Question

O número α E (0, π/2) é tal que tan α + cot α = 10/3
. Determine o valor das seis razões trigonométricas de α.

Answer 1

Explicação passo-a-passo:

$\sf tg~\alpha+cotg~\alpha=\dfrac{10}{3}$

Seja $\sf tg~\alpha=x$

$\sf x+\dfrac{1}{x}=\dfrac{10}{3}$

$\sf 3x^2+3=10x$

$\sf 3x^2-10x+3=0$

$\sf \Delta=(-10)^2-4\cdot3\cdot3$

$\sf \Delta=100-36$

$\sf \Delta=64$

$\sf x=\dfrac{-(-10)\pm\sqrt{64}}{2\cdot3}=\dfrac{10\pm8}{6}$

Como $\sf \alpha$ pertence ao 1° quadrante, sua tangente é positiva

$\sf x'=\dfrac{10+8}{6}~\Rightarrow~x'=\dfrac{18}{6}~\Rightarrow~\red{x'=3}$

$\sf x"=\dfrac{10-8}{6}~\Rightarrow~x"=\dfrac{2}{6}~\Rightarrow~\red{x'"=\dfrac{1}{3}}$

• Para $\sf x=3$:

$\sf \dfrac{sen~\alpha}{cos~\alpha}=3$

$\sf sen~\alpha=3\cdot cos~\alpha$

Pela relação fundamental da trigonometria:

$\sf sen^2~\alpha+cos^2~\alpha=1$

Substituindo $\sf sen~\alpha$ por $\sf 3\cdot cos~\alpha$:

$\sf (3\cdot cos~\alpha)^2+cos^2~\alpha=1$

$\sf 9\cdot cos^2~\alpha+cos^2~\alpha=1$

$\sf 10\cdot cos^2~\alpha=1$

$\sf cos^2~\alpha=\dfrac{1}{10}$

Como $\sf \alpha$ pertence ao 1° quadrante, sua cosseno é positivo

$\sf cos~\alpha=\sqrt{\dfrac{1}{10}}$

$\sf cos~\alpha=\dfrac{1}{\sqrt{10}}$

$\sf cos~\alpha=\dfrac{1}{\sqrt{10}}\cdot\dfrac{\sqrt{10}}{\sqrt{10}}$

$\sf cos~\alpha=\dfrac{\sqrt{10}}{10}$

Assim:

$\sf sen~\alpha=3\cdot\dfrac{\sqrt{10}}{10}$

$\sf sen~\alpha=\dfrac{3\sqrt{10}}{10}$

• $\sf cossec~\alpha=\dfrac{1}{sen~\alpha}$

$\sf cossec~\alpha=\dfrac{1}{\frac{3\sqrt{10}}{10}}$

$\sf cossec~\alpha=\dfrac{10}{3\sqrt{10}}$

$\sf cossec~\alpha=\dfrac{10}{3\sqrt{10}}\cdot\dfrac{\sqrt{10}}{\sqrt{10}}$

$\sf cossec~\alpha=\dfrac{10\sqrt{10}}{3\cdot10}$

$\sf cossec~\alpha=\dfrac{10\sqrt{10}}{30}$

$\sf cossec~\alpha=\dfrac{\sqrt{10}}{3}$

• $\sf sec~\alpha=\dfrac{1}{cos~\alpha}$

$\sf sec~\alpha=\dfrac{1}{\frac{\sqrt{10}}{10}}$

$\sf sec~\alpha=\dfrac{10}{\sqrt{10}}$

$\sf sec~\alpha=\dfrac{10}{\sqrt{10}}\cdot\dfrac{\sqrt{10}}{\sqrt{10}}$

$\sf sec~\alpha=\dfrac{10\sqrt{10}}{10}$

$\sf sec~\alpha=\sqrt{10}$

• $\sf cotg~\alpha=\dfrac{1}{tg~\alpha}$

$\sf cotg~\alpha=\dfrac{1}{3}$

As seis razões trigonométricas de α são:

