= Solution
An object at <matter-radiation equality> lies at comoving radial distance $\chi=\tau_0-\tau_{\rm eq}$ in the flat <FRW metric>. Its <angular diameter distance> is $D_A=a_{\rm eq}\chi$, while one <particle horizon> radius has physical size $a_{\rm eq}\tau_{\rm eq}$. Its small angular radius is consequently
$$
\theta\simeq\frac{\tau_{\rm eq}}{\tau_0-\tau_{\rm eq}}
\simeq\frac{\tau_{\rm eq}}{\tau_0}.
$$
The last step uses $\tau_{\rm eq}\ll\tau_0$. If $\Omega_\Lambda/\Omega_m\gg1$, the <matter-vacuum conformal age integral> tends to $I$: its integrand behaves as $y^{-1/2}$ at zero and $y^{-2}$ at infinity. Inserting the two <conformal times> gives
$$
\boxed{\theta\simeq\frac{2(\sqrt2-1)}I\,
\Omega_r^{1/2}\Omega_\Lambda^{1/6}\Omega_m^{-2/3}.}
$$
The negative matter exponent follows directly from $\Omega_m^{-1}/\Omega_m^{-1/3}$. This agrees with the original PDF; the TeX transcription drops the minus sign. If desired, $z=y^3$ evaluates the constant through the <beta function> and <gamma function>:
$$
I=\frac13\int_0^\infty z^{-5/6}(1+z)^{-1/2}dz
=\frac13B\left(\frac16,\frac13\right)
=\frac{\Gamma(1/6)\Gamma(1/3)}{3\sqrt\pi}.
$$
The infinite-upper-limit approximation requires vacuum dominance much stronger than in the illustrative parameters of part (a), so it is an asymptotic formula rather than a precision angular prediction for those values.
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