Question

[Nguyễn Trung Kiên]

A=\(\frac{1-2\sin\alpha.\cos\alpha}{\sin^2\alpha-\cos\alpha}\)Tính giá trị của A khi \(\tan\alpha=\frac{1}{3}\)

Answer 1

[Lê Song Phương]

Vì \(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}\), lại có \(\tan\alpha=\frac{1}{3}\)\(\Rightarrow\frac{\sin\alpha}{\cos\alpha}=\frac{1}{3}\)\(\Rightarrow\cos\alpha=3\sin\alpha\)

Mà \(\sin^2\alpha+\cos^2\alpha=1\)\(\Rightarrow\sin^2\alpha+9\sin^2\alpha=1\)\(\Rightarrow10\sin^2\alpha=1\)\(\Rightarrow\sin^2\alpha=\frac{1}{10}\)\(\Rightarrow\sin\alpha=\frac{\sqrt{10}}{10}\)

Ta có \(A=\frac{1-2\sin\alpha\cos\alpha}{\sin^2\alpha-\cos\alpha}=\frac{1-2\sin\alpha.3\sin\alpha}{\sin^2\alpha-3\sin\alpha}=\frac{1-6\sin^2\alpha}{\sin^2\alpha-3\sin\alpha}=\frac{1-\frac{6}{10}}{\frac{1}{10}-\frac{3\sqrt{10}}{10}}=\frac{\frac{4}{10}}{\frac{1-3\sqrt{10}}{10}}\)

\(=\frac{4}{1-3\sqrt{10}}=\frac{4\left(1+3\sqrt{10}\right)}{\left(1-3\sqrt{10}\right)\left(1+3\sqrt{10}\right)}=\frac{4\left(1+3\sqrt{10}\right)}{1^2-\left(3\sqrt{10}\right)^2}=\frac{4\left(1+3\sqrt{10}\right)}{1-90}=\frac{4+12\sqrt{10}}{-89}\)

Answer 2

[Nguyễn Nam Dương]

Đề sai nhé

Gía trị của \(1-2\sin.a.\cos\)không thể có giá trị \(\tan a=\frac{1}{3}\)

HT