Question

Sa se verifice egalitatea:  tg^2x + ctg^2x = 2 · (3 + cos4x)/(1 - cos4x)​

Answer 1

Răspuns:

$tg^2x+ctg^2x=\frac{2(3+cos4x)}{1-cos4x}$

$\frac{sin^2x}{cos^2x}+\frac{cos^2x}{sin^2x}=\frac{2(3+8cos^4x-8cos^2x+1}{cos0-cos4x}$

$\frac{sin^4x+cos^4x}{sin^2x*cos^2x}=\frac{2(4+8cos^4x-8cos^2x)}{-2sin\frac{4x}{2}*sin\frac{-4x}{2} }$

$\frac{sin^4x+cos^4x}{sin^2x*cos^2x}=\frac{4(1+2cos^4x-2cos^2x)}{sin^22x}$

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

$sin^4x+cos^4x=1+2cos^4x-2+2sin^2x$

$sin^4x=1+cos^4x-2+2sin^2x$

$sin^4x=1+(1-sin^2x)^2-2+2sin^2x$

$sin^4x=1+1-2sin^2x+sin^4x-2+2sin^2x$

$sin^4x=sin^4x$

"Egalitatea este adevarata"