5. Zadatak

Kako je $\sin(\alpha+\beta)=\sin(180^\circ-\gamma)=\sin\gamma$, to početni izraz možemo pisati kao

$$\begin{aligned} \frac{\operatorname{ctg}\gamma}{\operatorname{ctg}\alpha+\operatorname{ctg}\beta} &=\frac{\dfrac{\cos\gamma}{\sin\gamma}}{\dfrac{\sin(\alpha+\beta)}{\sin\alpha\sin\beta}}\\ &=\frac{\cos\gamma\sin\alpha\sin\beta}{\sin\gamma\sin\gamma}\\ &=\cos\gamma\cdot\frac{\sin\alpha}{\sin\gamma}\cdot\frac{\sin\beta}{\sin\gamma}\\ &=\frac{a^2+b^2-c^2}{2ab}\cdot\frac ac\cdot\frac bc, \end{aligned}$$

gdje smo u zadnjoj jednakosti primijenili poučak o sinusu i kosinusu. Na kraju,

$$\frac{a^2+b^2-c^2}{2ab}\cdot\frac ac\cdot\frac bc =\frac{a^2+b^2-c^2}{2c^2} =\frac{2015c^2-c^2}{2c^2}=1007.$$