Question

(Fuvest-SP) SE TG(X)=3/4 e x E ao 2°quadrante De: Cos(x)- sen(x) é :
A:7/5
B:-7/5
C:-2/5
D:1/3
E:1/5

Answer 1

Assumindo que

$\mathrm{tg\,}x=-\,\dfrac{3}{4}$

temos

$\dfrac{\mathrm{sen\,}x}{\cos x}=-\,\dfrac{3}{4}\\\\\\ 4\,\mathrm{sen\,}x=-3\cos x~~~~~~\mathbf{(i)}$


Elevando os dois lados ao quadrado, temos

$(4\,\mathrm{sen\,}x)^2=(-3\cos x)^2\\\\ 16\,\mathrm{sen^2\,}x=9\cos^2 x$


Mas $\cos^2 x=1-\mathrm{sen\,^2}x:$

$16\,\mathrm{sen^2\,}x=9\,(1-\mathrm{sen^2\,}x)\\\\ 16\,\mathrm{sen^2\,}x=9-9\,\mathrm{sen^2\,}x\\\\ 16\,\mathrm{sen^2\,}x+9\,\mathrm{sen^2\,}x=9\\\\ 25\,\mathrm{sen^2\,}x=9\\\\ \mathrm{sen^2\,}x=\dfrac{9}{25}\\\\\\ \mathrm{sen\,}x=\pm \sqrt{\dfrac{9}{25}}\\\\\\ \mathrm{sen\,}x=\pm \dfrac{3}{5}$


Como $x$ é do 2º quadrante, $\mathrm{sen\,}x>0.$ Portanto, desprezamos a raiz quadrada negativa, e obtemos

$\boxed{\begin{array}{c}\mathrm{sen\,}x=\dfrac{3}{5} \end{array}}$

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Substituindo em $\mathbf{(i)}$ o valor do seno encontrado,

$4\cdot \dfrac{3}{5}=-3\cos x\\\\\\ \dfrac{4\cdot 3}{5}=-3\cos x\\\\\\ \cos x=\dfrac{4\cdot \diagup\!\!\!\! 3}{5}\cdot \dfrac{1}{-\diagup\!\!\!\! 3}\\\\\\ \boxed{\begin{array}{c} \cos x=-\,\dfrac{4}{5} \end{array}}$

_______________

Calculando a expressão pedida:

$\cos x-\mathrm{sen\,}x=-\,\dfrac{4}{5}-\dfrac{3}{5}\\\\\\ \cos x-\mathrm{sen\,}x=\dfrac{-4-3}{5}\\\\\\ \boxed{\begin{array}{c}\cos x-\mathrm{sen\,}x=-\,\dfrac{7}{5} \end{array}}$


Resposta: alternativa $\text{B. }-7/5.$