4. Zadatak

Koristit ćemo adicijsku formulu za tangens,

$$\operatorname{tg}(x-y)=\frac{\operatorname{tg}x-\operatorname{tg}y}{1+\operatorname{tg}x\,\operatorname{tg}y},$$

tj. jednakost

$$\operatorname{tg}x\,\operatorname{tg}y=\frac{\operatorname{tg}x-\operatorname{tg}y}{\operatorname{tg}(x-y)}-1.$$

Uvrstimo $x=n+1$, $y=n$:

$$\operatorname{tg}(n+1)\,\operatorname{tg}n=\frac{\operatorname{tg}(n+1)-\operatorname{tg}n}{\operatorname{tg}1}-1.$$

Zbrajanjem se ponište svi članovi osim prvog i posljednjeg:

$$\begin{aligned} \sum_{n=1}^{2012}\operatorname{tg}n\,\operatorname{tg}(n+1) &=\frac{\operatorname{tg}2-\operatorname{tg}1}{\operatorname{tg}1}-1+\frac{\operatorname{tg}3-\operatorname{tg}2}{\operatorname{tg}1}-1+\cdots+\frac{\operatorname{tg}2013-\operatorname{tg}2012}{\operatorname{tg}1}-1\\ &=\frac{\operatorname{tg}2013-\operatorname{tg}1}{\operatorname{tg}1}-2012=\frac{\operatorname{tg}2013}{\operatorname{tg}1}-2013. \end{aligned}$$