Answer
$$\tan(30^\circ+\theta)=\frac{1+\sqrt3\tan\theta}{\sqrt3-\tan\theta}$$
Work Step by Step
$$X=\tan(30^\circ+\theta)$$
According to tangent sum identity:
$$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$$
That means
$$X=\frac{\tan30^\circ+\tan\theta}{1-\tan30^\circ\tan\theta}$$
$$X=\frac{\frac{1}{\sqrt3}+\tan\theta}{1-\frac{1}{\sqrt3}\tan\theta}$$
$$X=\frac{\frac{1+\sqrt3\tan\theta}{\sqrt3}}{\frac{\sqrt3-\tan\theta}{\sqrt3}}$$
$$X=\frac{1+\sqrt3\tan\theta}{\sqrt3-\tan\theta}$$
Overall,
$$\tan(30^\circ+\theta)=\frac{1+\sqrt3\tan\theta}{\sqrt3-\tan\theta}$$
