Question:
If $\tan x+\sec x=\sqrt{3}, 0 (a) $\frac{5 \pi}{6}$ (b) $\frac{2 \pi}{3}$ (c) $\frac{\pi}{6}$ (d) $\frac{\pi}{3}$
Solution:
(c) $\frac{\pi}{6}$
We have:
$\tan x+\sec x=\sqrt{3} \quad[0 $\Rightarrow \sec x+\tan x=\sqrt{3}$ $\Rightarrow \frac{1}{\cos x}+\frac{\sin x}{\cos x}=\sqrt{3}$ $\Rightarrow 1+\sin x=\sqrt{3} \cos x$ $\Rightarrow(1+\sin x)^{2}=(\sqrt{3} \cos x)^{2}$ $\Rightarrow 1+\sin ^{2} x+2 \sin x=3 \cos ^{2} x$ $\Rightarrow 1+\sin ^{2} x+2 \sin x=3\left(1-\sin ^{2} x\right)$ $\Rightarrow 4 \sin ^{2} x+2 \sin x=2$ $\Rightarrow 2 \sin ^{2} x+\sin x-1=0$ $\Rightarrow \sin x=-1, \frac{1}{2}$ Since $0 $\therefore \sin x=\frac{1}{2}$ $\therefore x=\frac{\pi}{6}$