Question

If tan x + sec x = $\sqrt{3}$, 0 < x < π, then x 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) $\frac{\mathrm{\pi }}{6}$

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