Question

[Lâm Anh]

cho \(\tan\)α = 2. Tính A = \(\dfrac{\sin\alpha+\cos\alpha}{\sin\alpha-\cos\alpha}\)

Answer 1

[Phạm Trần Hoàng Anh]

\(A=\dfrac{sin+cos}{sin-cos}\)

\(A=\dfrac{\dfrac{sin}{cos}+1}{\dfrac{sin}{cos}-1}\)

\(A=\dfrac{tan+1}{tan-1}\)

\(A=\dfrac{2+1}{2-1}\)

`A = 3`

Answer 2

[Nguyễn Lê Phước Thịnh]

\(tan\alpha=2\)

=>\(\dfrac{sin\alpha}{cos\alpha}=2\)

=>\(sin\alpha=2\cdot cos\alpha\)

\(A=\dfrac{sin\alpha+cos\alpha}{sin\alpha-cos\alpha}=\dfrac{2\cdot cos\alpha+cos\alpha}{2\cdot cos\alpha-cos\alpha}=\dfrac{2+1}{2-1}=3\)