Question Number 141423 by naka3546 last updated on 18/May/21
$$\mathrm{tan}\:\left({x}+{y}\right)=\:\frac{\mathrm{12}}{\mathrm{5}} \\ $$$$\mathrm{sin}\:\left({x}−{y}\right)\:=\:\frac{\mathrm{3}}{\mathrm{5}} \\ $$$${x}+{y}\:,\:{x}−{y}\:\:{are}\:\:{acute}\:\:{angles}\:. \\ $$$$\mathrm{tan}\:{x}\:\mathrm{tan}\:{y}\:=\:\:? \\ $$
Answered by mr W last updated on 18/May/21
$${u}=\mathrm{tan}\:{x},\:{v}=\mathrm{tan}\:{y} \\ $$$$\frac{{u}+{v}}{\mathrm{1}−{uv}}=\frac{\mathrm{12}}{\mathrm{5}}\:\Rightarrow{u}+{v}=\frac{\mathrm{12}\left(\mathrm{1}−{uv}\right)}{\mathrm{5}} \\ $$$$\mathrm{tan}\:\left({x}−{y}\right)=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\frac{{u}−{v}}{\mathrm{1}+{uv}}=\frac{\mathrm{3}}{\mathrm{4}}\:\Rightarrow{u}−{v}=\frac{\mathrm{3}\left(\mathrm{1}+{uv}\right)}{\mathrm{4}} \\ $$$${u}=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{12}\left(\mathrm{1}−{uv}\right)}{\mathrm{5}}+\frac{\mathrm{3}\left(\mathrm{1}+{uv}\right)}{\mathrm{4}}\right)=\frac{\mathrm{63}−\mathrm{33}{uv}}{\mathrm{40}} \\ $$$${u}=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{12}\left(\mathrm{1}−{uv}\right)}{\mathrm{5}}−\frac{\mathrm{3}\left(\mathrm{1}+{uv}\right)}{\mathrm{4}}\right)=\frac{\mathrm{33}−\mathrm{63}{uv}}{\mathrm{40}} \\ $$$${uv}=\frac{\mathrm{63}−\mathrm{33}{uv}}{\mathrm{40}}×\frac{\mathrm{33}−\mathrm{63}{uv}}{\mathrm{40}} \\ $$$$\mathrm{2079}\left({uv}\right)^{\mathrm{2}} −\mathrm{6658}\left({uv}\right)+\mathrm{2079}=\mathrm{0} \\ $$$${uv}=\frac{\mathrm{3329}\pm\mathrm{2600}}{\mathrm{2079}}=\frac{\mathrm{77}}{\mathrm{27}}\:{or}\:\:\frac{\mathrm{27}}{\mathrm{77}} \\ $$
Commented by naka3546 last updated on 25/May/21
$$\mathrm{Thank}\:\mathrm{you},\:\mathrm{sir}. \\ $$