Question

1-Sendo Cos(x)=-5/13 e x E ao 2º quadrante,determine:

A) Sen (x):
B) Tg (x):
C) Cotg (x):
D) Sec (x):
E) Cossec (x):

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Answer 1

$\cos x=-\,\dfrac{5}{13}$


a) $\mathrm{sen\,}x:$

Aqui, basta usarmos a Relação Trigonométrica Fundamental:

$\cos^2 x+\mathrm{sen^2\,}x=1\\\\ \mathrm{sen^2\,}x=1-\cos^2 x\\\\ \mathrm{sen^2\,}x=1-\left(-\dfrac{5}{13} \right )^{\!\!2}\\\\\\ \mathrm{sen^2\,}x=1-\dfrac{25}{169}\\\\\\ \mathrm{sen^2\,}x=\dfrac{169}{169}-\dfrac{25}{169}\\\\\\ \mathrm{sen^2\,}x=\dfrac{169-25}{169}\\\\\\ \mathrm{sen^2\,}x=\dfrac{144}{169}$

$\mathrm{sen\,}x=\pm \sqrt{\dfrac{144}{169}}\\\\\\ \mathrm{sen\,}x=\pm \dfrac{12}{13}$


Mas como $x$ é do 2º quadrante, $\mathrm{sen\,}x>0.$ Logo, desprezamos a raiz quadrada com sinal negativo, e obtemos

$\boxed{\begin{array}{c}\mathrm{sen\,}x=\dfrac{12}{13} \end{array}}$

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b) $\mathrm{tg\,}x:$

$\mathrm{tg\,}x=\dfrac{\mathrm{sen\,}x}{\cos x}\\\\\\ \mathrm{tg\,}x=\dfrac{\frac{12}{13}}{\left(-\frac{5}{13} \right )}\\\\\\ \mathrm{tg\,}x=\dfrac{12}{\diagup\!\!\!\!\! 13}\cdot \dfrac{\diagup\!\!\!\!\! 13}{-5}\\\\\\ \boxed{\begin{array}{c}\mathrm{tg\,}x=-\,\dfrac{12}{5} \end{array}}$

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c) $\mathrm{cotg\,}x:$

$\mathrm{cotg\,}x=\dfrac{1}{\mathrm{tg\,}x}\\\\\\ \mathrm{cotg\,}x=\dfrac{1}{\left(-\frac{12}{5} \right )}\\\\\\ \boxed{\begin{array}{c}\mathrm{cotg\,}x=-\,\dfrac{5}{12} \end{array}}$

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d) $\sec x:$

$\sec x=\dfrac{1}{\cos x}\\\\\\ \sec x=\dfrac{1}{\left(-\frac{5}{13}\right)}\\\\\\ \boxed{\begin{array}{c} \sec x=-\,\dfrac{13}{5} \end{array}}$

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e) $\mathrm{cossec\,}x:$

$\mathrm{cossec\,}x=\dfrac{1}{\mathrm{sen\,}x}\\\\\\ \mathrm{cossec\,}x=\dfrac{1}{\left(\frac{12}{13} \right )}\\\\\\ \boxed{\begin{array}{c}\mathrm{cossec\,}x=\dfrac{13}{12} \end{array}}$