Question

Solve the equation:-
${\tan }^2\theta +(1-\sqrt{3})\tan \theta -\sqrt{3}=0$

Answer 1

${\tan }^2\theta +\tan \theta -\sqrt{3}\tan \theta -\sqrt{3}=0$

$\Rightarrow \tan \theta (\tan \theta +1)-\sqrt{3}(\tan \theta +1)=0$

$\Rightarrow (\tan \theta +1)(\tan \theta -\sqrt{3})=0$

$\Rightarrow \tan \theta =-1,\tan \theta =\sqrt{3}$

$\Rightarrow \theta =-\frac{\pi }{4},\pi -\frac{\pi }{4},2\pi -\frac{\pi }{4},\theta =\frac{\pi }{3},\pi +\frac{\pi }{3}$

$\therefore \theta =-\frac{\pi }{4},\frac{3\pi }{4},\frac{7\pi }{4},\frac{\pi }{3},\frac{4\pi }{3}$