Question

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

Answer 1

\(\tan\alpha=2\Rightarrow\dfrac{\sin\alpha}{\cos\alpha}=2\Rightarrow\sin\alpha=2\cos\alpha\)

Ta có:

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

\(A=\dfrac{2\cos\alpha+\cos\alpha}{2\cos\alpha-\cos\alpha}\)

\(A=\dfrac{3\cos\alpha}{\cos\alpha}\)

\(A=3\)

Answer 2

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