= Solution
Initially suppose $0<\alpha<\pi/2$, the usual oscillatory range. The <oscillatory phase> is $n\Phi(\vartheta)$ with $\Phi=\sec\alpha\sin\vartheta-\vartheta$. Its <stationary point> is $\vartheta=\alpha$, since $\sec\alpha\cos\vartheta=1$. At that point
$$
\Phi(\alpha)=\tan\alpha-\alpha,\qquad \Phi''(\alpha)=-\tan\alpha.
$$
The <stationary phase method> therefore gives the <Debye Bessel asymptotic>
$$
\boxed{J_n(n\sec\alpha)=\sqrt{\frac{2}{\pi n\tan\alpha}}
\cos\left[n(\tan\alpha-\alpha)-\frac\pi4\right]+O(n^{-3/2})}.
$$
The error is additive for fixed positive $\alpha$ bounded away from $\pi/2$; this formula is not uniform as $\alpha\to0$, when the <stationary point> joins the endpoint and its curvature vanishes.
The printed condition $\alpha>0$ alone includes other trigonometric branches. All defined fixed real cases can be covered as follows. If $0<|\cos\alpha|<1$, put $\beta=\arccos|\cos\alpha|\in(0,\pi/2)$. Use the displayed formula with $\beta$ instead of $\alpha$, and multiply it by $1$ if $\sec\alpha>0$, or by $(-1)^n$ if $\sec\alpha<0$. This follows from the integer-order parity $J_n(-x)=(-1)^nJ_n(x)$, which is also obtained from the defining integral by $\vartheta\mapsto\pi-\vartheta$. If $|\cos\alpha|=1$, the argument is $\pm n$ and the turning-point answer below applies, with the same parity factor. If $\cos\alpha=0$, the stated argument is undefined.
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