Question

Dado tg x = 5/4, determine cossec x

Answer 1

$\large\begin{array}{l} \mathsf{tg\,x=\dfrac{5}{4}}\\\\ \mathsf{\dfrac{sen\,x}{cos\,x}=\dfrac{5}{4}}\\\\ \mathsf{4\,sen\,x=5\,cos\,x} \end{array}$


$\large\begin{array}{l} \textsf{Elevando os dois lados ao quadrado, obtemos}\\\\ \mathsf{(4\,sen\,x)^2=(5\,cos\,x)^2}\\\\ \mathsf{16\,sen^2\,x=25\,cos^2\,x\qquad\quad}\textsf{(mas }\mathsf{cos^2\,x=1-sen^2\,x}\textsf{)}\\\\ \mathsf{16\,sen^2\,x=25\cdot (1-sen^2\,x)}\\\\ \mathsf{16\,sen^2\,x=25-25\,sen^2\,x}\\\\ \mathsf{16\,sen^2\,x+25\,sen^2\,x=25}\\\\ \mathsf{41\,sen^2\,x=25}\end{array}$

$\large\begin{array}{l} \mathsf{sen^2\,x=\dfrac{25}{41}}\\\\ \mathsf{sen\,x=\pm\sqrt{\dfrac{25}{41}}}\\\\ \mathsf{sen\,x=\pm\,\dfrac{5}{\sqrt{41}}\qquad\quad}\left(\textsf{mas }\mathsf{sen\,x=\dfrac{1}{cossec\,x}}\right)\\\\ \mathsf{\dfrac{1}{cossec\,x}=\pm\,\dfrac{5}{\sqrt{41}}} \end{array}$

$\large\begin{array}{l} \mathsf{cossec\,x=\pm\,\dfrac{\sqrt{41}}{5}} \end{array}$


$\large\begin{array}{l} \textsf{Como a tangente de x \'e positiva, ent\~ao x pode ser do }\\\textsf{primeiro ou terceiro quadrantes.}\\\\ \bullet~~\textsf{Se x for do primeiro quadrante, ent\~ao}\\\\ \mathsf{cossec\,x>1}\\\\ \boxed{\begin{array}{c}\mathsf{cossec\,x=\dfrac{\sqrt{41}}{5}} \end{array}}\\\\\\ \bullet~~\textsf{Se x for do terceiro quadrante, ent\~ao}\\\\ \mathsf{cossec\,x<-1}\\\\ \boxed{\begin{array}{c}\mathsf{cossec\,x=-\,\dfrac{\sqrt{41}}{5}} \end{array}} \end{array}$


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$\large\textsf{Bons estudos! :-)}$