Question Number 176490 by mnjuly1970 last updated on 20/Sep/22
$$ \\ $$$$\:\:\:\:\:{solve}\:\:\:\left(\:{x},{y}\:\in\:\mathbb{R}\:\right) \\ $$$$\:\:\: \\ $$$$\:\:\:\:\:\:\begin{cases}{\:\mathrm{tan}\left({x}\:\right)\:+\:\mathrm{tan}\:\left({y}\:\right)=\mathrm{2}}\\{\:\mathrm{tan}\left(\mathrm{2}{x}\:\right)\:+\:\mathrm{tan}\left(\:\mathrm{2}{y}\:\right)\:=\:\mathrm{2}}\end{cases} \\ $$$$\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:−−−−−−−− \\ $$
Answered by ajfour last updated on 20/Sep/22
$$\frac{\mathrm{2tan}\:{x}}{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} {x}}+\frac{\mathrm{2tan}\:{y}}{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} {y}}=\mathrm{2} \\ $$$$\Rightarrow\:\frac{\mathrm{tan}\:{x}\left(\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} {y}\right)+\mathrm{tan}\:{y}\left(\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} {x}\right)}{\mathrm{1}+\left(\mathrm{tan}\:{x}\mathrm{tan}\:{y}\right)^{\mathrm{2}} −\mathrm{tan}\:^{\mathrm{2}} {x}−\mathrm{tan}\:^{\mathrm{2}} {y}}=\mathrm{1} \\ $$$$\Rightarrow\:\frac{\mathrm{2}−\mathrm{2tan}\:{x}\mathrm{tan}\:{y}}{\mathrm{1}−\mathrm{4}+\mathrm{2tan}\:{x}\mathrm{tan}\:{y}+\left(\mathrm{tan}\:{x}\mathrm{tan}\:{y}\right)^{\mathrm{2}} }=\mathrm{1} \\ $$$${call}\:\mathrm{tan}\:{x}={p},\:\:\:\mathrm{tan}\:{y}={q}\:\:\:,\:\&\:{pq}={m} \\ $$$${we}\:{then}\:{have}\:{p}+{q}=\mathrm{2}\:\:\:\& \\ $$$$\mathrm{2}−\mathrm{2}{m}=\mathrm{1}+\mathrm{2}{m}+{m}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:{m}^{\mathrm{2}} +\mathrm{4}{m}+\mathrm{4}=\mathrm{5} \\ $$$${m}=\pm\sqrt{\mathrm{5}}−\mathrm{2} \\ $$$${p},{q}\:=\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{2}\mp\sqrt{\mathrm{5}}} \\ $$$${choosing}\:−\:{sign} \\ $$$${p},{q}=\mathrm{1}\pm\sqrt{\mathrm{3}−\sqrt{\mathrm{5}}} \\ $$$${while}\:{if}\:{we}\:{choose}\:+\:{sign} \\ $$$${p},{q}=\mathrm{1}\pm\sqrt{\mathrm{3}+\sqrt{\mathrm{5}}} \\ $$$${And}\:{when} \\ $$$${p}=\mathrm{tan}\:{x} \\ $$$${x}=\mathrm{tan}^{−\mathrm{1}} {p}+{m}\pi\:\:\:\:\forall\:{m}\in\mathbb{Z} \\ $$
