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Question-148289




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

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