Trigonometria

[Kapizany]

Ae pessoal será q alguém pode me ajudar nessa questão ta bem complicada,
Supondo que Sec x + Tg x=22/7 e que Cossec x + Cotg x=m/n, com m e n primos entre si, o valor de m+n é:
a)40
b)42
c)44
d)46
e)48
A resposta é a letra c,
Obrigado!

[danjr5]

Condição I:

\(\sec x + \tan x = \frac{22}{7}\)

\(\frac{1}{\cos x} + \frac{\sin x}{\cos x} = \frac{22}{7}\)

\(\frac{1 + \sin x}{\cos x} = \frac{22}{7}\)

\(\left ( \frac{1 + \sin x}{\cos x} \right )^2 = \left ( \frac{22}{7} \right )^2\)

\(\frac{(1 + \sin x)^2}{\cos^2 x} = \frac{484}{49}\)

\(\frac{(1 + \sin x)^2}{1 - \sin^2 x} = \frac{484}{49}\)

\(\frac{(1 + \sin x)^2}{(1 + \sin x)(1 - \sin x)} = \frac{484}{49}\)

\(\frac{1 + \sin x}{1 - \sin x} = \frac{484}{49}\)

\(49 + 49 \cdot \sin x = 484 - 484 \cdot \sin x\)

\(533 \cdot \sin x = 435\)

\(\fbox{\sin x = \frac{435}{533}}\)


Condição II:

\(cossec \ x + cotan \ x = \frac{m}{n}\)

\(\frac{1}{\sin x} + \frac{\cos x}{\sin x} = \frac{m}{n}\)

\(\frac{1 + \cos x}{\sin x} = \frac{m}{n}\)

\(\left ( \frac{1 + \cos x}{\sin x} \right )^2 = \left ( \frac{m}{n} \right )^2\)

\(\frac{(1 + \cos x)^2}{\sin^2 x} = \frac{m^2}{n^2}\)

\(\frac{(1 + \cos x)^2}{1 - \cos^2 x} = \frac{m^2}{n^2}\)

\(\frac{(1 + \cos x)^2}{(1 + \cos x)(1 - \cos x)} = \frac{m^2}{n^2}\)

\(\frac{1 + \cos x}{1 - \cos x} = \frac{m^2}{n^2}\)


Da terceira linha da condição I temos que:

\(\frac{1 + \sin x}{\cos x} = \frac{22}{7}\)

\(22 \cdot \cos x = 7 + 7 \cdot \sin x\)

\(22 \cdot \cos x = 7 + 7 \cdot \frac{435}{533}\)

\(533 \cdot 22 \cdot \cos x = 7 \cdot 533 + 7 \cdot 435\)

\(533 \cdot 22 \cdot \cos x = 6776 \;\; \div (22\)

\(533 \cdot \cos x = 308\)

\(\fbox{\cos x = \frac{308}{533}}\)


Agora podemos prosseguir com a condição II:

\(\frac{1 + \cos x}{1 - \cos x} = \frac{m^2}{n^2}\)

\(\frac{1 + \frac{308}{533}}{1 - \frac{308}{533}} = \frac{m^2}{n^2}\)

\(\frac{m^2}{n^2} = \frac{\frac{533 + 308}{533}}{\frac{533 - 308}{533}}\)

\(\frac{m^2}{n^2} = \frac{841}{533} \div \frac{225}{533}\)

\(\frac{m^2}{n^2} = \frac{841}{533} \times \frac{533}{225}\)

\(\left ( \frac{m}{n} \right )^2 = \frac{841}{225}\)

\(\frac{m}{n} = \sqrt{\frac{841}{225}}\)

\(\frac{m}{n} = \frac{29}{15}\)

\(m + n = 29 + 15\)

\(\fbox{\fbox{m + n = 44}}\)