Bài 1:
Ta có: \(\tan\alpha+\cot\alpha=3\)
\(\Leftrightarrow\dfrac{\sin\alpha}{\cos\alpha}+\dfrac{\cos\alpha}{\sin\alpha}=3\)
\(\Leftrightarrow\dfrac{\sin^2\alpha}{\cos\alpha.\sin\alpha}+\dfrac{\cos^2\alpha}{\sin\alpha.\cos\alpha}=3\)
\(\Leftrightarrow\dfrac{\sin^2\alpha+\cos^2\alpha}{\sin\alpha.\cos\alpha}=3\)
\(\Leftrightarrow\dfrac{1}{\sin\alpha.\cos\alpha}=3\)
\(\Leftrightarrow\dfrac{1}{\sin\alpha.\cos\alpha}=\dfrac{3\left(\sin\alpha.\cos\alpha\right)}{\sin\alpha.\cos\alpha}\)
\(\Leftrightarrow1=3\left(\sin\alpha.\cos\alpha\right)\)
\(\Leftrightarrow\sin\alpha.\cos\alpha=\dfrac{1}{3}\)
Vậy \(\tan\alpha+\cot\alpha=3\) thì \(\sin\alpha.\cos\alpha=\dfrac{1}{3}\)
Chứng minh
\(AB^2=AC^2+BC^2-2AC.BC.\cos C\)
Kẻ \(AH\perp BC\)
Ta có: \(VP=\)\(AC^2+BC^2-2AC.BC.\cos C\)
\(=AC^2+BC^2-2AC.BC.\dfrac{CH}{AC}\)
\(=AC^2+BC^2-2BC.CH\)
\(=AH^2+HC^2+BH^2+HC^2+2BH.CH-2BH.CH-2CH^2\)
\(=AH^2+BH^2\)
\(=AB^2=VT\)
Vậy đẳng thức được chứng minh.