1. Zadatak

Svaki pribrojnik u zbroju $A$ je oblika

$$\frac{\sin(4n+1)^\circ}{\cos(2n)^\circ\cos(2n+1)^\circ}$$

za $n=0,1,\ldots,44$.

Uočimo da je

$$\begin{aligned} \frac{\sin(4n+1)^\circ}{\cos(2n)^\circ\cos(2n+1)^\circ} &=\frac{\sin\left((2n)^\circ+(2n+1)^\circ\right)} {\cos(2n)^\circ\cos(2n+1)^\circ}\\ &=\frac{\sin(2n)^\circ\cos(2n+1)^\circ+\cos(2n)^\circ\sin(2n+1)^\circ} {\cos(2n)^\circ\cos(2n+1)^\circ}\\ &=\operatorname{tg}(2n)^\circ+\operatorname{tg}(2n+1)^\circ. \end{aligned}$$

Time zbroj $A$ sada postaje

$$A=\operatorname{tg}(0^\circ)+\operatorname{tg}(1^\circ)+\cdots+\operatorname{tg}(88^\circ)+\operatorname{tg}(89^\circ).$$

Kako je

$$\operatorname{tg}(k)^\circ+\operatorname{tg}(180-k)^\circ =\operatorname{tg}(k)^\circ-\operatorname{tg}(k)^\circ=0$$

slijedi da je

$$A+B=0.$$