ANSWER:
A.
[tex]\frac{\sqrt[]{3}\tan x+1}{\sqrt[]{3}-\tan x}[/tex]STEP-BY-STEP EXPLANATION:
We have the following equation:
[tex]tan\mleft(x+\frac{\pi}{6}\mright)[/tex]We solve to calculate its equivalence:
[tex]\begin{gathered} \tan x=\frac{\sin x}{\cos x} \\ \text{therefore:} \\ \frac{\sin (x+\frac{\pi}{6})}{\cos (x+\frac{\pi}{6})} \\ \sin (s+t)=\sin s\cdot\cos t+\cos s\cdot\sin t \\ \cos (s+t)=\cos s\cos t-\sin s\cdot\sin t \\ \text{ replacing:} \\ \frac{\sin x\cdot\cos\frac{\pi}{6}+\cos x\cdot\sin\frac{\pi}{6}}{\cos x\cos\frac{\pi}{6}-\sin x\cdot\sin\frac{\pi}{6}}=\frac{\frac{\sqrt[]{3}}{2}\sin x+\frac{1}{2}\cos x}{\frac{\sqrt[]{3}}{2}\cos x-\frac{1}{2}\sin x} \\ \cos x=\frac{1}{\sec x} \\ \frac{\frac{\sqrt[]{3}}{2}\sin x+\frac{1}{2}\frac{1}{\sec x}}{\frac{\sqrt[]{3}}{2}\frac{1}{\sec x}-\frac{1}{2}\sin x}=\frac{\frac{1+\sqrt[]{3}\cdot\sin x\cdot\sec x}{2\sec x}}{\frac{\sqrt[]{3}-\sin x\cdot\sec x}{2\sec x}}=\frac{1+\sqrt[]{3}\cdot\sin x\cdot\sec x}{\sqrt[]{3}-\sin x\cdot\sec x} \\ \sin x\cdot\sec x=\tan x \\ \text{ replacing} \\ \frac{1+\sqrt[]{3}\cdot\tan x}{\sqrt[]{3}-\tan x}=\frac{\sqrt[]{3}\tan x+1}{\sqrt[]{3}-\tan x} \end{gathered}[/tex]