Commented by Tawa11 last updated on 20/Sep/22
$$\mathrm{I}\:\mathrm{have}\:\mathrm{subscribe}\:\mathrm{too}. \\ $$
Commented by Tawa11 last updated on 20/Sep/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Commented by ajfour last updated on 20/Sep/22
https://youtu.be/GmIVICE0YrQ
Commented by mnjuly1970 last updated on 20/Sep/22
$$\:\:\:{thanks}\:{alot}\:\:{sir}\:{ajfor} \\ $$$$\:\:\:\:{your}\:\:{youtube}\:{is}\:{very}\:{very} \\ $$$${excellent}…{grateful}\:{sir}… \\ $$
Commented by ajfour last updated on 20/Sep/22
$${thanks}\:{for}\:{viewing},\:{appreciating}! \\ $$
Answered by behi834171 last updated on 20/Sep/22
![tgx=m,tgy=n tg(x+y)=(2/(1−mn)) tg(2x+2y)=(2/(1−((2m)/(1−m^2 )).((2n)/(1−n^2 )))) tg(2x+2y)=((2×(2/(1−mn)))/(1−(4/((1−mn)^2 ))))=((4(1−mn))/((1−mn)^2 −4)) ⇒((1−m^2 −n^2 +m^2 n^2 )/(1−n^2 −m^2 +m^2 n^2 −4mn))=((2−2mn)/(m^2 n^2 −2mn−3))⇒ m^2 n^2 −2mn−3−m^4 n^2 +2m^3 n+3m^2 −m^2 n^4 +2mn^3 +3n^2 +m^4 n^4 −2m^3 n^3 −3m^2 n^2 = 2−2n^2 −2m^2 +2m^2 n^2 −8mn−2mn+ 4mn^3 +4m^3 n−2m^3 n^3 +8mn⇒ −2m^2 n^2 −2mn−3−m^4 n^2 +2m^3 n+3m^2 −m^2 n^4 +2mn^3 +3n^2 +m^4 n^4 −2m^3 n^3 −3m^2 n^2 = 2−2n^2 −2m^2 +2m^2 n^2 −2mn+4mn^3 + 4m^3 n−2m^3 n^3 ⇒ −4m^2 n^2 −5−m^4 n^2 −2m^3 n+5m^2 −m^2 n^4 −2mn^3 +5m^2 +m^4 n^4 =0⇒ m^4 n^4 −2m^3 n−2mn^3 −m^4 n^2 −m^2 n^4 −4m^2 n^2 +5m^2 +5n^2 −5=0 ⇒t^4 −3t(m^2 +n^2 )−t^2 (m^2 +n^2 )−4t^2 + +5(m^2 +n^2 )−5=0 ⇒t^4 −3t(4−2t)−t^2 (4−2t)−4t^2 + +5(4−2t)−5=0⇒ t^4 −12t+6t^2 −4t^2 +2t^3 −4t^2 +20−10t−5=0 ⇒t^4 +2t^3 −2t^2 −22t+15=0⇒ t=mn=0.67, t=mn=2.16 ⇒ { ((m+n=2)),((mn=0.67 or 2.16)) :} ⇒z^2 −2z+[0.67 or 2.16]=0 ⇒z−1=±0.57 ⇒ { ((tgx=1.57 or 0.43)),((tgy=0.43 or 1.57)) :}](https://www.tinkutara.com/question/Q176548.png)
$${tgx}={m},{tgy}={n} \\ $$$${tg}\left({x}+{y}\right)=\frac{\mathrm{2}}{\mathrm{1}−{mn}} \\ $$$${tg}\left(\mathrm{2}{x}+\mathrm{2}{y}\right)=\frac{\mathrm{2}}{\mathrm{1}−\frac{\mathrm{2}{m}}{\mathrm{1}−{m}^{\mathrm{2}} }.