Question Number 28624 by Joel578 last updated on 28/Jan/18
Commented by Joel578 last updated on 28/Jan/18
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\:\mathrm{the}\:\mathrm{given}\:\mathrm{expression} \\ $$
Commented by Tinkutara last updated on 28/Jan/18
1?
Commented by matikzone.tk last updated on 28/Jan/18
$${piye}\:{kabare}\:{pak}\:{joel}? \\ $$
Commented by Joel578 last updated on 29/Jan/18
$${baik} \\ $$
Answered by moxhix last updated on 28/Jan/18
$$\mathrm{show}:\:\mathrm{if}\:\:\mathrm{0}<\alpha,\beta,\gamma<\pi,\:\alpha+\beta+\gamma=\pi \\ $$$$\mathrm{then}\:{tan}\alpha+{tan}\beta+{tan}\gamma={tan}\alpha{tan}\beta{tan}\gamma\:. \\ $$$$// \\ $$$${tan}\alpha=−{tan}\left(\pi−\alpha\right)=−{tan}\left(\beta+\gamma\right) \\ $$$$\:\:\:\:=−\frac{{tan}\beta+{tan}\gamma}{\mathrm{1}−{tan}\beta{tan}\gamma} \\ $$$${tan}\alpha\left(\mathrm{1}−{tan}\beta{tan}\gamma\right)=−{tan}\beta−{tan}\gamma \\ $$$$\therefore\:{tan}\alpha+{tan}\beta+{tan}\gamma={tan}\alpha{tan}\beta{tan}\gamma\:. \\ $$$$\left(\therefore\frac{{tan}\alpha+{tan}\beta+{tan}\gamma}{{tan}\alpha{tan}\beta{tan}\gamma}=\mathrm{1}\right) \\ $$$$// \\ $$$$\mathrm{now}\:\mathrm{we}\:\mathrm{set}\:\alpha=\frac{\mathrm{2}\pi}{\mathrm{2017}},\:\beta=\frac{\mathrm{1006}\pi}{\mathrm{2017}},\:\gamma=\frac{\mathrm{1009}\pi}{\mathrm{2017}}\:. \\ $$
Commented by Joel578 last updated on 29/Jan/18
$${thank}\:{you}\:{very}\:{much} \\ $$