Question

[Tran kha]

Giúp e với mọi người

Answer 1

[Khôi Bùi]

Có : 0 < \(\alpha< \dfrac{\pi}{2}\Rightarrow cos\alpha>0\)  \(\Rightarrow cos\alpha=\sqrt{1-\left(\dfrac{3}{4}\right)^2}=\dfrac{\sqrt{7}}{4}\)

\(\Rightarrow sin2\alpha=2sin\alpha.cos\alpha=2.\dfrac{3}{4}.\dfrac{\sqrt{7}}{4}=\dfrac{3\sqrt{7}}{8}\) 

\(cos2\alpha=1-2sin^2\alpha=1-2.\left(\dfrac{3}{4}\right)^2=-\dfrac{1}{8}\)

Ta có : \(A=\dfrac{cot2\alpha}{tan2\alpha+sin2\alpha}=\dfrac{1}{sin2\alpha+\dfrac{sin^22\alpha}{cos2\alpha}}=\dfrac{1}{\dfrac{3\sqrt{7}}{8}+\dfrac{\dfrac{63}{64}}{-\dfrac{1}{8}}}=\dfrac{8}{-63+3\sqrt{7}}\)

Answer 2

[Khôi Bùi]

Sửa lại : \(A=\dfrac{cot2\alpha}{tan2\alpha+sin2\alpha}=\dfrac{1}{tan^22\alpha+\dfrac{sin^22\alpha}{cos2\alpha}}=\dfrac{1}{\left(-3\sqrt{7}\right)^2+\left(\dfrac{3\sqrt{7}}{8}\right)^2:-\dfrac{1}{8}}=\dfrac{8}{441}\)