Question

A expressao (Foto) é igual a:
a) 1 + 2cos2 (X)
b)3 + 2cos2 (X)
c)3 + 2sen2 (X)
d)1
e)1 + 2sen2 (X)

Answer 1

Resposta:

Opção E)

Explicação passo-a-passo:

$\frac{sen^2(x)-1}{tg^2(x)+1}+\frac{cossen^2(x)+1}{cotg^2(x)+1}$

$\frac{sen^2(x)-sen^2(x)-cos^2(x)}{tg^2(x)+1}+\frac{\frac{1}{sen^2(x)}+1 }{\frac{cos^2(x)}{sen^2(x)}+1 }$

$\frac{\frac{1}{cos^2(x)}-1 }{\frac{sen^2(x)}{cos^2(x)}+1 } +\frac{\frac{1+sen^2(x)}{sen^2(x)} }{\frac{cos^2(x)+sen^2(x)}{sen^2(x)} }$

$\frac{\frac{1-cos^2(x)}{cos^2(x)} }{\frac{sen^2(x)+cos^2(x)}{cos^2(x)} }+\frac{sen^2(x)(1+sen^2(x))}{sen^2(x)(cos^2(x)+sen^2(x))}$

$\frac{1-cos^2(x)}{sen^2(x)+cos^2(x)}+\frac{1+sen^2(x)}{1}$

$1-cos^2(x)+1+sen^2(x)=1+2sen^2(x)$

Opção E)

Espero ter ajudado!!

Qualquer duvida, comente.