Question

Oi! Quero a resolução desse exercício de matemática , se alguém puder me ajudar ficarei muuuito feliz ! Então vamos lá : Calculeo valor da expressão :
X=cossec a - sen a /sec a - cos a. Sabendo que Cos a= 1/2 , 0

Answer 1

Temos o seguinte:

$X = \dfrac{csc( \alpha ) - sin ( \alpha )}{sec( \alpha ) - cos( \alpha )}$

Logo temos:

$csc( \alpha ) - sin( \alpha ) = \frac{1}{sin( \alpha )} -sin( \alpha ) = \frac{1-sin^2( \alpha )}{sin( \alpha )} \\ \\ sec( \alpha ) - cos( \alpha ) = \frac{1}{cos( \alpha )} -cos( \alpha ) = \frac{1 - cos^2( \alpha )}{cos( \alpha )}$

Usando:

$\boxed{sin^2(\alpha) + cos^2(\alpha) = 1} \\ \\ sin^2( \alpha ) = 1 - (\frac{1}{2} )^2 \\ \\ sin^2( \alpha ) = 1 - \frac{1}{4} \\ \\ sin( \alpha ) = \sqrt{ \frac{3}{4} } = \frac{ \sqrt{3} }{2}$

Logo:

$\frac{1-sin^2( \alpha )}{sin( \alpha )} = \dfrac{1 - ( \frac{ \sqrt{3} }{2} )^2}{ \frac{ \sqrt{3} }{2} } = \dfrac{1 - \frac{3}{4} }{ \frac{ \sqrt{3} }{2} } = \frac{1}{4} * \frac{2}{ \sqrt{3} } = \frac{ 1 }{2 \sqrt{3} } = \frac{ \sqrt{3} }{6} \\ \\ \\ \frac{1 - cos^2( \alpha )}{cos( \alpha )} = \dfrac{1 - ( \frac{1}{2} )^2}{ \frac{1}{2} } = \dfrac{1 - \frac{1}{4} }{ \frac{1}{2} } = \frac{3}{4} * \frac{2}{1} = \frac{3}{2}$

Assim:

$X = \dfrac{csc( \alpha ) - sin ( \alpha )}{sec( \alpha ) - cos( \alpha )} = \dfrac{ \frac{ \sqrt{3} }{6} }{ \frac{3}{2} } = \frac{ \sqrt{3} }{6} * \frac{2}{3} = \boxed{ \frac{ \sqrt{3} }{9} }$

Answer 2

$\mathrm{x=\dfrac{\csc{a}-\sin{a}}{\sec{a}-\cos{a}}\ \ \bigg\| \ \ \cos{a}=\dfrac{1}{2}}\\\\\\ \textbf{Desenvolvendo a express\~ao:}\\\\ \mathrm{x=\dfrac{\bigg(\dfrac{1}{\sin{a}}-\sin{a}\bigg)}{\bigg(\dfrac{1}{\cos{a}}-\cos{a}\bigg)}=\dfrac{\bigg(\dfrac{1-\sin^2{a}}{\sin{a}}\bigg)}{\bigg(\dfrac{1-\cos^2{a}}{\cos{a}}\bigg)}}\\\\\\ \mathrm{Lembrando\ que\ \to\ \boxed{\mathrm{\sin^2{a}+\cos^2{a}=1}}}$

$\mathrm{x=\dfrac{\bigg(\dfrac{cos^2{a}}{\sin{a}}\bigg)}{\bigg(\dfrac{\sin^2{a}}{\cos{a}}\bigg)}=\dfrac{\cos^2{a}}{\sin{a}}\times\dfrac{\cos{a}}{\sin^2{a}}=\dfrac{\cos^3{a}}{\sin^3{a}}}\\\\\\ \mathbf{Calculando\ \sin{a}:}\\\\ \mathrm{\sin{a}=\pm\sqrt{1-\cos^2{a}}=\pm\sqrt{1-\bigg(\dfrac{1}{2}\bigg)^2}}\\\\\\ \mathrm{\sin{x}=\pm\sqrt{\dfrac{4}{4}-\dfrac{1}{4}}=\pm\sqrt{\dfrac{3}{4}}=\pm\dfrac{\sqrt{3}}{2}}\\\\\\ \mathrm{Como\ 0\ \textless \ a\ \textless \ \dfrac{\pi}{2}\ \to\ \boxed{\mathrm{\sin{a}=\dfrac{\sqrt{3}}{2}}}}$

$\textbf{Voltando \`a express\~ao:}\\\\ \mathrm{x=\dfrac{\cos^3{a}}{\sin^3{a}}=\dfrac{\bigg(\dfrac{1}{2}\bigg)^3}{\bigg(\dfrac{\sqrt{3}}{2}\bigg)^3}=\dfrac{\bigg(\dfrac{1}{8}\bigg)}{\bigg(\dfrac{3\sqrt{3}}{8}\bigg)}}\\\\\\ \mathrm{x=\dfrac{1}{8}\times\dfrac{8}{3\sqrt{3}}=\dfrac{1}{3\sqrt{3}}\times\dfrac{\sqrt{3}}{\sqrt{3}}=\dfrac{\sqrt{3}}{9}}\\\\\\ \mathbf{Resposta\ \to\ \boxed{\mathbf{x=\dfrac{\sqrt{3}}{9}}}}$