[TEX]49/tan^2x + tan^2y + cot^2(x+y) = 1[/TEX]
có [TEX]cotg(x+y)=\frac{1-tgxtgy}{tgx+tgy}[/TEX]
[TEX]\Leftrightarrow (tgx+tgy).cotg(x+y)=1-tgxtgy[/TEX]
[TEX]\Leftrightarrow cotg(x+y).tgx+cotg(x+y).tgy=1-tgxtgy[/TEX]
[TEX]\Leftrightarrow tgxtgy+tgx+cotg(x+y).tgy=1[/TEX](1)
thay bt (1) vào pt trên ta dc
[TEX]tan^2x + tan^2y + cot^2(x+y) =tgxtgy+tgx+cotg(x+y).tgy[/TEX]
[TEX]\Leftrightarrow 2tgxtgy+2tgxcot(x+y)+2tgy.cot(x+y)=2tg^2x+2tg^2y+2cot^2(x+y)[/TEX]
[TEX]\Leftrightarrow (tgx-tgy)^2+[tgx-cot(x+y)]^2+[tgy-cot(x+y)]^2=0[/TEX]
dấu '=' xảy ra \Leftrightarrow[TEX]tgx=tgy=cot(x+y)[/TEX] \Rightarrow[TEX]tg^2x=tg^2y=cot^2(x+y)[/TEX]
mà [TEX]tan^2x + tan^2y + cot^2(x+y) = 1[/TEX] \Rightarrow [TEX]tg^2x=tg^2y=cot^2(x+y)=\frac{1}{3}[/TEX]
\Rightarrow [TEX]tgx=tgy=cot(x+y)=\frac{+-\sqr{3}}{3}[/TEX]
\Rightarrow [TEX]tgx=\frac{\sqr{3}}{3}}[/TEX]\Leftrightarrow[TEX]x=\frac{\pi}{6}+k\pi[/TEX]
[TEX]tgx=\frac{-\sqr{3}}{3}[/TEX]\Leftrightarrow [TEX]x=\frac{-\pi}{6}+k\pi[/TEX]
tt với [TEX]tgy=\frac{\sqr{3}}{3}[/TEX]
và [TEX]tgy=\frac{-\sqr{3}}{3}[/TEX] ok