Question

Which of the following are solutions to 2tanx/1-tan^2x=sqrt 3?

Answer 1

The other answer is incorrect

Answer 2

Answer:

A, D

Step-by-step explanation:

$(\frac{2tanx}{1-tan^{2}x}) =\sqrt{3}$ (1)

Assume that tan x = y. Replace this into the equation (1), we have:

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

=> 2y = $\sqrt{3} . (1 -y^{2} ) =- \sqrt{3} .y^{2} + \sqrt{3}$

=> $\sqrt{3} .y^{2} + 2y - \sqrt{3} = 0$

=> $\sqrt{3} . y^{2} - 1.y + 3.y - \sqrt{3} =0$

=> $(\sqrt{3} . y.y - 1.y) + (\sqrt{3}.\sqrt{3} .y - \sqrt{3}) =0$

=> $y.(\sqrt{3} y - 1) + \sqrt{3} (\sqrt{3} y - 1) = 0$

=> $(y + \sqrt{3} ).(\sqrt{3} .y -1)=0$

=> $y + \sqrt{3} = 0$ or $\sqrt{3} y -1 =0$

If $y + \sqrt{3} = 0$

=> y = - √3

=> tan x = - √3

=> x = $\frac{2\pi }{3}$ + n.$\pi$ with n is an integral

If $\sqrt{3} y -1 =0$

=> y = 1/√3

=> tan x = 1/√3

=> x = $\frac{\pi }{6} + n.\pi$ n is an integral

So that A, D are answers.