Let $m$ be the slope of the tangent to the curve $y=e^{x} \cos x$
Then, $m=\frac{d y}{d x}=e^{x}(\cos x-\sin x)$ diff w.r.t $x$
$\Rightarrow \frac{ dm }{ dx }= e ^{ x }(\cos x -\sin x )+ e ^{ x }(-\cos x -\sin x )$ $=-2 e^{x} \sin x$
and $\frac{d^{2} m}{d x^{2}}=-2 e^{x}(\sin x+\cos x)$
Put $\frac{ dm }{ dx }=0$
$\Rightarrow \sin x =0$
$\Rightarrow x =0, \pi, 2 \pi$
Clearly $\frac{d^{2} m}{d x^{2}} >0$ for $x=\pi$
Thus, $y$ is mininum at $x=\pi$
Hence, the value of $\alpha=\pi$.