Question Number 107932 by bemath last updated on 13/Aug/20
$$\:\:\:\frac{\mathbb{B}{e}\mathbb{M}{ath}}{\bullet} \\ $$$${Given}\:\begin{cases}{\mathrm{tan}\:\left({x}−{y}\right)=\frac{\mathrm{3}}{\mathrm{4}}}\\{\mathrm{tan}\:{x}\:=\:\mathrm{2}\:}\end{cases} \\ $$$${find}\:\:\mathrm{tan}\:{y}\:? \\ $$
Commented by mnjuly1970 last updated on 13/Aug/20
$$\:{excellent}. \\ $$
Commented by mr W last updated on 13/Aug/20
$$\mathrm{tan}\:\left({x}−{y}\right)=\frac{\mathrm{tan}\:{x}−\mathrm{tan}\:{y}}{\mathrm{1}+\mathrm{tan}\:{x}\:\mathrm{tan}\:{y}}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\frac{\mathrm{2}−\mathrm{tan}\:{y}}{\mathrm{1}+\mathrm{2}\:\mathrm{tan}\:{y}}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\mathrm{10}\:\mathrm{tan}\:{y}=\mathrm{5} \\ $$$$\Rightarrow\mathrm{tan}\:{y}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by bemath last updated on 13/Aug/20
$${santuyy}…. \\ $$
Answered by bemath last updated on 13/Aug/20
$$\:\frac{\boldsymbol{\mathcal{B}}{e}\boldsymbol{\mathcal{M}}{ath}}{°\bullet°} \\ $$$$\:\:\:{y}\:=\:{x}−\left({x}−{y}\right) \\ $$$$\mathrm{tan}\:{y}\:=\:\mathrm{tan}\:\left({x}−\left({x}−{y}\right)\right) \\ $$$$\mathrm{tan}\:{y}\:=\:\frac{\mathrm{tan}\:{x}−\mathrm{tan}\:\left({x}−{y}\right)}{\mathrm{1}+\mathrm{tan}\:{x}\:\mathrm{tan}\:\left({x}−{y}\right)} \\ $$$$\mathrm{tan}\:{y}\:=\:\frac{\mathrm{2}−\frac{\mathrm{3}}{\mathrm{4}}}{\mathrm{1}+\mathrm{2}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)}\:=\:\frac{\frac{\mathrm{5}}{\mathrm{4}}}{\frac{\mathrm{10}}{\mathrm{4}}}\:=\:\mathrm{0}.\mathrm{5}\: \\ $$