Question

demonstrati:
tg^(2)x+ctg^(2)x=4(sin2x)^(-2) -2.​

Answer 1

Răspuns:

$tg^2x+ctg^2x=4sin(2x)^-^2-2=\frac{sin^2x}{cos^2x}+\frac{cos^2x}{sin^2x}=4*\frac{1}{sin(2x)^2} -2$

Amplificam cu sin^2(x) si cos^2(x) partea stanga a ecuatiei:

$\frac{sin^4x+cos^4x}{sin^2x*cos^2x}=\frac{4}{(2sinx*cosx)^2}-2$

$=\frac{(sin^2x+cos^2x)-2sinx*cosx}{sin^2x*cos^2x}=\frac{4}{4sin^2x*cos^2x}-2$

$=\frac{1-2sinx*cosx}{sin^2x*cos^2x}=\frac{1}{sin^2x*cos^2x} -\frac{2sin^2x*cos^2x}{sin^2x*cos^2x}$

$=>\frac{1-2sinx*cosx}{sin^2x*cos^2x}=\frac{1-2sin^2x*cos^2x}{sin^2x*cos^2x}$ ''Adevarat"