$7 \cdot 24$ If $\operatorname{tg} x=\frac{2 a b}{a^{2}-b^{2}}$, where $a>b>0$ and $0^{\circ}<x<90^{\circ}$, then $\sin x$ equals
(A) $\frac{a}{b}$.
(B) $\frac{b}{a}$.
(C) $\frac{\sqrt{a^{2}-b^{2}}}{2 a}$.
(D) $\frac{\sqrt{a^{2}-b^{2}}}{2 a b}$.
(E) $\frac{2 a b}{a^{2}+b^{2}}$.
(23rd American High School Mathematics Examination, 1972)
[Solution] As shown in the figure, angle $x$ is the acute angle opposite the side of length $2 a b$ in a right-angled triangle, with the other side being of length $a^{2}-b^{2}$. Therefore, the hypotenuse is
$$
\begin{aligned}
& \sqrt{(2 a b)^{2}+\left(a^{2}-b^{2}\right)^{2}} \\
= & \sqrt{a^{4}+2 a^{2} b^{2}+b^{4}}=a^{2}+b^{2},
\end{aligned}
$$

Thus, according to the definition of sine, $\sin x=\frac{2 a b}{a^{2}+b^{2}}$.
Therefore, the answer is $(E)$.