 Solution

Set $z=e^{i\theta }$. Then $d\theta =dz/(iz)$ and $\tan \theta =(z^2-1)/[i(z^2+1)]$, so the requested integral isThe denominator factors asFor $0<a<1$, its two inner roots are $z=\pm r$, $r=\sqrt{(1-a)/(1+a)}<1$, and the other two roots have modulus $1/r>1$. There is also a simple pole at zero. Their residues areThe latter values follow by differentiating the quadratic product, or from the supplied hint. The sum of the enclosed residues isThe residue theorem therefore givesAt the angles where the tangent is undefined, the original reciprocal has a removable zero limit. Its continuous extension gives the same integral; the transformed rational integrand has no pole on the unit circle.