Question

Cho hệ phương trình: \(\left\{{}\begin{matrix}secx+tanx=\dfrac{22}{7}\\cscx+cotx=\dfrac{m}{n}\end{matrix}\right.\), với \(\dfrac{m}{n}\) tối giản.

Tính \(S=m+n\).

Answer 1

\(\dfrac{1}{cosx}+\dfrac{sinx}{cosx}=\dfrac{1+sinx}{cosx}=\dfrac{\left(sin\dfrac{x}{2}+cos\dfrac{x}{2}\right)^2}{\left(cos\dfrac{x}{2}-sin\dfrac{x}{2}\right)\left(cos\dfrac{x}{2}+sin\dfrac{x}{2}\right)}\)

\(=\dfrac{cos\dfrac{x}{2}+sin\dfrac{x}{2}}{cos\dfrac{x}{2}-sin\dfrac{x}{2}}=\dfrac{1+tan\dfrac{x}{2}}{1-tan\dfrac{x}{2}}=\dfrac{22}{7}\)

\(\Rightarrow tan\dfrac{x}{2}=\dfrac{15}{29}\)

\(\dfrac{1}{sinx}+\dfrac{cosx}{sinx}=\dfrac{1+cosx}{sinx}=\dfrac{1+2cos^2\dfrac{x}{2}-1}{2sin\dfrac{x}{2}cos\dfrac{x}{2}}=\dfrac{cos\dfrac{x}{2}}{sin\dfrac{x}{2}}=\dfrac{1}{tan\dfrac{x}{2}}=\dfrac{29}{15}\)

\(\Rightarrow m=29;n=15\)