= Solution
Write $A=r^2+a^2$, $s=\sin\theta$, $\Sigma=A-a^2s^2$, and $\Delta=A-2Mr$. Use <ingoing Kerr coordinates>, so $dt=dv-A\,dr/\Delta$ and $d\phi=d\chi-a\,dr/\Delta$. To see the cancellations without expanding every term, write the <Kerr metric> in the equivalent form
$$
ds^2=-\frac\Delta\Sigma(dt-a s^2d\phi)^2+\frac\Sigma\Delta dr^2+\Sigma d\theta^2+\frac{s^2}\Sigma(A\,d\phi-a\,dt)^2.
$$
The combinations become
$$
dt-a s^2d\phi=dv-a s^2d\chi-\frac\Sigma\Delta dr,\qquad A\,d\phi-a\,dt=A\,d\chi-a\,dv.
$$
The first square contributes $-\Sigma dr^2/\Delta$, cancelling the explicit radial term. Expanding the remaining terms gives
$$
\boxed{\begin{aligned}
ds^2={}&-\left(1-\frac{2Mr}\Sigma\right)dv^2+2\,dv\,dr-\frac{4Mar s^2}\Sigma\,dv\,d\chi-2a s^2\,dr\,d\chi\\
&+\Sigma\,d\theta^2+\left(A+\frac{2Ma^2r s^2}\Sigma\right)s^2d\chi^2.
\end{aligned}}
$$
There is no denominator $\Delta$ in this <Lorentzian metric>. At the outer horizon $r_+=M+\sqrt{M^2-a^2}$, $\Sigma>0$ and the components are smooth. The determinant is $-\Sigma^2\sin^2\theta$, so away from the usual polar-coordinate degeneracy the metric is nondegenerate and extends across $r_+$. The axis can be covered by regular angular charts. Thus the <Boyer-Lindquist coordinates> are singular there, while the <ingoing Kerr coordinates> are regular at the future horizon.
For physical nonextremality the invariant parameter condition is $M>|a|$, $M>0$. The printed $M>a$ is sufficient when the rotation orientation has been chosen so that $a\geq0$; without that convention it needs the absolute value.
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