Question

Find the general solution of ${\tan }^2\theta +(1-\sqrt{3})\tan \theta -\sqrt{3}=0$.

Answer 1

Given,

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

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

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

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

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

$\theta =n\pi +\frac{\pi }{3}$ or, $\theta =n\pi -\frac{\pi }{4}$ for $n$ being integer.