Question

SOCORRO PFVR!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
Se x∈├]0,π/2┤[, então (cos^2 x-sen^2 x)/((cosx+senx)^2-1) é sempre igual a:
a) cotg2x
b) cotgx
c) tgx
d) tg2x
e) 1

Answer 1

Resposta:

Item a)

Explicação passo-a-passo:

Sabemos que $cos^{2}x-sen^{2}x=cos2x$

$\dfrac{cos^{2}x-sen^{2}x}{(cosx+senx)^{2}-1}=\dfrac{cos2x}{cos^{2}x+2senx.cosx+sen^{2}x-1}\\\\\\Rearrumando\\\\\dfrac{cos^{2}x-sen^{2}x}{(cosx+senx)^{2}-1}=\dfrac{cos2x}{2senx.cosx+cos^{2}x+sen^{2}x-1}\\\\\\onde \\\\\\cos^{2}x+sen^{2}x=1\\\\\\\dfrac{cos^{2}x-sen^{2}x}{(cosx+senx)^{2}-1}=\dfrac{cos2x}{2senx.cosx+1-1}\\\\\\\dfrac{cos^{2}x-sen^{2}x}{(cosx+senx)^{2}-1}=\dfrac{cos2x}{2senx.cosx}\\\\\\\dfrac{cos^{2}x-sen^{2}x}{(cosx+senx)^{2}-1}=\dfrac{cos2x}{sen2x}$

$\dfrac{cos^{2}x-sen^{2}x}{(cosx+senx)^{2}-1}=cotg2x$