n+1)\frac{\pi }{12};n\in I$

$\tan \theta +\tan (\theta +\frac{\pi }{3})+\tan (\theta +\frac{2\pi }{3})=3$

$\Rightarrow \tan \theta +\frac{\tan \theta +\tan \frac{\pi }{3}}{1-\tan \theta .\tan \frac{\pi }{3}}+\frac{\tan \theta +\tan \frac{2\pi }{3}}{1-\tan \theta .\tan \frac{2\pi }{3}}=3$

$\Rightarrow \tan \theta +\frac{\tan \theta +\sqrt{3}}{1-\sqrt{3}\tan \theta }+\frac{\tan \theta -\sqrt{3}}{1-\tan \theta (-\sqrt{3})}=3$

$\Rightarrow \tan \theta +\frac{\tan \theta +\sqrt{3}}{1-\sqrt{3}\tan \theta }+\frac{\tan \theta -\sqrt{3}}{1+\sqrt{3}\tan \theta }=3$

$\Rightarrow \frac{\tan \theta +(\tan \theta +\sqrt{3})(1+\sqrt{3}\tan \theta )+(\tan \theta -\sqrt{3})(1-\sqrt{3}\tan \theta )}{(1-\sqrt{3}\tan \theta )(1+\sqrt{3}\tan \theta )}=3$

$\Rightarrow \frac{\tan \theta +\tan \theta +\sqrt{3}{\tan }^2\theta +\sqrt{3}+3\tan \theta +\tan \theta -\sqrt{3}{\tan }^2\theta -\sqrt{3}+3\tan \theta }{1^2-(\sqrt{3}\tan \theta {)}^2}=3$

$\Rightarrow \tan \theta +\frac{8\tan \theta }{14-3{\tan }^2\theta }=3$

$\Rightarrow \tan \theta (1-3{\tan }^2\theta )+8\tan \theta =3(1-3{\tan }^2\theta )$

$\Rightarrow \tan \theta -3{\tan }^3\theta +8\tan \theta =3-9{\tan }^2\theta$

$\Rightarrow 9\tan \theta -3{\tan }^3\theta =3-9{\tan }^2\theta$

$\Rightarrow 9\tan \theta +9{\tan }^2\theta =3+3{\tan }^3\theta$

$\Rightarrow 9\tan \theta (1+\tan \theta )=3(1+{\tan }^3\theta )$

$\Rightarrow 9\tan \theta (1+\tan \theta )=3(1+\tan \theta )(1+{\tan }^2\theta -\tan \theta )$

$\Rightarrow (1+\tan \theta )(1+{\tan }^2\theta -\tan \theta -3\tan \theta )=0$

$\Rightarrow \tan \theta =-1$ and $1+{\tan }^2\theta -4\tan \theta =0$

$=\tan (-{45}^o)$ ${\tan }^2\theta -4\tan \theta +1=0$

$\tan \theta =\tan \left(\frac{-\pi }{4}\right)\tan \theta =\frac{+4\pm \sqrt{16-4}}{2}$

$\theta =n\pi +\left(\frac{-\pi }{4}\right)$ $\tan \theta =\frac{4\pm 2\sqrt{3}}{2}$

$\tan \theta =2\pm \sqrt{3}$

$\tan \theta =\tan \left(\frac{\pi }{12}\right)=2\pm \sqrt{3}$

$=\tan \left(\frac{5\pi }{12}\right)=2+\sqrt{3}$

$\Rightarrow \theta =n\pi +\left(\frac{5\pi }{12}\right)=n\pi +\frac{\pi }{12}$

Combine all the solution we get

$\theta =(4n+1)\frac{\pi }{12};n\in I$