Question

Considerando que tg x +cotg x =2 , o valor numérico da expressão trigonométrica
E = sen x ∙ cos x é
a) 1 b) 0,5 c) 0,6 d) 0,8 e) 2

Answer 1

$\large\boxed{\begin{array}{l}\sf tg(x)+cotg(x)=2\\\sf\dfrac{sen(x)}{cos(x)}+\dfrac{1}{\frac{sen(x)}{cos(x)}}=2\\\\\sf\dfrac{sen(x)}{cos(x)}+\dfrac{cos(x)}{sen(x)}=2\\\\\sf\dfrac{sen^2(x)+cos^2(x)}{sen(x)\cdot cos(x)}=2\\\\\sf\dfrac{1}{sen(x)\cdot cos(x)}=2\\\sf 2sen(x)\cdot cos(x)=1\\\sf sen(x)\cdot cos(x)=\dfrac{1}{2}\\\sf portanto\\\sf E=\dfrac{1}{2}=0,5\\\huge\boxed{\boxed{\boxed{\boxed{\sf\maltese~alternativa~b}}}} \end{array}}$

Answer 2

Resposta:

Letra B

Explicação passo a passo:

$tgx=\frac{senx}{cosx} \\\\cotgx=\frac{cosx}{senx} \\\\sen^2x+cos^2x=1\\\\tgx+cotgx = 2\\\\\frac{senx}{cosx}+\frac{cosx}{senx} =2 \\\\\frac{sen^2x+cos^2x}{senxcosx} =2\\\\\frac{1}{senxcosx} =2\\\\Invertendo~~as~~ frac_{\!\!\!,}\tilde os\\\\senxcosx=\frac{1}{2} =0,5$