# The value of the summation is =?

Mar 21, 2018

The Right Option is $\left(4\right) : - 4 - 2 \sqrt{3}$.

#### Explanation:

$| \left(0 , \cos x , - \sin x\right) , \left(\sin x , 0 , \cos x\right) , \left(\cos x , \sin x , 0\right) | = 0$.

Expanding by the first raw, we have,

$\therefore 0 - \cos x \left(0 - {\cos}^{2} x\right) - \sin x \left({\sin}^{2} x - 0\right) = 0$.

$\therefore {\cos}^{3} x - {\sin}^{3} x = 0 , \mathmr{and} , {\sin}^{3} x = {\cos}^{3} x \ldots \ldots \left(\star\right)$.

We must have ${\cos}^{3} x \ne 0 , \because , \text{ otherwise } \cos x = 0$, and, then,

$\left(\star\right) \Rightarrow {\sin}^{3} x = {\cos}^{3} x = 0 , \text{ so that, } \sin x = 0$.

$\Rightarrow {\cos}^{2} x + {\sin}^{2} x = 0 + 0 = 0$, a contradiction.

So, dividing $\left(\star\right) \text{ by "cos^3x!=0, "we get, } {\tan}^{3} x = 1$.

:. tanx=1, &, x in [0,2pi], x=pi/4,or, x=pi/4+pi=5/4pi.

$\therefore S = \left\{\frac{\pi}{4} , 5 \frac{\pi}{4}\right\} \Rightarrow \forall x \in S , \tan x = 1$.

$\therefore {\sum}_{x \in S} \tan \left(\frac{\pi}{3} + x\right)$,

$= \tan \left(\frac{\pi}{3} + \frac{\pi}{4}\right) + \tan \left(\frac{\pi}{3} + 5 \frac{\pi}{4}\right)$,

$= \frac{\tan \left(\frac{\pi}{3}\right) + \tan \left(\frac{\pi}{4}\right)}{1 - \tan \left(\frac{\pi}{3}\right) \tan \left(\frac{\pi}{4}\right)}$

$+ \frac{\tan \left(\frac{\pi}{3}\right) + \tan \left(5 \frac{\pi}{4}\right)}{1 - \tan \left(\frac{\pi}{3}\right) \tan \left(5 \frac{\pi}{4}\right)}$,

$= \frac{\sqrt{3} + 1}{1 - \sqrt{3}} + \frac{\sqrt{3} + 1}{1 - \sqrt{3}}$,

$= \frac{2 \left(\sqrt{3} + 1\right)}{1 - \sqrt{3}} \times \frac{1 + \sqrt{3}}{1 + \sqrt{3}}$,

$= \frac{2}{1 - 3} {\left(1 + \sqrt{3}\right)}^{2}$.

$\Rightarrow {\sum}_{x \in S} \tan \left(\frac{\pi}{3} + x\right) = - \left(1 + 2 \sqrt{3} + 3\right)$,

showing that the right option is $\left(4\right) : - 4 - 2 \sqrt{3}$.

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