Question

The smallest value of θ satisfying the equation $\sqrt{3}\left(\cot \mathrm{\theta }+\tan \mathrm{\theta }\right)=4$ is
(a) $2\pi /3$
(b) $\pi /3$
(c) $\pi /6$
(d) $\pi /12$

Answer 1

(c) $\pi /6$

Given:

$$\begin{gathered} \sqrt{3}(\cot \theta +\tan \theta )=4 \\ \Rightarrow \sqrt{3}\left(\frac{\cos \theta }{\sin \theta }+\frac{\sin \theta }{\cos \theta }\right)=4 \\ \Rightarrow \sqrt{3}({\cos }^2\theta +{\sin }^2\theta )=4\sin \theta \cos \theta \\ \Rightarrow \sqrt{3}=2\sin 2\theta [\sin 2\theta =2\sin \theta \cos \theta ] \\ \Rightarrow \sin 2\theta =\frac{\sqrt{3}}{2} \\ \Rightarrow \sin 2\theta =\sin \frac{\mathrm{\pi }}{3} \\ \Rightarrow 2\theta =n\mathrm{\pi }+(-1{)}^{\mathrm{n}}\frac{\mathrm{\pi }}{3},n\in Z \\ \Rightarrow \theta =\frac{n\mathrm{\pi }}{2}+(-1{)}^n\frac{\mathrm{\pi }}{6},n\in Z \end{gathered}$$
To obtain the smallest value of $\theta$, we will put $n=0$ in the above equation.
Thus, we have:
$\theta =\frac{\mathrm{\pi }}{6}$
Hence, the smallest value of $\theta$ is $\frac{\mathrm{\pi }}{6}$.