1. Given $\frac{\sin (\beta+\gamma) \cdot \sin (\gamma+\alpha)}{\cos \alpha \cdot \cos \beta}=\frac{4}{9}$. Then the value of $\frac{\sin (\beta+\gamma) \cdot \sin (\gamma+\alpha)}{\cos (\alpha+\beta+\gamma) \cdot \cos \gamma}$ is $\qquad$ .
$$
\begin{array}{l}
=1 \cdot \frac{4}{5} . \\
\sin (\beta+\gamma) \cdot \sin (\gamma+\alpha) \\
=\frac{1}{2}[\cos (\alpha-\beta)-\cos (\alpha+\beta+2 \gamma)], \\
\cos \alpha \cdot \cos \beta=\frac{1}{2}[\cos (\alpha+\beta)+\cos (\alpha-\beta)], \\
\cos (\alpha+\beta+\gamma) \cdot \cos \gamma \\
=\frac{1}{2}[\cos (\alpha+\beta+2 \gamma)+\cos (\alpha+\beta)] .
\end{array}
$$

Let $a=\cos (\alpha-\beta), b=\cos (\alpha+\beta+2 \gamma) \cdot c=\cos (\alpha+\beta)$, then the known conditions become $A=\frac{a-b}{a+c}=\frac{4}{9}$.
$$
\therefore B=\frac{\sin (\beta+\gamma) \cdot \sin (\gamma+\alpha)}{\cos (\alpha+\beta+\gamma) \cdot \cos \gamma}=\frac{a-b}{b+c} \text {. }
$$

Then $\frac{1}{B}=\frac{b+c}{a-b}=\frac{(a+c)-(a-b)}{a-b}=\frac{1}{A}-1=\frac{5}{4}$. Hence $B=\frac{4}{5}$.