2. If the real number $\alpha$ satisfies $\cos \alpha=\tan \alpha$, then $\frac{1}{\sin \alpha}+\cos ^{4} \alpha=$ $\qquad$
2. 2 .

From the condition, we know that $\cos ^{2} \alpha=\sin \alpha$.
$$
\begin{array}{l}
\text { Then } \frac{1}{\sin \alpha}+\cos ^{4} \alpha=\frac{\cos ^{2} \alpha+\sin ^{2} \alpha}{\sin \alpha}+\sin ^{2} \alpha \\
=(1+\sin \alpha)+\left(1-\cos ^{2} \alpha\right) \\
=2+\sin \alpha-\cos ^{2} \alpha=2 .
\end{array}
$$