Question

Sabendo que Sen a = -3/5 e Cotg b = 24/7, sendo que a está no 3° quadrante e 0 < b < pi/2, Calcule:
E = (sen a . sec b - tg a) / (cotg a . cotg b + cossec b)​

Answer 1

a∈ 3ºQ

b∈ 1ºQ

tgb= 7/24

$\cos(a) = - \sqrt{1 - { \sin}^{2}a} \\ = - \sqrt{1 - {( - \frac{3}{5} )}^{2} }$

$- \sqrt{1 - \frac{9}{25} } = - \sqrt{ \frac{25 - 9}{25} } \\ = - \sqrt{ \frac{16}{25} } = - \frac{4}{5} .$

${ \sec }^{2}b =1 + { \tan }^{2} b \\ { \sec }^{2}b = 1 + { (\frac{7}{24})}^{2} = 1 + \frac{49}{576} \\ { \sec }^{2}b = \frac{625}{576}$

$\sec(b) = \frac{25}{24}$

$\tan(a) = \frac{ \sin(a) }{ \cos(a) }= \\( -\frac{3}{5}) \div (-\frac{4}{5}) \\ = ( - \frac{3}{5} ).( - \frac{5}{4} ) = \frac{3}{4}$

$\cot(a) = \frac{4}{3}$

${ \csc }^{2}b = 1 + { \cot }^{2}b \\ = 1 + { (\frac{24}{7}) }^{2} = 1 + \frac{576}{49 } \\ = \frac{49 + 576}{49} = \frac{625}{49}$

$\csc(b) = \frac{25}{7}$

E= (sen a . sec b - tg a) / (cotg a . cotg b + cossec b)

$E = \frac{( - \frac{3}{5}. \frac{25}{24} - \frac{3}{4})}{ \frac{4}{3}. \frac{24}{7} + \frac{25}{7}} \\ E = \frac{ (- \frac{5}{8} - \frac{3}{4}) }{ \frac{32}{7} + \frac{25}{7} }$

$E = \frac{ \frac{ - 5 - 6}{8} }{ \frac{57}{7} } = \frac{ - \frac{11}{8} }{ \frac{57}{7} } \\ E = ( -\frac{11}{8}). \frac{7}{57} = - \frac{77}{456}$