Question Number 148289 by saly last updated on 26/Jul/21
Commented by saly last updated on 26/Jul/21
$$\:{hepl}\:{me} \\ $$
Answered by mr W last updated on 26/Jul/21
$$\gamma=\frac{\pi}{\mathrm{2}}−\left(\alpha+\beta\right) \\ $$$$\mathrm{tan}\:\gamma=\frac{\mathrm{1}}{\mathrm{tan}\:\left(\alpha+\beta\right)} \\ $$$$\mathrm{tan}\:\alpha\:\mathrm{tan}\:\beta+\mathrm{tan}\:\beta\:\mathrm{tan}\:\gamma+\mathrm{tan}\:\alpha\mathrm{tan}\:\gamma \\ $$$$=\mathrm{tan}\:\alpha\:\mathrm{tan}\:\beta+\frac{\mathrm{tan}\:\alpha+\mathrm{tan}\:\beta}{\mathrm{tan}\:\left(\alpha+\beta\right)} \\ $$$$=\mathrm{tan}\:\alpha\:\mathrm{tan}\:\beta+\frac{\mathrm{tan}\:\alpha+\mathrm{tan}\:\beta}{\frac{\mathrm{tan}\:\alpha+\mathrm{tan}\:\beta}{\mathrm{1}−\mathrm{tan}\:\alpha\mathrm{tan}\:\beta}} \\ $$$$=\mathrm{tan}\:\alpha\:\mathrm{tan}\:\beta+\mathrm{1}−\mathrm{tan}\:\alpha\mathrm{tan}\:\beta \\ $$$$=\mathrm{1} \\ $$
Commented by puissant last updated on 26/Jul/21
$${good}\:{sir}\:{nice}.. \\ $$