Question

Trigonometry help me ​

Answer 1

Answer:

$\theta = \frac{\pi}{6}$

Step-by-step explanation:

$tan^ 2 \theta - ( \sqrt 3 + \frac{1}{\sqrt3}}) tan \theta + 1 = 0\\\\tan \theta - ( \sqrt 3 + \frac{1}{\sqrt3}}) +\frac{1}{ tan \theta } = 0$          $[ \ divide \ by \ tan \theta \ on \ both \ sides \ ]$

$tan\theta + \frac{1}{ tan \theta }- ( \sqrt 3 + \frac{1}{\sqrt3}}) = 0\\\\\frac{tan^2 \theta + 1}{ tan \theta } - ( \sqrt 3 + \frac{1}{\sqrt3}}) = 0\\\\\frac{sec ^2 \theta}{ \frac{sin \theta }{cos \theta}} - ( \sqrt 3 + \frac{1}{\sqrt3}}) = 0$              $[ \tan ^ 2\theta + 1 = sec ^2 \theta \ , \ tan \theta = \frac{sin \theta }{cos \theta } \ ]$

$\frac{sec^2 \theta }{sin \theta \times sec \theta } - ( \sqrt 3 + \frac{1}{\sqrt3}}) = 0$                  $[\ \frac{sin \theta }{cos \theta } = sin \theta \times sec \theta \ ]$

$\frac{sec \theta }{sin \theta } - ( \sqrt 3 + \frac{1}{\sqrt3}}) = 0$

$sec \theta \ cosec \theta - ( \sqrt 3 + \frac{1}{\sqrt3}}) = 0$              $[ \ \frac{1}{sin \theta } = cosec \theta \ , \ \frac{ sec \theta }{sin \theta } = sec \theta cosec \theta \ ]$

$sec \theta \ cosec \theta - \sqrt 3 - \frac{1}{\sqrt3}} = 0\\\\\frac{\sqrt 3\ sec \theta \ cosec \theta - 3 - 1}{\sqrt3} = 0\\\\\sqrt 3 sec \theta cosec \theta - 4 = 0$              

$\sqrt3 \frac{1}{cos \theta } \frac{1}{sin \theta } - 4 = 0\\\\\frac{\sqrt3 - 4sin \theta cos \theta} { sin \theta cos \theta } = 0$                      

$\sqrt 3 - 2sin 2\theta = 0$                                  $[ \ sin 2 \theta = 2 sin \theta cos \theta \ ]$

$2sin 2 \theta = \sqrt3\\\\sin 2 \theta = \frac{\sqrt3 }{2} \\\\2 \theta = sin^{-1} (\frac{\sqrt3}{2})\\\\2 \theta = 60^{ \circ} = \frac{ \pi}{3}\\\\\theta = \frac{\pi} {6}$