Question

a expressao (foto) é igual a

a) sen2 x - 1

b) -1

c) senx + 1

d) 1

e) cos2x + 1

Answer 1

Resposta:

1

Explicação passo-a-passo:

Considerando a lei fundamental $sen^2x+cos^2x=1$

$\frac{senx}{cossecx}+\frac{2cosx}{secx}-\frac{cotg^2x}{1+cotg^2x}= \\\frac{senx}{\frac{1}{senx} }+\frac{2cosx}{\frac{1}{cosx}}-\frac{\frac{cos^2x}{sen^2x}}{1+\frac{cos^2x}{sen^2x}}=senx.senx+2cosx.cosx-\frac{\frac{cos^2x}{sen^2x} }{\frac{sen^2x +cos^2x}{sen^2x}}=\\ sen^2x+2cos^2x-\frac{cos^2x.sen^2x}{sen^2x.(sen^2x+cos^2x)}=sen^2x+2cos^2x-\frac{cos^2x.sen^2x}{sen^2x}\\ sen^2x+2cos^2x-cos^2x = sen^2x+cos^2x=1$