Question

The general solution of $\tan \alpha +2\tan 2\alpha +4\tan 4\alpha +8\cot 8\alpha =\sqrt{3}$ is
$(\text{where}n\in Z)$

• A. $2n\pi \pm \frac{\pi }{3}$
• B. $n\pi \pm \frac{\pi }{6}$
• C. $n\pi +\frac{\pi }{6}$
• D. $n\pi +(-1{)}^n\frac{\pi }{3}$

Answer 1

The correct option is C $n\pi +\frac{\pi }{6}$

$\tan \alpha +2\tan 2\alpha +4\tan 4\alpha +8\cot 8\alpha =\sqrt{3}$
We know that,
$\cot \alpha -\tan \alpha =2\cot 2\alpha$
Now,
$\tan \alpha =\cot \alpha -2\cot 2\alpha 2\tan 2\alpha =2(\cot 2\alpha -2\cot 4\alpha )4\tan 4\alpha =4(\cot 4\alpha -2\cot 8\alpha )8\cot 8\alpha =8\cot 8\alpha \Rightarrow \tan \alpha +2\tan 2\alpha +4\tan 4\alpha +8\cot 8\alpha =\sqrt{3}\Rightarrow \cot \alpha =\sqrt{3}\Rightarrow \tan \alpha =\frac{1}{\sqrt{3}}\therefore \alpha =n\pi +\frac{\pi }{6},n\in Z$