Question

Um arco θ do terceiro quadrante apresenta tg θ= 1/3.
Obtenha o valor de $\frac{sen\theta  + cos\theta}{1 + sen\theta}$

Answer 1

$\dfrac{\mathrm{sen\,}\theta+\cos \theta}{1+\mathrm{sen\,}\theta}$


Multiplicando o numerador e o denominador por $\dfrac{1}{\cos \theta},$

$=\dfrac{\left(\mathrm{sen\,}\theta+\cos \theta \right )\cdot \frac{1}{\cos \theta}}{\left(1+\mathrm{sen\,}\theta \right )\cdot \frac{1}{\cos \theta}}\\\\\\ =\dfrac{\frac{\mathrm{sen\,}\theta}{\cos \theta}+\frac{\cos \theta}{\cos \theta}}{\frac{1}{\cos \theta}+\frac{\mathrm{sen\,}\theta}{\cos \theta}}\\\\\\ =\dfrac{\mathrm{tg\,}\theta+1}{\sec \theta+\mathrm{tg\,}\theta}~~~~~~\mathbf{(i)}$

________

Vamos encontrar a secante, usando a identidade trigonométrica:

$\sec^2 \theta=1+\mathrm{tg^2\,}\theta\\\\ \sec^2 \theta=1+\left(\dfrac{1}{3} \right )^{\!2}\\\\\\ \sec^2 \theta=1+\dfrac{1}{9}\\\\\\ \sec^2 \theta=\dfrac{9}{9}+\dfrac{1}{9}\\\\\\ \sec^2 \theta=\dfrac{9+1}{9}\\\\\\ \sec^2 \theta=\dfrac{10}{9}$

$\sec \theta=\pm \sqrt{\dfrac{10}{9}}\\\\\\ \sec \theta=\pm \dfrac{\sqrt{10}}{3}$


Como $\theta$ é do 3º quadrante, a secante é negativa. Portanto,

$\sec \theta=-\,\dfrac{\sqrt{10}}{3}$

________

Substituindo em $\mathbf{(i)}$ os valores da tangente e da secante, fica

$=\dfrac{\frac{1}{3}+1}{-\frac{\sqrt{10}}{3}+\frac{1}{3}}\\\\\\ =\dfrac{\left(\frac{1}{3}+1 \right )\cdot 3}{\left(-\frac{\sqrt{10}}{3}+\frac{1}{3} \right )\cdot 3}\\\\\\ =\dfrac{1+3}{-\sqrt{10}+1}\\\\\\ =\dfrac{4}{1-\sqrt{10}}$


poderia parar aqui. Mas caso queira racionalizar o denominador (não é necessário)

$=\dfrac{4}{1-\sqrt{10}}\cdot \dfrac{1+\sqrt{10}}{1+\sqrt{10}}\\\\\\ =\dfrac{4\cdot \big(1+\sqrt{10}\big)}{\big(1-\sqrt{10}\big)\cdot \big(1+\sqrt{10}\big)}\\\\\\ =\dfrac{4\cdot \big(1+\sqrt{10}\big)}{1^2-\big(\sqrt{10}\big)^{2}}\\\\\\ =\dfrac{4\cdot \big(1+\sqrt{10}\big)}{1-10}\\\\\\ =\dfrac{4\cdot \big(1+\sqrt{10}\big)}{-9}\\\\\\\\ \therefore~~\boxed{\begin{array}{c}\dfrac{\mathrm{sen\,}\theta+\cos \theta}{1+\mathrm{sen\,}\theta}=-\,\dfrac{4}{9}\big(1+\sqrt{10}\big) \end{array}}$


Dúvidas? Comente.


Bons estudos! :-)