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Question Number 8287 by lepan last updated on 06/Oct/16
Show that tan(α+β)=((tanα+tanβ)/(1−tanαtanβ)).
$${Show}\:{that}\:{tan}\left(\alpha+\beta\right)=\frac{{tan}\alpha+{tan}\beta}{\mathrm{1}−{tan}\alpha{tan}\beta}. \\ $$
Answered by ridwan balatif last updated on 06/Oct/16
tan(α+β)=((sin(α+β))/(cos(α+β)))                         =((sinαcosβ+sinβcosα)/(cosαcosβ−sinαsinβ))                         =(((sinαcosβ+sinβcosα).((1/(cosαcosβ))))/((cosαcosβ−sinαsinβ).((1/(cosαcosβ)))))                        =((tanα+tanβ)/(1−tanαtanβ))
$$\mathrm{tan}\left(\alpha+\beta\right)=\frac{\mathrm{sin}\left(\alpha+\beta\right)}{\mathrm{cos}\left(\alpha+\beta\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{sin}\alpha\mathrm{cos}\beta+\mathrm{sin}\beta\mathrm{cos}\alpha}{\mathrm{cos}\alpha\mathrm{cos}\beta−\mathrm{sin}\alpha\mathrm{sin}\beta} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\left(\mathrm{sin}\alpha\mathrm{cos}\beta+\mathrm{sin}\beta\mathrm{cos}\alpha\right).\left(\frac{\mathrm{1}}{\mathrm{cos}\alpha\mathrm{cos}\beta}\right)}{\left(\mathrm{cos}\alpha\mathrm{cos}\beta−\mathrm{sin}\alpha\mathrm{sin}\beta\right).\left(\frac{\mathrm{1}}{\mathrm{cos}\alpha\mathrm{cos}\beta}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{tan}\alpha+\mathrm{tan}\beta}{\mathrm{1}−\mathrm{tan}\alpha\mathrm{tan}\beta} \\ $$$$ \\ $$

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