Question Number 187491 by cortano12 last updated on 18/Feb/23
$$\:\:\begin{cases}{\mathrm{sin}\:{x}+\mathrm{sin}\:{y}={a}}\\{\mathrm{cos}\:{x}+\mathrm{cos}\:{y}={b}}\end{cases} \\ $$$$\:\:\:\:\mathrm{tan}\:{x}+\mathrm{tan}\:{y}=? \\ $$
Answered by horsebrand11 last updated on 18/Feb/23
![(∗) tan (((x−y)/2))=(a/b) ⇒tan (x+y)=((2ab)/(b^2 −a^2 )) ⇒cos (x+y)=((b^2 −a^2 )/(b^2 +a^2 )) ⇒cos (x−y)=((a^2 +b^2 −2)/2) ⇒cos x cos y = (((a^2 +b^2 )^2 −4a^2 )/(4(a^2 +b^2 ))) (∴) tan x+tan y= tan (x+y)[1−tan x tan y ] = tan (x+y) [((cos (x+y))/(cos x cos y)) ] = (((2ab)/(b^2 −a^2 )))(((b^2 −a^2 )/(b^2 +a^2 )))(((4(a^2 +b^2 ))/((a^2 +b^2 )^2 −4a^2 ))) = ((8ab)/((a^2 +b^2 )^2 −4a^2 ))](https://www.tinkutara.com/question/Q187493.png)
$$\:\left(\ast\right)\:\mathrm{tan}\:\left(\frac{{x}−{y}}{\mathrm{2}}\right)=\frac{{a}}{{b}} \\ $$$$\:\:\Rightarrow\mathrm{tan}\:\left({x}+{y}\right)=\frac{\mathrm{2}{ab}}{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} } \\ $$$$\:\:\Rightarrow\mathrm{cos}\:\left({x}+{y}\right)=\frac{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} +{a}^{\mathrm{2}} } \\ $$$$\:\:\Rightarrow\mathrm{cos}\:\left({x}−{y}\right)=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}}{\mathrm{2}} \\ $$$$\:\Rightarrow\mathrm{cos}\:{x}\:\mathrm{cos}\:{y}\:=\:\frac{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{4}{a}^{\mathrm{2}} }{\mathrm{4}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)} \\ $$$$\:\left(\therefore\right)\:\mathrm{tan}\:{x}+\mathrm{tan}\:{y}=\:\mathrm{tan}\:\left({x}+{y}\right)\left[\mathrm{1}−\mathrm{tan}\:{x}\:\mathrm{tan}\:{y}\:\right] \\ $$$$\:\:=\:\mathrm{tan}\:\left({x}+{y}\right)\:\left[\frac{\mathrm{cos}\:\left({x}+{y}\right)}{\mathrm{cos}\:{x}\:\mathrm{cos}\:{y}}\:\right] \\ $$$$\:\:=\:\left(\frac{\mathrm{2}{ab}}{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }\right)\left(\frac{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} +{a}^{\mathrm{2}} }\right)\left(\frac{\mathrm{4}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)}{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{4}{a}^{\mathrm{2}} }\right) \\ $$$$\:=\:\frac{\mathrm{8}{ab}}{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{4}{a}^{\mathrm{2}} }\: \\ $$