Solution:

We may assume that the edges of the cube have length $1$. Denote $M = A_{1}B \cap KB_{1}$ and set $KB = x$. Since $\triangle KBM \sim \triangle A_{1}MB_{1}$ it follows that $\frac{MB}{A_{1}M} = x$. Now using the identity $A_{1}M + MB = \sqrt{2}$, we get $MB = \frac{x \sqrt{2}}{x+1}$.

Denote by $V$ the volume of the tetrahedron $KBCB_{1}$. Then
$$V = \frac{1}{3} BB_{1} \cdot S_{KBC} = \frac{x}{6}$$
On the other hand, the altitude of $KBCB_{1}$ through $B$ is equal to $h = BM \sin 60^{\circ} = \frac{x \sqrt{6}}{2(x+1)}$. Since $B_{1}K = CK = \sqrt{x^{2}+1}$ and $B_{1}C = \sqrt{2}$, we have $S_{B_{1}KC} = \frac{\sqrt{2x^{2}+1}}{2}$. Therefore
$$V = \frac{h \cdot S_{B_{1}KC}}{3} = \frac{x \sqrt{6(2x^{2}+1)}}{12(x+1)}$$
This and (1) imply that
$$\frac{x}{6} = \frac{x \sqrt{6(2x^{2}+1)}}{12(x+1)}$$
and we obtain easily that $x = \frac{1}{2}$.

Denote by $L$ the foot of the perpendicular from $B$ to $KC$. Then $KC \perp BL$ and $KC \perp BB_{1}$ which shows that $KC \perp B_{1}L$. Hence $\text{= B 1 LB}$. We get from $\triangle KBC$ that $BL = \frac{1}{\sqrt{5}}$ and therefore $\tan \alpha = \sqrt{5}$.