Răspuns:
[tex]tg^2x+ctg^2x=4sin(2x)^-^2-2=\frac{sin^2x}{cos^2x}+\frac{cos^2x}{sin^2x}=4*\frac{1}{sin(2x)^2} -2[/tex]
Amplificam cu sin^2(x) si cos^2(x) partea stanga a ecuatiei:
[tex]\frac{sin^4x+cos^4x}{sin^2x*cos^2x}=\frac{4}{(2sinx*cosx)^2}-2[/tex]
[tex]=\frac{(sin^2x+cos^2x)-2sinx*cosx}{sin^2x*cos^2x}=\frac{4}{4sin^2x*cos^2x}-2[/tex]
[tex]=\frac{1-2sinx*cosx}{sin^2x*cos^2x}=\frac{1}{sin^2x*cos^2x} -\frac{2sin^2x*cos^2x}{sin^2x*cos^2x}[/tex]
[tex]=>\frac{1-2sinx*cosx}{sin^2x*cos^2x}=\frac{1-2sin^2x*cos^2x}{sin^2x*cos^2x}[/tex] ''Adevarat"
