= Solution
Opposite directions replace $\cos\theta$ by $-\cos\theta$. The even term cancels from the numerator and the odd term from the denominator. For $(P,Q)\ne(0,0)$, the <angular asymmetry of a chiral charged-current interaction> is
$$
\boxed{A(\theta)=\frac{H_2\cos\theta}{H_1(1+\cos^2\theta)}=\frac{8P^2Q^2}{(P^2+Q^2)^2}\frac{\cos\theta}{1+\cos^2\theta}.}
$$
Here $\theta+\pi$ denotes the antipodal direction in the stated angular law. If $P=Q=0$, both rates vanish and the asymmetry is undefined.
The overall <Fermi constant>, energy-dependent prefactor and common normalization cancel. Thus this observable constrains the relative <vector current> and <axial current> couplings of the <weak charged current> rather than only their total strength. Pure <vector currents> or pure <axial currents> give zero asymmetry; $P=\pm Q$ gives the maximal coefficient, $A=2\cos\theta/(1+\cos^2\theta)$. The inequality $4P^2Q^2\leq(P^2+Q^2)^2$ ensures $|A|\leq1$.
It does not uniquely identify the handedness: the formula is insensitive to $P\mapsto-P$, $Q\mapsto-Q$, and interchanging $P,Q$. In particular equally strong purely left- and right-handed <weak charged currents> at both vertices are indistinguishable in this unpolarized measurement. Additional spin-sensitive observables would be needed to remove that degeneracy. The cancellation and constraints are within the massless parton approximation specified for the calculation.
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