Question

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

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (C)

We have, $\sec \theta +\tan \theta =\sqrt{3}...(i)$
$\Rightarrow \sec \theta -\tan \theta =\frac{1}{\sqrt{3}}$
$\left[\because {\sec }^2\theta -{\tan }^2\theta =1\right]$
By solving (i) and (ii), we get
$\tan \theta =\frac{1}{2}\left(\sqrt{3}-\frac{1}{\sqrt{3}}\right)=\frac{1}{\sqrt{3}}$
$\therefore \tan \theta =\tan \left(\frac{\pi }{6}\right)$
$\Rightarrow \theta =n\pi +\frac{\pi }{6}$
$\therefore$ Solutions for $0\le \theta \le 2\pi$ are $\frac{\pi }{6}$ and $\frac{7\pi }{6}$.
Hence, there are two solutions.