## Trigonometry (11th Edition) Clone

$$\tan(30^\circ+\theta)=\frac{1+\sqrt3\tan\theta}{\sqrt3-\tan\theta}$$
$$X=\tan(30^\circ+\theta)$$ 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, $$\tan(30^\circ+\theta)=\frac{1+\sqrt3\tan\theta}{\sqrt3-\tan\theta}$$