Question

Если tg^4a+ctg^4a=14 найдите tga+ctga

Answer 1

$tg^4a+ctg^4a=14\\\\\\(tga+ctga)^2=tg^2a+ctg^2a+2\underbrace {tga\cdot ctga}_{1}=tg^2a+ctg^2a+2\; \; \Rightarrow \\\\tg^2a+ctg^2a=(tga+ctga)^2-2\; \; \Rightarrow \\\\(tg^2a+ctg^2a)^2=\Big ((tga+ctga)^2-2\Big )^2\\\\\underbrace {tg^4a+ctg^4a}_{14}+2\cdot \underbrace {tg^2a\cdot ctg^2a}_{1}=(tga+ctga)^4-4(tga+ctga)^2+4\\\\16=(tga+ctga)^4-4(tga+ctga)^2+4\\\\(tga+ctga)^4-4(tga+ctga)^2-12=0\\\\Teorema\; Vieta:\; \; (tga+ctga)^2=-2<0\; \; ne\; podxodit\\\\(tga+ctga)^2=6\; \; \Rightarrow \; \; tga+ctga=\pm \sqrt6\\\\Otvet:\; \; tga+ctga=-\sqrt6\; \; \; ili\; \; \; tga+ctga=\sqrt6\; .$

$P.S.\qquad tga\cdot ctga=\frac{sina}{cosa}\cdot \frac{cosa}{sina}=1\\\\tg^2a\cdot ctg^2a=(tga\cdot ctga)^2=1^2=1$