Question

(UFMS) Dado:Cos(x)=4/5 ao 1º Quadrante calcule o valor de Y:12 (Sen(x)-Cossec(x)
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1-tg (x)

Answer 1

$x$ é do 1º quadrante e

$\cos x=\dfrac{4}{5}$


$\bullet~~$ Encontrando $\mathrm{sen\,}x:$

Usando a Relação Trigonométrica Fundamental,

$\cos^2 x+\mathrm{sen^2\,} x=1\\\\ \mathrm{sen^2\,} x=1-\cos^2 x\\\\ \mathrm{sen^2\,} x=1-\left(\dfrac{4}{5} \right )^{\!\!2}\\\\\\ \mathrm{sen^2\,} x=1-\dfrac{16}{25}\\\\\\ \mathrm{sen^2\,} x=\dfrac{25}{25}-\dfrac{16}{25}\\\\\\ \mathrm{sen^2\,} x=\dfrac{25-16}{25}\\\\\\ \mathrm{sen^2\,} x=\dfrac{9}{25}$

$\mathrm{sen\,} x=\pm \sqrt{\dfrac{9}{25}}\\\\\\ \mathrm{sen\,} x=\pm\,\dfrac{3}{5}$


Como $x$ é do 1º quadrante, $\mathrm{sen\,} x>0.$ Portanto, desprezamos o sinal negativo, e obtemos

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

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$\bullet~~$ Avaliando a expressão dada:

$y=\dfrac{12\,(\mathrm{sen\,}x-\mathrm{cossec\,}x)}{1-\mathrm{tg\,}x}\\\\\\ y=\dfrac{12\,\big(\mathrm{sen\,}x-\frac{1}{\mathrm{sen\,}x}\big)}{1-\frac{\mathrm{sen\,}x}{\cos x}}\\\\\\ y=\dfrac{12\,\big(\frac{\mathrm{sen^2\,}x}{\mathrm{sen\,}x}-\frac{1}{\mathrm{sen\,}x}\big)}{\frac{\cos x}{\cos x}-\frac{\mathrm{sen\,}x}{\cos x}}\\\\\\ y=\dfrac{12\,\big(\frac{\mathrm{sen^2\,}x-1}{\mathrm{sen\,}x}\big)}{\frac{\cos x-\mathrm{sen\,}x}{\cos x}}$

$y=12\cdot \dfrac{\mathrm{sen^2\,}x-1}{\mathrm{sen\,}x}\cdot \dfrac{\cos x}{\cos x-\mathrm{sen\,}x}\\\\\\ y=-12\cdot \dfrac{1-\mathrm{sen^2\,}x}{\mathrm{sen\,}x}\cdot \dfrac{\cos x}{\cos x-\mathrm{sen\,}x}\\\\\\ y=-12\cdot \dfrac{\cos^2 x}{\mathrm{sen\,}x}\cdot \dfrac{\cos x}{\cos x-\mathrm{sen\,}x}\\\\\\ y=-12\cdot \dfrac{\cos^3 x}{\mathrm{sen\,}x\,(\cos x-\mathrm{sen\,}x)}$


Substituindo os valores conhecidos, temos

$y=-12\cdot \dfrac{\big(\frac{4}{5}\big)^3}{\frac{3}{5}\cdot \big(\frac{4}{5}-\frac{3}{5}\big)}\\\\\\ y=-12\cdot \dfrac{\big(\frac{4}{5}\big)^3}{\frac{3}{5}\cdot \frac{1}{5}}\\\\\\ y=-12\cdot \dfrac{\big(\frac{4^3}{5^3}\big)}{\frac{3}{5^2}}\\\\\\ y=-12\cdot \dfrac{4^3}{5^3}\cdot \dfrac{5^2}{3}\\\\\\ y=-12\cdot \dfrac{4^3}{5}\cdot \dfrac{1}{3}\\\\\\ y=-4\cdot \dfrac{64}{5}\\\\\\ \boxed{\begin{array}{c}y=-\,\dfrac{256}{5} \end{array}}$