Question

The number of distinct solutions of $\sec \theta +\tan \theta =\sqrt{3},0\le \theta \le 2\pi is$

• A. 3
• B. 5
• C. 4
• D. none of these

Answer 1

Answer: (D)

$\sec \theta +\tan \theta =\sqrt{3}$
$\Rightarrow 1+\sin \theta =\sqrt{3}\cos \theta (\cos \theta \ne 0)$
$\Rightarrow \sqrt{3}\cos \theta -\sin \theta =1$
$\Rightarrow \frac{\sqrt{3}}{2}\cos \theta -\frac{1}{2}\sin \theta =1$
$\Rightarrow \cos \theta \cos \frac{\pi }{6}-\sin \theta \sin \frac{\pi }{6}=\cos \frac{\pi }{3}$
$\Rightarrow \cos \left(\theta +\frac{\pi }{6}\right)=\cos \frac{\pi }{3}$
$\Rightarrow \theta +\frac{\pi }{6}=2n\pi \pm \frac{\pi }{3}(n\in Z)$
$\Rightarrow$ $\theta =2n\pi -\frac{\pi }{6}\pm \frac{\pi }{3}(n\in Z)$
= $2n\pi +\frac{\pi }{6}$ or $2n\pi -\frac{\pi }{2}$
But we reject the value $\theta =2n\pi -\frac{\pi }{2}$
$\because \cos \theta =0$ for this values of $\theta$
$\theta =2n\pi +\frac{\pi }{6},n\in I$
Thus, $\theta =\frac{\pi }{6}$ is the only value of $\theta \in [0,2\pi ]$