Rafaela,
sabendo que [tex]\sec x = \frac{1}{\cos x}[/tex] podemos encontrar o valor do cosseno de x, veja:
[tex]\\ \sec x = \frac{1}{\cos x} \\\\ \frac{7}{3} = \frac{1}{\cos x} \\\\ 7 \cdot \cos x = 3 \\\\ \boxed{\cos x = \frac{3}{7}}[/tex]
Já que temos o valor do cosseno podemos encontrar o valor do seno de x através da relação [tex]\sin^2 x + \cos^2 x = 1[/tex]
[tex]\\ \sin^2 x + \cos^2 x = 1 \\\\ \sin^2 x + \left ( \frac{3}{7} \right )^2 = 1 \\\\ \sin^2 x = 1 - \frac{9}{49} \\\\ \sin^2 x = \frac{40}{49} \\\\ \sin x = \pm \frac{\sqrt{40}}{\sqrt{49}} \rightarrow \; \text{Como x pertence ao quarto quadrante,\\ o valor do seno de x e negativo!} \\\\ \sin x = - \frac{\sqrt{4 \cdot 10}}{7} \\\\ \boxed{\sin x = - \frac{2\sqrt{10}}{7}}[/tex]
Por fim, encontramos o valor da tangente de x.
[tex]\\ \tan x = \frac{\sin x}{\cos x} \\\\ \tan x = \frac{\frac{- 2\sqrt{10}}{7}}{\frac{3}{7}} \\\\ \tan x = \frac{- 2\sqrt{10}}{7} \div \frac{3}{7} \\\\ \tan x = \frac{- 2\sqrt{10}}{7} \times \frac{7}{3} \\\\ \boxed{\boxed{\tan x = \frac{- 2\sqrt{10}}{3}}}[/tex]
Já ia esquecendo o valor de "m"!
Segue,
[tex]\\ cotg \; x = \frac{1}{\tan \; x} \\\\ cotg \; x \cdot \tan x = 1 \\\\ 1,5m \cdot \frac{- 2\sqrt{10}}{3} = 1 \\\\ - 3m \cdot \sqrt{10} = 3 \;\;\;\; \div(- 3 \\\\ m \cdot \sqrt{10} = - 1 \\\\ m = \frac{- 1}{\sqrt{10}} \times \frac{\sqrt{10}}{\sqrt{10}} \\\\ \boxed{\boxed{m = \frac{- \sqrt{10}}{10}}}[/tex]
