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Question Number 111267 by I want to learn more last updated on 03/Sep/20

Answered by Her_Majesty last updated on 03/Sep/20

tanα−tanβ=((sin(α−β))/(cosαcosβ)); cosαcosβ=((cos(α−β)+cos(α+β))/2)  (a/b)=((tan10°)/(tan50°−tan40°))=(((sin10°)/(cos10°))/((sin10°)/(cos50°cos40°)))=((cos50°cos40°)/(cos10°))=  =(((cos10°+cos90°)/2)/(cos10°))=(((cos10°)/2)/(cos10°))=(1/2)

$${tan}\alpha−{tan}\beta=\frac{{sin}\left(\alpha−\beta\right)}{{cos}\alpha{cos}\beta};\:{cos}\alpha{cos}\beta=\frac{{cos}\left(\alpha−\beta\right)+{cos}\left(\alpha+\beta\right)}{\mathrm{2}} \\ $$$$\frac{{a}}{{b}}=\frac{{tan}\mathrm{10}°}{{tan}\mathrm{50}°−{tan}\mathrm{40}°}=\frac{\frac{{sin}\mathrm{10}°}{{cos}\mathrm{10}°}}{\frac{{sin}\mathrm{10}°}{{cos}\mathrm{50}°{cos}\mathrm{40}°}}=\frac{{cos}\mathrm{50}°{cos}\mathrm{40}°}{{cos}\mathrm{10}°}= \\ $$$$=\frac{\frac{{cos}\mathrm{10}°+{cos}\mathrm{90}°}{\mathrm{2}}}{{cos}\mathrm{10}°}=\frac{\frac{{cos}\mathrm{10}°}{\mathrm{2}}}{{cos}\mathrm{10}°}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Commented by I want to learn more last updated on 03/Sep/20

Thanks ma, i appreciate

$$\mathrm{Thanks}\:\mathrm{ma},\:\mathrm{i}\:\mathrm{appreciate} \\ $$

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