= Solution
Let $s(\theta)$ be the $2\pi$-periodic ramp represented by $\theta$ on $(-\pi,\pi)$. The <Fourier series of a periodic sawtooth> gives
$$
s(\theta)=2\sum_{n=1}^\infty\frac{(-1)^{n+1}}n\sin(n\theta).
$$
Both boundary data are multiples of this odd function, so their angular means and cosine coefficients vanish. Consequently $A_0=C_0=A_n=C_n=0$. Put $h_n=2(-1)^{n+1}/n$. Matching the sine coefficients at the two radii gives $B_na^n+D_na^{-n}=h_n\mu_a$ and $B_nb^n+D_nb^{-n}=h_n\mu_b$. Solving this two-by-two <linear system> yields
$$
\boxed{B_n=h_n\frac{\mu_bb^n-\mu_aa^n}{b^{2n}-a^{2n}},\qquad
D_n=h_n\frac{a^nb^n(\mu_ab^n-\mu_ba^n)}{b^{2n}-a^{2n}}.}
$$
Equivalently, <two-boundary annular harmonic interpolation> expresses the same solution as
$$
\boxed{T(r,\theta)=\sum_{n=1}^\infty h_n\frac{\mu_a\sinh(n\log(b/r))+\mu_b\sinh(n\log(r/a))}{\sinh(n\log(b/a))}\sin(n\theta).}
$$
Every term is a <harmonic function>, and the radial factors enforce the specified boundary values. For a radius strictly between $a$ and $b$, their geometric decay makes the series smooth. At the boundary discontinuity $\theta=\pi$, the limiting Fourier value is the mean of the two one-sided values, zero. The boundary data are imposed away from that exceptional point, as stated.
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