Đặt sina=a; cosa=b

Theo đề, ta có: \(\left\{{}\begin{matrix}a+b=1.4\\a^2+b^2=1\end{matrix}\right.\Leftrightarrow ab=\dfrac{1.4^2-1}{2}=0.48\)

=>a,b là các nghiệm của pt là:

\(x^2-1.4x+0.48=0\)

=>x=0,6 hoặc x=0,8

=>(a,b)=(0,6;0,8) hoặc (a,b)=(0,8;0,6)

TH1: a=0,6; b=0,8

tan a=a/b=3/4

TH2: a=0,8; b=0,6

tan a=a/b=4/3

sin²α + cos²α = 1

⇒ cosα = \(\sqrt{1-sin^2\alpha}\) = \(\sqrt{1-\left(\dfrac{2}{5}\right)^2}\) =\(\sqrt{\dfrac{21}{25}}\) = \(\dfrac{\sqrt{21}}{5}\)

tanα = \(\dfrac{sin\alpha}{cos\alpha}\) = \(\dfrac{2}{5}:\dfrac{\sqrt{21}}{25}\) = \(\dfrac{10}{\sqrt{21}}\)

cot α = \(\dfrac{\sqrt{21}}{10}\)

Có \(\tan\alpha=\frac{\sin\alpha}{\cos\alpha};\cot\alpha=\frac{\cos\alpha}{\sin\alpha}\)

\(\Rightarrow\frac{\sin\alpha}{\cos\alpha}+\frac{\cos\alpha}{\sin\alpha}=8\)

\(\Leftrightarrow\sin^2\alpha+\cos^2\alpha=8\sin\alpha.\cos\alpha\)

\(\Leftrightarrow\sin\alpha.\cos\alpha=\frac{1}{8}\Leftrightarrow2\sin\alpha.\cos\alpha=\frac{1}{4}\)

Có \(\sin^2\alpha+\cos^2\alpha=1\Leftrightarrow\sin^2\alpha+2\sin\alpha.\cos\alpha+\cos^2\alpha=1+2\sin\alpha.\cos\alpha\)

\(\Leftrightarrow\left(\sin\alpha+\cos\alpha\right)^2=1+\frac{1}{4}=\frac{5}{4}\)

\(\Leftrightarrow\sin\alpha+\cos\alpha=\frac{\sqrt{5}}{2}\)

như tek ko hay, xét tek náy cho độc