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If

Option 1)
$4+2\sqrt{3}$
Option 2)
$-2+\sqrt{3}$
Option 3)
$-2-\sqrt{3}$
Option 4)
$-4 - 2\sqrt{3}$

Answers (2)

best_answer

As we learnt in

Cramer's rule for solving system of linear equations -

When $\Delta =0$ and $\Delta _{1}=\Delta _{2}=\Delta _{3}=0$ ,

then the system of equations has infinite solutions.

- wherein

$a_{1}x+b_{1}y+c_{1}z=d_{1}$

$a_{2}x+b_{2}y+c_{2}z=d_{2}$

$a_{3}x+b_{3}y+c_{3}z=d_{3}$

and

$\Delta =\begin{vmatrix} a_{1} &b_{1} &c_{1} \\ a_{2} & b_{2} &c_{2} \\ a_{3}&b _{3} & c_{3} \end{vmatrix}$

$\Delta _{1},\Delta _{2},\Delta _{3}$ are obtained by replacing column 1,2,3 of $\Delta$ by $\left ( d_{1},d_{2},d_{3} \right )$ column

$\Rightarrow \begin{vmatrix} 0 &\cos x &-\sin x \\ \sin x& 0 & \cos x\\ \cos x& \sin x &0 \end{vmatrix} =0$

$\Rightarrow 0 (0 - \sin x \cos x)- \cos x(0 - \ cos^{2}x)- \sin x (\sin^{2}x) = 0$

$\Rightarrow \sin^{3}x - \cos^{3}x = 0$

$\Rightarrow (\sin x - \cos x) (1 + \sin x \cos x) = 0$

$\Rightarrow\therefore \sin x - \cos x = 0$

$\Rightarrow\therefore \sin x - \cos x = 0$

$\therefore \tan x= 1$

$\therefore x= \frac{\pi }{4}\ or\ \frac{5\pi}{4}$

Now, $\tan(\frac{\pi}{4}+x)= \frac{1+\tan x }{1-\tan x}$

Now Put $x= \frac{\pi }{3}$

$= \frac{1+\tan\frac{\pi}{3} }{1-\tan\frac{\pi}{3}}$

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

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

$\sum \tan (\frac{\pi }{3}+x) = 2- \sqrt{3} -2-\sqrt{3}=-4-2\sqrt{3}$

Option 1)

$4+2\sqrt{3}$

This option is incorrect.

Option 2)

$-2+\sqrt{3}$

This option is incorrect.

Option 3)

$-2-\sqrt{3}$

This option is incorrect.

Option 4)

$-4 - 2\sqrt{3}$

This option is correct.

Posted by

prateek

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