Question

Calcule o valor numérico da expressão:
E = [tg² 3x + tg (x + 5°)] : [tg 6x . tg (4x + 50°)], para x = 40°

Answer 1

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                                                                                                                              $\sf tg(3x)\bigg|_{x=40^\circ}=tg(3\cdot40^\circ)=tg(120^\circ)=-\sqrt{3}\\\sf tg(x+5^\circ)\bigg|_{x=40^\circ}=tg(40^\circ+5^\circ)=tg(45^\circ)=1\\\sf tg(6x)\bigg|_{x=40^\circ}=tg(6\cdot40^\circ)=tg(240^\circ)=\sqrt{3}\\\sf tg(4x+50^\circ)\bigg|_{x=40^\circ}=tg(4\cdot40^\circ+50^\circ)=tg(210^\circ)=\dfrac{\sqrt{3}}{3}$

$\sf E=\left[tg^2(3x)+tg(x+5^\circ)\right]:\left[tg(6x)\cdot tg(4x+50^\circ)\right]\bigg|_{x=40^\circ}\\\sf E=\left[(-\sqrt{3})^2+1\right]:\left[\sqrt{3}+\dfrac{\sqrt{3}}{3}\right]\\\sf E=\left[3+1\right]:\left[\dfrac{3\sqrt{3}+\sqrt{3}}{3}\right]\\\sf E=4:\dfrac{4\sqrt{3}}{3}\\\sf E=\backslash\!\!\!4\cdot\dfrac{3}{\backslash\!\!\!4\sqrt{3}}\\\sf E=\dfrac{3}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{\sqrt{3}}\\\sf E=\dfrac{\backslash\!\!\!3\sqrt{3}}{\backslash\!\!\!3}$

$\huge\boxed{\boxed{\boxed{\boxed{\sf E=\sqrt{3}\checkmark}}}}\\\huge\boxed{\boxed{\boxed{\boxed{\sf Espero~ter~ajudado~^\star~_{\cup}~^\star}}}}$

Answer 2

Oie, Td Bom?!

E = [tan(3x)² + tan(x + 5°)] ÷ [tan(6x) . tan(4x + 50°)]

• tan(3x) = tan(3 . 40°) = tan(120°) = - √3

• tan(x + 5°) = tan(40° + 5°) = tan(45°) = 1

• tan(6x) = tan(6 . 40°) = tan(240°) = √3

• tan(4x + 50°) = tan(4 . 40° + 50°) = tan(160° + 50°) = tan(210°) = √3/3

E = [(- √3)² + 1] ÷ [√3 . √3/3]

E = [√[3]² + 1] ÷ 3/3

E = [3 + 1] ÷ 1

E = 4 ÷ 1

E = 4

Att. Makaveli1996