11. Given that $a, b$ are integers, and satisfy $4 \cos 70^{\circ}=\sqrt{a+b \csc 70^{\circ}}$. Find all possible values of $a, b$.
11. Solution: By the triple angle formula, we have $\sin 3 \cdot 70^{\circ}=3 \sin 70^{\circ}-4 \sin ^{3} 70^{\circ}=-\frac{1}{2}$.

Thus, $16 \cos ^{2} 70^{\circ} \cdot \sin 70^{\circ}=16 \sin 70^{\circ}-16 \sin ^{3} 70^{\circ}=4 \sin 70^{\circ}-2$, which means
$16 \cos ^{2} 70^{\circ}=4-2 \csc 70^{\circ}$, and since $4-2 \csc 70^{\circ}>0$, we have $4 \cos 70^{\circ}=\sqrt{4-2 \csc 70^{\circ}}$.
Therefore, $a=4, b=-2$ satisfy the conditions. Suppose there exists another pair $\left(a_{0}, b_{0}\right)$ that satisfies the given conditions, then $a_{0}, b_{0} \in \mathbf{Z}$, such that $a_{0}+b_{0} \csc 70^{\circ}=4-2 \csc 70^{\circ}$, i.e., $\left(b_{0}+2\right) \csc 70^{\circ}=4-a_{0}$.
Since $a_{0} \neq 4, b_{0} \neq-2$, we have $\sin 70^{\circ}=\frac{b_{0}+2}{4-a_{0}}$.
And $\sin 70^{\circ}$ is an irrational number. Otherwise, we can assume $\sin 70^{\circ}=\frac{p}{q},(p, q)=1, p, q \in \mathbf{N}_{+}$. Substituting into the triple angle formula and rearranging gives $8 p^{3}-$ $6 p q^{2}-q^{3}=0$. This implies $\left\{\begin{array}{l}p \mid q^{3} \\ q^{2} \mid 8 p^{3}\end{array}\right.$, and since $(p, q)=1$, we have $\left\{\begin{array}{l}p=1 \\ q=1 \text { or } 2\end{array}\right.$, substituting these values does not hold.
Therefore, the pair $(a, b)$ that satisfies the conditions is unique.