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 | ||
$$\:\left(\ast\right)\:\mathrm{tan}\:\left(\frac{{x}−{y}}{\mathrm{2}}\right)=\frac{{a}}{{b}} \\ $$$$\:\:\Rightarrow\mathrm{tan}\:\left({x}+{y}\right)=\frac{\mathrm{2}{ab}}{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} } \\ $$$$\:\:\Rightarrow\mathrm{cos}\:\left({x}+{y}\right)=\frac{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} +{a}^{\mathrm{2}} } \\ $$$$\:\:\Rightarrow\mathrm{cos}\:\left({x}−{y}\right)=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}}{\mathrm{2}} \\ $$$$\:\Rightarrow\mathrm{cos}\:{x}\:\mathrm{cos}\:{y}\:=\:\frac{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{4}{a}^{\mathrm{2}} }{\mathrm{4}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)} \\ $$$$\:\left(\therefore\right)\:\mathrm{tan}\:{x}+\mathrm{tan}\:{y}=\:\mathrm{tan}\:\left({x}+{y}\right)\left[\mathrm{1}−\mathrm{tan}\:{x}\:\mathrm{tan}\:{y}\:\right] \\ $$$$\:\:=\:\mathrm{tan}\:\left({x}+{y}\right)\:\left[\frac{\mathrm{cos}\:\left({x}+{y}\right)}{\mathrm{cos}\:{x}\:\mathrm{cos}\:{y}}\:\right] \\ $$$$\:\:=\:\left(\frac{\mathrm{2}{ab}}{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }\right)\left(\frac{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} +{a}^{\mathrm{2}} }\right)\left(\frac{\mathrm{4}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)}{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{4}{a}^{\mathrm{2}} }\right) \\ $$$$\:=\:\frac{\mathrm{8}{ab}}{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{4}{a}^{\mathrm{2}} }\: \\ $$ | ||