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$tg\frac{\pi }{8}=\sqrt{\frac{\sqrt2-1}{\sqrt2+1}}\\\\\frac{1+tg^2x+2tgx}{(1+\sqrt2)tgx}=\frac{(1+tgx)^2}{(1+\sqrt2)tgx}\\\\(1+tg\frac{\pi}{8})^2=(1+\sqrt{\frac{\sqrt2-1}{\sqrt2+1}})^2=(1+\frac{\sqrt{\sqrt2-1}}{\sqrt{\sqrt2+1}})^2=\\\\=\frac{(\sqrt{\sqrt2+1}+\sqrt{\sqrt2-1})^2}{\sqrt2+1}=\frac{\sqrt2+1+2\sqrt{(\sqrt2-1)(\sqrt2+1)}+\sqrt2-1}{\sqrt2+1}=\\\\=\frac{2\sqrt2+2\sqrt{2-1}}{\sqrt2+1}=\frac{2\sqrt2+2}{\sqrt2+1}=\frac{2(\sqrt2+1)}{\sqrt2+1}=2$

$(1+\sqrt2)tg\frac{\pi}{8}=(1+\sqrt2)\cdot \frac{\sqrt{\sqrt2-1}}{\sqrt{\sqrt2+1}}=\sqrt{\sqrt2+1}\cdot \sqrt{\sqrt2-1}=\\\\=\sqrt{(\sqrt2+1)(\sqrt2-1)}=\sqrt{(\sqrt2)^2-1^2}=\sqrt{2-1}=1\\\\\\\frac{1+tg^2\frac{\pi }{8}+2tg\frac{\pi }{8}}{(1+\sqrt2)tg\frac{\pi }{8}}=\frac{2}{1}=2$