If tan x + sec x =

(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

$\therefore \sin x=\frac{1}{2}$

$\therefore x=\frac{\pi}{6}$