Question

If $\tan ^2 \theta-(1+\sqrt{3}) \tan \theta+\sqrt{3}=0$, then the general value of $\theta$ is

(A) $n \pi+\frac{\pi}{4}, n \pi+\frac{\pi}{3}$

(B) $n \pi-\frac{\pi}{4}, n \pi+\frac{\pi}{3}$

(C) $n \pi+\frac{\pi}{4}, n \pi-\frac{\pi}{3}$

(D) $n \pi-\frac{\pi}{4}, n \pi-\frac{\pi}{3}$

Answer 1

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

$\begin{array}{lc}\Rightarrow & (\tan \theta-\sqrt{3})(\tan \theta-1)=0 \\\Rightarrow & \theta=n \pi+\frac{\pi}{3}, n \pi+\frac{\pi}{4}\end{array}$

So, The correct option will be (A).