Question Number 21141 by oyshi last updated on 14/Sep/17
$${if}\:\mathrm{sin}\:\alpha\mathrm{sin}\:\beta−\mathrm{cos}\:\alpha\mathrm{cos}\:\beta+\mathrm{1}=\mathrm{0}\: \\ $$$${so}\:{proof}\:{that}\:\mathrm{1}+\mathrm{cot}\:\alpha\mathrm{tan}\:\beta=\mathrm{0} \\ $$
Answered by $@ty@m last updated on 15/Sep/17
$$\mathrm{sin}\:\alpha\mathrm{sin}\:\beta−\mathrm{cos}\:\alpha\mathrm{cos}\:\beta+\mathrm{1}=\mathrm{0}\: \\ $$$$\mathrm{cos}\:\alpha\mathrm{cos}\:\beta−\mathrm{sin}\:\alpha\mathrm{sin}\:\beta=\mathrm{1} \\ $$$$\mathrm{cos}\:\left(\alpha+\beta\right)=\mathrm{1} \\ $$$$\alpha+\beta=\mathrm{0} \\ $$$$\alpha=−\beta \\ $$$$\mathrm{tan}\alpha=\mathrm{tan}\:\left(−\beta\right) \\ $$$$\mathrm{tan}\alpha=−\mathrm{tan}\:\beta \\ $$$$\mathrm{1}=\frac{−\mathrm{tan}\:\beta}{\mathrm{tan}\:\alpha} \\ $$$$\mathrm{1}=−\mathrm{cot}\:\alpha\mathrm{tan}\:\beta \\ $$$$\mathrm{1}+\mathrm{cot}\:\alpha\mathrm{tan}\:\beta=\mathrm{0} \\ $$