Question

Considerando que sen X = $\frac{2}{3}$ e x ∈ II Q, o valor de $\frac{tg X +cotgX}{secX + cossec X}$ é:

a) sen X

b) $\frac{\sqrt{5}}{3}$

c) cos²X

d) -3 (2 + $\sqrt{5}$)

e) -6 + $\sqrt{5}$)

Answer 1

Primeiro vamos calcular o cos x


$\mathsf{\sin^{2} x+\cos^{2} x=1}\\\\\\ \mathsf{\left ( \dfrac{2}{3} \right )^{2}+\cos^{2} x=1}\\\\\\ \mathsf{\cos^{2} x=1- \dfrac{4}{9}}\\\\\\ \mathsf{\cos^{2} x=\dfrac{5}{9}}\\\\\\ \mathsf{\cos x=\dfrac{\sqrt{5}}{3}}$

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Agora vamos trabalhar com a a expressão procurando simplifica-la ao máximo


$\mathsf{\dfrac{\tan x+\cot x}{\sec x+\csc x}}\\\\\\\\ \mathsf{=\dfrac{\dfrac{\sin x}{\cos x}+\dfrac{\cos x}{\sin x}}{\dfrac{1}{\cos x}+\dfrac{1}{\sin x}}}\\\\\\\\ \mathsf{=\dfrac{\dfrac{\sin^{2} x+\cos^{2} x}{\sin x\cdot \cos x}}{\dfrac{\sin x+\cos x}{\sin x\cdot \cos x}}}\\\\\\\\ \mathsf{=\dfrac{\sin^{2} x+\cos^{2} x}{\sin x+\cos x}}\\\\\\\\ \mathsf{=\dfrac{1}{\sin x+\cos x}}$

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Agora já podemos substituir os valores de seno e cosseno e encontrar o resultado


$\mathsf{=\dfrac{1}{\dfrac{2}{3}-\dfrac{\sqrt{5}}{3}}}\\\\\\\\ \mathsf{=\dfrac{1}{\dfrac{2-\sqrt{5}}{3}}}\\\\\\\\ \mathsf{=\dfrac{3}{2-\sqrt{5}}}\\\\\\\\ \mathsf{=\dfrac{3}{2-\sqrt{5}}\cdot \dfrac{(2+\sqrt{5})}{(2+\sqrt{5})}}$


$\mathsf{=\dfrac{6+3\cdot\sqrt{5}}{4-5}}\\\\\\\\ \mathsf{=\dfrac{6+3\cdot\sqrt{5}}{-1}}\\\\\\\\ \mathsf{-6-3\cdot\sqrt{5}}\\\\\\ \begin{array}{c} \boxed{\mathsf{-3\cdot (2+\sqrt{5})}}\end{array}$


Alternativa: D


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