Question

(UNI-RIO) Sen(x) √3/2 e x E ao 1º Quadrante calcule: y=sec(x)-Cossec(x)
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1-Tg(x)

Answer 1

Sabemos que

Se $\mathrm{sen\,}x=\dfrac{\sqrt{3}}{2},$ e $x$ é do 1º quadrante, então

$\cos x=\dfrac{1}{2}$

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$y=\dfrac{\sec(x)-\mathrm{cossec}(x)}{1-\mathrm{tg}(x)}\\\\\\ =\dfrac{\frac{1}{\cos x}-\frac{1}{\mathrm{sen\,}x}}{1-\frac{\mathrm{sen\,}x}{\cos x}}\\\\\\ =\dfrac{\frac{\mathrm{sen\,}x}{\mathrm{sen\,}x\cos x}-\frac{\cos x}{\mathrm{sen\,}x\cos x}}{1-\frac{\mathrm{sen\,}x}{\cos x}}\\\\\\ =\dfrac{\left(\frac{\mathrm{sen\,}x-\cos x}{\mathrm{sen\,}x\cos x} \right )}{\left(\frac{\cos x-\mathrm{sen\,}x}{\cos x} \right )}\\\\\\ =\dfrac{\mathrm{sen\,}x-\cos x}{\mathrm{sen\,}x\,\mathrm{c\!\diagup\!\!\!\! os\,}x}\cdot \dfrac{\,\mathrm{c\!\diagup\!\!\!\! os\,}x}{\cos x-\mathrm{sen\,}x}\\\\\\ =(\mathrm{sen\,}x-\cos x)\cdot \dfrac{1}{\cos x-\mathrm{sen\,}x}\\\\\\ =\dfrac{\mathrm{sen\,}x-\cos x}{\cos x-\mathrm{sen\,}x}$

$=\dfrac{-1\cdot (\cos x-\mathrm{sen\,}x)}{\cos x-\mathrm{sen\,}x}\\\\\\ =-1$