\frac{\mathrm{2}{n}}{\mathrm{1}−{n}^{\mathrm{2}} }} \\ $$$${tg}\left(\mathrm{2}{x}+\mathrm{2}{y}\right)=\frac{\mathrm{2}×\frac{\mathrm{2}}{\mathrm{1}−{mn}}}{\mathrm{1}−\frac{\mathrm{4}}{\left(\mathrm{1}−{mn}\right)^{\mathrm{2}} }}=\frac{\mathrm{4}\left(\mathrm{1}−{mn}\right)}{\left(\mathrm{1}−{mn}\right)^{\mathrm{2}} −\mathrm{4}} \\ $$$$\Rightarrow\frac{\mathrm{1}−{m}^{\mathrm{2}} −{n}^{\mathrm{2}} +{m}^{\mathrm{2}} {n}^{\mathrm{2}} }{\mathrm{1}−{n}^{\mathrm{2}} −{m}^{\mathrm{2}} +{m}^{\mathrm{2}} {n}^{\mathrm{2}} −\mathrm{4}{mn}}=\frac{\mathrm{2}−\mathrm{2}{mn}}{{m}^{\mathrm{2}} {n}^{\mathrm{2}} −\mathrm{2}{mn}−\mathrm{3}}\Rightarrow \\ $$$${m}^{\mathrm{2}} {n}^{\mathrm{2}} −\mathrm{2}{mn}−\mathrm{3}−{m}^{\mathrm{4}} {n}^{\mathrm{2}} +\mathrm{2}{m}^{\mathrm{3}} {n}+\mathrm{3}{m}^{\mathrm{2}} \\ $$$$−{m}^{\mathrm{2}} {n}^{\mathrm{4}} +\mathrm{2}{mn}^{\mathrm{3}} +\mathrm{3}{n}^{\mathrm{2}} +{m}^{\mathrm{4}} {n}^{\mathrm{4}} −\mathrm{2}{m}^{\mathrm{3}} {n}^{\mathrm{3}} −\mathrm{3}{m}^{\mathrm{2}} {n}^{\mathrm{2}} = \\ $$$$\mathrm{2}−\mathrm{2}{n}^{\mathrm{2}} −\mathrm{2}{m}^{\mathrm{2}} +\mathrm{2}{m}^{\mathrm{2}} {n}^{\mathrm{2}} −\mathrm{8}{mn}−\mathrm{2}{mn}+ \\ $$$$\mathrm{4}{mn}^{\mathrm{3}} +\mathrm{4}{m}^{\mathrm{3}} {n}−\mathrm{2}{m}^{\mathrm{3}} {n}^{\mathrm{3}} +\mathrm{8}{mn}\Rightarrow \\ $$$$−\mathrm{2}{m}^{\mathrm{2}} {n}^{\mathrm{2}} −\mathrm{2}{mn}−\mathrm{3}−{m}^{\mathrm{4}} {n}^{\mathrm{2}} +\mathrm{2}{m}^{\mathrm{3}} {n}+\mathrm{3}{m}^{\mathrm{2}} \\ $$$$−{m}^{\mathrm{2}} {n}^{\mathrm{4}} +\mathrm{2}{mn}^{\mathrm{3}} +\mathrm{3}{n}^{\mathrm{2}} +{m}^{\mathrm{4}} {n}^{\mathrm{4}} −\mathrm{2}{m}^{\mathrm{3}} {n}^{\mathrm{3}} −\mathrm{3}{m}^{\mathrm{2}} {n}^{\mathrm{2}} = \\ $$$$\mathrm{2}−\mathrm{2}{n}^{\mathrm{2}} −\mathrm{2}{m}^{\mathrm{2}} +\mathrm{2}{m}^{\mathrm{2}} {n}^{\mathrm{2}} −\mathrm{2}{mn}+\mathrm{4}{mn}^{\mathrm{3}} + \\ $$$$\mathrm{4}{m}^{\mathrm{3}} {n}−\mathrm{2}{m}^{\mathrm{3}} {n}^{\mathrm{3}} \Rightarrow \\ $$$$−\mathrm{4}{m}^{\mathrm{2}} {n}^{\mathrm{2}} −\mathrm{5}−{m}^{\mathrm{4}} {n}^{\mathrm{2}} −\mathrm{2}{m}^{\mathrm{3}} {n}+\mathrm{5}{m}^{\mathrm{2}} −{m}^{\mathrm{2}} {n}^{\mathrm{4}} \\ $$$$−\mathrm{2}{mn}^{\mathrm{3}} +\mathrm{5}{m}^{\mathrm{2}} +{m}^{\mathrm{4}} {n}^{\mathrm{4}} =\mathrm{0}\Rightarrow \\ $$$${m}^{\mathrm{4}} {n}^{\mathrm{4}} −\mathrm{2}{m}^{\mathrm{3}} {n}−\mathrm{2}{mn}^{\mathrm{3}} −{m}^{\mathrm{4}} {n}^{\mathrm{2}} −{m}^{\mathrm{2}} {n}^{\mathrm{4}} \\ $$$$−\mathrm{4}{m}^{\mathrm{2}} {n}^{\mathrm{2}} +\mathrm{5}{m}^{\mathrm{2}} +\mathrm{5}{n}^{\mathrm{2}} −\mathrm{5}=\mathrm{0} \\ $$$$\Rightarrow{t}^{\mathrm{4}} −\mathrm{3}{t}\left({m}^{\mathrm{2}} +{n}^{\mathrm{2}} \right)−{t}^{\mathrm{2}} \left({m}^{\mathrm{2}} +{n}^{\mathrm{2}} \right)−\mathrm{4}{t}^{\mathrm{2}} + \\ $$$$+\mathrm{5}\left({m}^{\mathrm{2}} +{n}^{\mathrm{2}} \right)−\mathrm{5}=\mathrm{0} \\ $$$$\Rightarrow{t}^{\mathrm{4}} −\mathrm{3}{t}\left(\mathrm{4}−\mathrm{2}{t}\right)−{t}^{\mathrm{2}} \left(\mathrm{4}−\mathrm{2}{t}\right)−\mathrm{4}{t}^{\mathrm{2}} + \\ $$$$+\mathrm{5}\left(\mathrm{4}−\mathrm{2}{t}\right)−\mathrm{5}=\mathrm{0}\Rightarrow \\ $$$${t}^{\mathrm{4}} −\mathrm{12}{t}+\mathrm{6}{t}^{\mathrm{2}} −\mathrm{4}{t}^{\mathrm{2}} +\mathrm{2}{t}^{\mathrm{3}} −\mathrm{4}{t}^{\mathrm{2}} +\mathrm{20}−\mathrm{10}{t}−\mathrm{5}=\mathrm{0} \\ $$$$\Rightarrow{t}^{\mathrm{4}} +\mathrm{2}{t}^{\mathrm{3}} −\mathrm{2}{t}^{\mathrm{2}} −\mathrm{22}{t}+\mathrm{15}=\mathrm{0}\Rightarrow \\ $$$${t}={mn}=\mathrm{0}.\mathrm{67},\:{t}={mn}=\mathrm{2}.\mathrm{16} \\ $$$$\Rightarrow\begin{cases}{{m}+{n}=\mathrm{2}}\\{{mn}=\mathrm{0}.\mathrm{67}\:{or}\:\mathrm{2}.\mathrm{16}}\end{cases} \\ $$$$\Rightarrow\boldsymbol{{z}}^{\mathrm{2}} −\mathrm{2}\boldsymbol{{z}}+\left[\mathrm{0}.\mathrm{67}\:\:\boldsymbol{{or}}\:\mathrm{2}.\mathrm{16}\right]=\mathrm{0} \\ $$$$\Rightarrow\boldsymbol{{z}}−\mathrm{1}=\pm\mathrm{0}.\mathrm{57} \\ $$$$\Rightarrow\begin{cases}{\boldsymbol{{tgx}}=\mathrm{1}.\mathrm{57}\:\boldsymbol{{or}}\:\:\mathrm{0}.\mathrm{43}}\\{\boldsymbol{{tgy}}=\mathrm{0}.\mathrm{43}\:\:\boldsymbol{{or}}\:\:\mathrm{1}.\mathrm{57}}\end{cases} \\ $$
Commented by Tawa11 last updated on 22/Sep/22
$$\mathrm{Great}\:\mathrm{sir}. \\ $$