Question

TanA=3. Tính

B=(sinA-cosA)/ (sin^3A+3.cos^3A+2sinA

Answer 1

\(\tan A=\frac{\sin A}{\cos A}=3\Rightarrow\sin A=3\cos A\Leftrightarrow\sin^3A=27\cos^3A\)

\(B=\frac{\sin A-\cos A}{\sin^3A+3\cos^3A+2\sin A}=\frac{3\cos A-\cos A}{27\cos^3A+27\cos^3A+6\cos A}=\frac{1}{27\cos^2A+3}\) (*)

Ta có công thức:\(1+\tan^2A=\frac{1}{\cos^2A}\Leftrightarrow\cos^2A=\frac{1}{1+\tan^2A}=\frac{1}{1+3^2}=\frac{1}{10}\)

thay vào (*) ta được: \(B=\frac{1}{27\cos^2A+3}=\frac{1}{27\times\frac{1}{10}+3}=\frac{10}{57}\)