The answer is no, there are transcendental numbers $\theta$ which aren't rational multiples of $\pi$ such that $\tan(\theta)$ is algebraic. Indeed, since $\tan$ is onto $\mathbb R$, there exists a value of $\theta$ for which $\tan(\theta)=\sqrt{2}$. On one hand, by Lindemann-Weierstrass theorem, $\theta$ has to be transcendental, for otherwise $e^{i\theta}$ would be transcendental and the equality
$$-\frac{i}{2}\frac{e^{i\theta}-e^{-i\theta}}{e^{i\theta}+e^{-i\theta}}=\tan(\theta)=\sqrt{2}$$
would be impossible. So now we want to argue that $\theta$ is not a rational multiple of $\pi$. Let me only outline the solution.
Suppose $\theta$ was a rational multiple of $\pi$. Then, because $\tan$ is $\pi$-periodic, the sequence $\tan(2^n\theta)$ for $n=0,1,2,\dots$ would take only finitely many values (since it only depends on $n$ modulo the denominator of $\theta/\pi$). However, by repeatedly using the tangent of double angle formula, we can derive $\tan(2^n\theta)=\frac{p_n}{q_n}\sqrt{2}$, where the sequences $p_n,q_n$ are defined by $p_0=q_0=1,p_{n+1}=2p_nq_n,q_{n+1}=q_n^2-2p_n^2$ and we can easily check that $p_n,q_n$ are nonzero and relatively prime. But then the values of $\tan(2^n\theta)$ are all clearly pairwise distinct, since $p_n$ is strictly increasing, which is a contradiction.
Just for fun, we can now conclude that $\theta$ is not an algebraic multiple of $\pi$ either, for if $\theta=\alpha\pi$, then $\alpha$ would be irrational and we would deduce that $e^{i\theta}=(e^{i\pi})^\alpha=(-1)^\alpha$ is transcendental by the Gelfond-Schneider theorem.
