Question

[Thảo Vũ]

cho cosα =\(\frac{3}{5}\). Tính giá trị củ biểu thức A=\(\frac{tan\alpha+\cot\alpha+1}{\tan\alpha-\cot\alpha+1}\)

Answer 1

[Nguyễn Việt Lâm]

\(tan^2a=\frac{sin^2a}{cos^2a}=\frac{1-cos^2a}{cos^2a}=\frac{1-\left(\frac{3}{5}\right)^2}{\left(\frac{3}{5}\right)^2}=\frac{16}{9}\Rightarrow\left[{}\begin{matrix}tana=\frac{4}{3}\\tana=-\frac{4}{3}\end{matrix}\right.\)

Với \(tana=\frac{4}{3}\Rightarrow cota=\frac{3}{4}\)

\(A=\frac{\frac{4}{3}+\frac{3}{4}+1}{\frac{4}{3}-\frac{3}{4}+1}=\frac{37}{19}\)

Với \(tana=-\frac{4}{3}\Rightarrow cota=-\frac{3}{4}\)

\(A=\frac{-\frac{4}{3}-\frac{3}{4}+1}{-\frac{4}{3}+\frac{3}{4}+1}=-\frac{13}{5}\)