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)

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.

N neha

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