6.44 $A$ is an acute angle, $\lg (1+\sin A)=m, \lg \frac{1}{1-\sin A}=n$, then the value of $\lg \cos A$ is
(A) $m+\frac{1}{n}$.
(B) $m-n$.
(C) $\frac{1}{2}\left(m+\frac{1}{n}\right)$.
(D) $\frac{1}{2}(m-n)$.
(E) $2(m-n)$.
(Beijing Junior High School Mathematics Competition, 1983)
[Solution] Given $\lg (1+\sin A)=m$, and
$$
\begin{array}{rlrl} 
& \lg \frac{1}{1-\sin A}= & \lg (1-\sin A)^{-1}=-\lg (1-\sin A), \\
\therefore \quad & \lg (1-\sin A)=-n . \\
\therefore \quad & \lg \cos A & =\frac{1}{2} \lg \cos ^{2} A=\frac{1}{2} \lg \left(1-\sin ^{2} A\right) \\
& =\frac{1}{2} \lg (1+\sin A)(1-\sin A) \\
& =\frac{1}{2}[\lg (1+\sin A)+\lg (1-\sin A)] \\
& =\frac{1}{2}(m-n) .
\end{array}
$$

Therefore, the answer is $(D)$.