Question

100 marks question answer fast​

Answer 1

$Let \: \: \tan( \theta) = t</p><p>$

So, we have :-

=> t² + 2√3 t = 1

=> t² + (√3 - 2)t + (√3 + 2)t - 1 = 0

=> (t + √3 - 2)(t + √3 + 2) = 0

So, t = 2 - √3 or t = -(2 + √3)

$\tan( \theta) = 2 - \sqrt{3} \: \: or \: \: \tan( \theta) = - (2 + \sqrt{3} ) \\ = > \theta = n\pi + \frac{\pi}{12} \: \: or \: \: \theta = n\pi - \frac{ 5\pi}{12}$

Hope, it'll help you.....

Answer 2

Answer:

The answer is D

Step-by-step explanation:

t² + 2√3 t = 1

=> t² + (√3 - 2)t + (√3 + 2)t - 1 = 0

=> (t + √3 - 2)(t + √3 + 2) = 0

So, t = 2 - √3 or t = -(2 + √3)

\tan( \theta) = 2 - \sqrt{3} \: \: or \: \: \tan( \theta) = - (2 + \sqrt{3} ) \\ = > \theta = n\pi + \frac{\pi}{12} \: \: or \: \: \theta = n\pi - \frac{ 5\pi}{12}