• $\sf \red{sen~\alpha=\dfrac{3\sqrt{10}}{10}}$

• $\sf \red{cos~\alpha=\dfrac{\sqrt{10}}{10}}$

• $\sf \red{tg~\alpha=3}$

• $\sf \red{sec~\alpha=\sqrt{10}}$

• $\sf \red{cossec~\alpha=\dfrac{\sqrt{10}}{3}}$

• $\sf \red{cotg~\alpha=\dfrac{1}{3}}$

Para $\sf x=\dfrac{1}{3}$:

$\sf \dfrac{sen~\alpha}{cos~\alpha}=\dfrac{1}{3}$

$\sf cos~\alpha=3\cdot sen~\alpha$

Pela relação fundamental da trigonometria:

$\sf sen^2~\alpha+cos^2~\alpha=1$

Substituindo $\sf cos~\alpha$ por $\sf 3\cdot sen~\alpha$:

$\sf sen^2~\alpha+(3\cdot sen~\alpha)^2=1$

$\sf sen^2~\alpha+9\cdot sen^2~\alpha=1$

$\sf 10\cdot sen^2~\alpha=1$

$\sf sen^2~\alpha=\dfrac{1}{10}$

Como $\sf \alpha$ pertence ao 1° quadrante, seu seno é positivo

$\sf sen~\alpha=\sqrt{\dfrac{1}{10}}$

$\sf sen~\alpha=\dfrac{1}{\sqrt{10}}$

$\sf sen~\alpha=\dfrac{1}{\sqrt{10}}\cdot\dfrac{\sqrt{10}}{\sqrt{10}}$

$\sf sen~\alpha=\dfrac{\sqrt{10}}{10}$

Assim:

$\sf cos~\alpha=3\cdot\dfrac{\sqrt{10}}{10}$

$\sf cos~\alpha=\dfrac{3\sqrt{10}}{10}$

• $\sf cossec~\alpha=\dfrac{1}{sen~\alpha}$

$\sf cossec~\alpha=\dfrac{1}{\frac{\sqrt{10}}{10}}$

$\sf cossec~\alpha=\dfrac{10}{\sqrt{10}}$

$\sf cossec~\alpha=\dfrac{10}{\sqrt{10}}\cdot\dfrac{\sqrt{10}}{\sqrt{10}}$

$\sf cossec~\alpha=\dfrac{10\sqrt{10}}{10}$

$\sf cossec~\alpha=\sqrt{10}$

• $\sf sec~\alpha=\dfrac{1}{cos~\alpha}$

$\sf sec~\alpha=\dfrac{1}{\frac{3\sqrt{10}}{10}}$

$\sf sec~\alpha=\dfrac{10}{3\sqrt{10}}$

$\sf sec~\alpha=\dfrac{10}{3\sqrt{10}}\cdot\dfrac{\sqrt{10}}{\sqrt{10}}$

$\sf sec~\alpha=\dfrac{10\sqrt{10}}{3\cdot10}$

$\sf sec~\alpha=\dfrac{10\sqrt{10}}{30}$

$\sf sec~\alpha=\dfrac{\sqrt{10}}{3}$

• $\sf cotg~\alpha=\dfrac{1}{tg~\alpha}$

$\sf cotg~\alpha=\dfrac{1}{3}$

As seis razões trigonométricas de α são:

• $\sf \red{sen~\alpha=\dfrac{\sqrt{10}}{10}}$

• $\sf \red{cos~\alpha=\dfrac{3\sqrt{10}}{10}}$

• $\sf \red{tg~\alpha=\dfrac{1}{3}}$

• $\sf \red{sec~\alpha=\dfrac{\sqrt{10}}{3}}$

• $\sf \red{cossec~\alpha=\sqrt{10}}$

• $\sf \red{cotg~\alpha=3}$