Question

If $S=\left\{xϵ\right[0,2\pi ]:\left|\begin{array}{ccc}0& \cos x& -\sin x\\ \sin x& 0& \cos x\\ \cos x& \sin x& 0\end{array}\right|=0\}$, then $\sum _{xϵS}\tan \left(\frac{\pi }{3}+x\right)$ is equal to

• A. $4+2\sqrt{3}$
• B. $-2-\sqrt{3}$
• C. $-4-2\sqrt{3}$
• D. $-2+\sqrt{3}$

Answer 1

Answer: (C)

$-4-2\sqrt{3}$

$0(0-\cos x)-\cos x(0-{\cos }^{2}x)-\sin x({\sin }^2-0)=0$
$\Rightarrow {\cos }^{3}x-{\sin }^{3}x=0$
$\Rightarrow {\tan }^{3}x=1$
$\Rightarrow \tan x=1$
$\sum \frac{\sqrt{3}+\tan x}{1-\sqrt{3}}$

$=\sum \frac{\sqrt{3}+1}{1-\sqrt{3}}\times \frac{1+\sqrt{3}}{1+\sqrt{3}}$

$=\sum \frac{1+3+2\sqrt{3}}{-2}$

$=\sum \frac{2^2}{-2}-\frac{2\sqrt{3}}{2}$

$=\sum -2-\sqrt{3}$.

This will happen twice. once at $\frac{\pi }{4}$ and again at $\frac{5\pi }{4}$

so the answer must be option C