Question

If tan θ + sec θ = $\sqrt{3}$, 0<θ<π, then θ is equal to
(a) $\frac{5\mathrm{\pi }}{6}$
(b) $\frac{2\mathrm{\pi }}{3}$
(c) $\frac{\mathrm{\pi }}{6}$
(d) $\frac{\mathrm{\pi }}{3}$

Answer 1

(c) $\frac{\mathrm{\pi }}{6}$

$$\begin{gathered} \mathrm{We}\mathrm{have}: \\ \tan \theta +sec\theta =\sqrt{3}\left[0<\mathrm{\theta }<\mathrm{\pi }\right] \\ \Rightarrow sec\theta +\tan \theta =\sqrt{3} \\ \Rightarrow \frac{1}{\cos \theta }+\frac{\sin \theta }{\cos \theta }=\sqrt{3} \\ \Rightarrow 1+\sin \theta =\sqrt{3}\cos \theta \end{gathered}$$
$$\begin{gathered} \Rightarrow {\left(1+\sin \theta \right)}^2={\left(\sqrt{3}\cos \theta \right)}^2 \\ \Rightarrow 1+{\sin }^2\theta +2\sin \theta =3{\cos }^2\theta \\ \Rightarrow 1+{\sin }^2\theta +2\sin \theta =3(1-{\sin }^2\theta ) \\ \Rightarrow 4{\sin }^2\theta +2\sin \theta =2 \\ \Rightarrow 2{\sin }^2\theta +\sin \theta -1=0 \\ \Rightarrow \sin \theta =-1,\frac{1}{2} \\ \mathrm{Since}0<\theta <\mathrm{\pi },\sin \theta \mathrm{cannot}\mathrm{be}\mathrm{negative}. \\ \therefore \mathrm{sin\theta }=\frac{1}{2} \\ \therefore \theta =\frac{\mathrm{\pi }}{6} \end{gathered}$$