Solve $\tan (x)+\sec (x)=\sqrt{3}, x \in[0,2 \pi]$

Question

Aicharya · Accepted Answer

Given trigonometric equation is $\begin{aligned} & \tan x+\sec x=\sqrt{3}, x \in[0,2 \pi] \ & \therefore \quad \sec x-\tan x=\frac{1}{\sqrt{3}} \end{aligned}$ So, $2 \tan x=\sqrt{3}-\frac{1}{\sqrt{3}}$ $\Rightarrow 2 \tan x=\frac{2}{\sqrt{3}} \Rightarrow \tan x=\frac{1}{\sqrt{3}} \Rightarrow x=\frac{\pi}{6}$ or $\frac{7 \pi}{6}$ Hence, option (b) is correct.

Solve \(\tan (x)+\sec (x)=\sqrt{3}, x \in[0,2 \pi]\)

Solution