←Vrati se na zadatke1. ZadatakIzračunaj tg192∘+tg48∘1+tg168∘⋅tg408∘.\frac{\operatorname{tg}192^\circ+\operatorname{tg}48^\circ}{1+\operatorname{tg}168^\circ\cdot\operatorname{tg}408^\circ}.1+tg168∘⋅tg408∘tg192∘+tg48∘.RješenjeBudući da je tg192∘=tg(12∘+180∘)=tg12∘,tg408∘=tg(48∘+360∘)=tg48∘,tg168∘=tg(−12∘+180∘)=−tg12∘,\begin{aligned} \operatorname{tg}192^\circ&=\operatorname{tg}(12^\circ+180^\circ)=\operatorname{tg}12^\circ,\\ \operatorname{tg}408^\circ&=\operatorname{tg}(48^\circ+360^\circ)=\operatorname{tg}48^\circ,\\ \operatorname{tg}168^\circ&=\operatorname{tg}(-12^\circ+180^\circ)=-\operatorname{tg}12^\circ, \end{aligned}tg192∘tg408∘tg168∘=tg(12∘+180∘)=tg12∘,=tg(48∘+360∘)=tg48∘,=tg(−12∘+180∘)=−tg12∘, zadani izraz je jednak tg12∘+tg48∘1−tg12∘⋅tg48∘=tg(12∘+48∘)=tg60∘=3.\frac{\operatorname{tg}12^\circ+\operatorname{tg}48^\circ}{1-\operatorname{tg}12^\circ\cdot\operatorname{tg}48^\circ} =\operatorname{tg}(12^\circ+48^\circ) =\operatorname{tg}60^\circ=\sqrt3.1−tg12∘⋅tg48∘tg12∘+tg48∘=tg(12∘+48∘)=tg60∘=3.
