Question

tgα+ctgα=√8 0<α<$\frac{\pi }{4}$
Найти: tgα-ctgα

Answer 1

$0<a<\frac{\pi }{4}\; \; \to \; \; 0<tga<1\; ,\; \; 0<ctga<1\; \; i\; \; \underline {ctga>tga}\\\\tga+ctga=\sqrt8\\\\(tga+ctga)^2=8\\\\tg^2a+2\, \underbrace {tga\cdot ctga}_{1}+ctg^2a=8\\\\tg^2a+2+ctg^2a=8\\\\tg^2a+ctg^2a=6\\\\\\(tga-ctga)^2=tg^2a+ctg^2a-2=6-2=4\; \; \to \\\\(tga-ctga)=2\; \; \; ili\; \; \; (tga-ctga)=-2\\\\ctga>tga\; \; \Rightarrow \; \; \underline {tga-ctga=-2}$