Question

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

• A. $n\mathrm{\pi }+\frac{\mathrm{\pi }}{4},\mathrm{n\pi }+\frac{\mathrm{\pi }}{3}$
• B. $n\mathrm{\pi }-\frac{\mathrm{\pi }}{4},\mathrm{n\pi }+\frac{\mathrm{\pi }}{3}$
• C. $n\mathrm{\pi }+\frac{\mathrm{\pi }}{4},\mathrm{n\pi }-\frac{\mathrm{\pi }}{3}$
• D. $n\mathrm{\pi }-\frac{\mathrm{\pi }}{4},\mathrm{n\pi }-\frac{\mathrm{\pi }}{3}$

Answer 1

The correct option is A

$n\mathrm{\pi }+\frac{\mathrm{\pi }}{4},\mathrm{n\pi }+\frac{\mathrm{\pi }}{3}$

Explanation for correct option

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

$$\begin{gathered} \Rightarrow {\tan }^2\left(\theta \right)-\tan \left(\theta \right)-\sqrt{3}\tan \left(\theta \right)+\sqrt{3}=0 \\ \Rightarrow \tan \left(\theta \right)\left(\tan \left(\theta \right)-1\right)-\sqrt{3}\left(\tan \left(\theta \right)-1\right)=0 \\ \Rightarrow \left(\tan \left(\theta \right)-\sqrt{3}\right)\left(\tan \left(\theta \right)-1\right)=0 \\ \therefore \tan \left(\theta \right)-\sqrt{3}=0|\tan (\theta )-1=0 \\ \Rightarrow \tan (\theta )=\sqrt{3}|\tan \left(\theta \right)=1 \\ \Rightarrow \theta =n\pi +\frac{\pi }{3}|\theta =n\pi +\frac{\pi }{4} \end{gathered}$$

Hence, option A is correct.