20.43 If $0^{\circ}<\alpha<90^{\circ}$, then the sum of the radii of the incircle and circumcircle of $\triangle A B C$ with sides $\sin \alpha, \cos \alpha, \operatorname{tg} \alpha \operatorname{ctg} \alpha$ is 
(A) $\frac{\sin \alpha+\cos \alpha}{2}$.
(B) $\frac{\operatorname{tg} \alpha+\operatorname{ctg} \alpha}{2}$.
(C) $2 \sin \alpha \cos \alpha$.
(D) $\frac{1}{\sin \alpha \cos \alpha}$.
(Anhui Province, China Junior High School Mathematics Competition, 1997)
[Solution] From $\sin ^{2} \alpha+\cos ^{2} \alpha=1$ and $\operatorname{tg} \alpha \operatorname{ctg} \alpha=1$, we know that $\triangle A B C$ is a right triangle with the hypotenuse length of 1, so the circumradius $R=\frac{1}{2}$ of $\triangle A B C$.
Also, the inradius $r=\frac{\sin \alpha+\cos \alpha-1}{2}$ of the right triangle $\triangle A B C$,
$\therefore$
$$
R+r=\frac{\sin \alpha+\cos \alpha}{2}
$$

Therefore, the answer is $(A)$.