Question

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

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

Answer 1

Answer: (C)

$0\left(0-cosx\right)-cosx\left(0-co{s}^{2}x\right)-sinx\left(si{n}^{2}x-0\right)=0$
$co{s}^{3}x-si{n}^{3}x=0$
$ta{n}^3=1\Rightarrow tanx=1$
$\sum \frac{\sqrt{3}+tanx}{1-\sqrt{3}}$
$\sum \frac{\sqrt{3}+1}{1-\sqrt{3}}\times \frac{1+\sqrt{3}}{1+\sqrt{3}}\Rightarrow \sum \frac{1+3+2\sqrt{3}}{-2}=\sum \frac{4^2}{-2}-\frac{2\sqrt{3}}{3}$
$\sum -2-\sqrt{3}$