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Q.
The number of solutions of the given equation
$\tan \,\theta+\sec\, \theta=\sqrt{3}$, where $0 \leq \theta \leq 2 \pi$ is
Trigonometric Functions
Solution:
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 $0 \leq \theta \leq 2 \pi$ are $\frac{\pi}{6}$ and $\frac{7 \pi}{6}$. Hence, there are two solutions.