$6 \cdot 36 \lg (\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}})$ is equal to
(A) $\frac{1}{2}$.
(B) 1 .
(C) $2 \sqrt{3}$.
(D) $\frac{1}{2} \lg 6$.
(E) $2 \sqrt{5}$.
(China Beijing Junior High School Mathematics Competition, 1984)
[Solution]
$$
\begin{aligned}
& \lg (\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}) \\
= & \frac{1}{2} \lg (\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}})^{2} \\
= & \frac{1}{2} \lg [3+\sqrt{5}+2 \sqrt{(3+\sqrt{5})(3-\sqrt{5})}+3-\sqrt{5}] \\
= & \frac{1}{2} \lg 10=\frac{1}{2} .
\end{aligned}
$$

Therefore, the correct choice is (A).