11. Given that $\alpha$ is an interior angle of a triangle, and satisfies $\frac{1}{\sin \alpha}+\frac{1}{\cos \alpha}=2$, then $\alpha=(\quad$.
A. $\pi+\frac{1}{2} \arcsin \frac{1-\sqrt{5}}{2}$
B. $\frac{\pi}{2}-\frac{1}{2} \arcsin \frac{1-\sqrt{5}}{2}$
C. $\frac{1}{2} \arcsin \frac{1-\sqrt{5}}{2}$
D. $\frac{\pi}{2}+\frac{1}{2} \arcsin \frac{1-\sqrt{5}}{2}$
11. A. The original equation can be transformed into $\sin \alpha+\cos \alpha=\sin 2 \alpha$, squaring gives $\sin ^{2} 2 \alpha-\sin 2 \alpha-1=0$, solving yields $\sin 2 \alpha=\frac{1-\sqrt{5}}{2}<0$. Given that $\alpha$ is an obtuse angle, the general solution is $\alpha=\frac{1}{2}\left[k \pi+(-1)^{k} \arcsin \frac{1-\sqrt{5}}{2}\right]$. Taking $k=0$, $\alpha=-\frac{1}{2} \operatorname{arc} \sin \frac{1-\sqrt{5}}{2}<0$ is discarded; taking $k=1$ is also discarded; taking $k=2$ satisfies.