5. Given $\tan A_{1} \cdot \tan A_{2} \cdot \tan A_{3} \cdots \tan A_{n}=1, \sin A_{1} \cdot \sin A_{2} \cdots \sin A_{n}$ the maximum value is ( ).
A. $2^{-n}$
B. $2^{-\frac{n}{3}}$
C. $2^{-\frac{n}{2}}$
D. $2^{-\frac{n}{3}}$
5. C.

First, when $A_{1}=A_{2}=A_{3}=\cdots=A_{n}=\frac{\pi}{4}$, we have $\tan A_{1} \cdot \tan A_{2} \cdots \tan A_{n}=1$, and $\sin A_{1} \cdot$ $\sin A_{2} \cdots \sin A_{n}=2^{-\frac{n}{2}}$. On the other hand, $\sec ^{2} A_{1} \cdot \sec ^{2} A_{2} \cdots \sec ^{2} A_{n}=\left(1+\tan A_{1}^{2}\right)\left(1+\tan ^{2} A_{2}\right) \cdots$ $\left(1+\tan ^{2} A_{n}\right) \geqslant\left(2 \tan A_{1}\right)\left(2 \tan A_{2}\right) \cdots\left(2 \tan A_{n}\right)=2^{n}$, so $\left|\sec A_{1} \cdot \sec A_{2} \cdots \sec A_{n}\right| \geqslant 2^{\frac{n}{2}}$, thus $\left|\cos A_{1} \cos A_{2} \cdots \cos A_{n}\right| \leqslant 2^{-\frac{n}{2}}$. Given the condition $\tan A_{1} \tan A_{2} \cdots \tan A_{n}=1$, we know $\sin A_{1} \sin A_{2} \cdots \sin A_{n}=$ $\cos A_{1} \cdot \cos A_{2} \cdots \cos A_{n}$, hence $\sin A_{1} \sin A_{2} \cdots \sin A_{n} \leqslant\left|\cos A_{1} \cos A_{2} \cdots \cos A_{n}\right| \leqslant 2^{-\frac{n}{2}}$.