AMC 12 Mock Exam 2 Problem 7

Rewrite the equation as $\sin\theta=2(1-\cos\theta)$. Using $\sin\theta=2\sin\frac\theta2\cos\frac\theta2$ and $1-\cos\theta=2\sin^2\frac\theta2$, this becomes $2\sin\frac\theta2\cos\frac\theta2=4\sin^2\frac\theta2$. Since $\sin\frac\theta2\ne0$, $\tan\frac\theta2=\frac12$, and $$\tan\theta=\frac{2\cdot\frac12}{1-\frac14}=\boxed{\tfrac43}.$$ (Check: $\sin\theta=\frac45$, $\cos\theta=\frac35$, and $\frac45+\frac65=2$.)