Question

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

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

Answer 1

Answer: (D)

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