Question

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

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

Answer 1

The correct option is D none of these

$sec\theta +tan\theta =\sqrt{3}$
$\Rightarrow 1+sin\theta =\sqrt{3}cos\theta$
$\Rightarrow \sqrt{3}cos\theta -sin\theta =1$
$\Rightarrow \frac{\sqrt{3}}{2}cos\theta -\frac{1}{2}sin\theta =\frac{1}{2}$
$\Rightarrow cos\frac{\pi }{6}cos\theta -sin\frac{\pi }{6}sin\theta =\frac{1}{2}$
$\Rightarrow cos(\frac{\pi }{6}+\theta )=cos\frac{\pi }{3}$
therefore general solution is given by,
$\theta =2n\pi -\frac{\pi }{6}\pm \frac{\pi }{3}$
$\Rightarrow \theta =2n\pi +\frac{\pi }{6}$ and $\theta =2n\pi -\frac{\pi }{2}$
but $\theta =2n\pi -\frac{\pi }{2}$ , which does not satisfy the given equation.
Hence solution in $0\le \theta \le 3\pi$ is,
$\theta =\frac{\pi }{6}and\frac{13\pi }{6}$
Hence, option 'D' is correct.