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mylen [45]
3 years ago
6

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

Mathematics
2 answers:
Lesechka [4]3 years ago
8 0

The other answer is incorrect

Tanya [424]3 years ago
3 0

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.

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