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C A H B
Gỉa sử \(\Delta ABC\)cân tại C, kẻ \(CH⊥AB\)
Ta có VT= \(\cos^2A=\frac{AH^2}{AC^2};\cos^2B=\frac{BH^2}{BC^2}\Rightarrow\cos^2A+\cos^2B=\frac{AH^2}{AC^2}+\frac{BH^2}{BC^2}=2.\frac{AH^2}{AC^2}\)do \(\hept{\begin{cases}AH=BH\\AC=BC\end{cases}}\)
\(\sin^2A=\frac{CH^2}{CA^2};\sin^2B=\frac{CH^2}{CB^2}\Rightarrow\sin^2A+\sin^2B=2.\frac{CH^2}{CA^2}\)
\(\Rightarrow\frac{\cos^2A+\cos^2B}{\sin^2A+\sin^2B}=\frac{2.\frac{AH^2}{AC^2}}{2.\frac{CH^2}{AC^2}}=\frac{AH^2}{CH^2}\)
Ta có VP =\(\frac{1}{2}\left(\cot^2A+\cot^2B\right)=\frac{1}{2}.\left(\frac{AH^2}{CH^2}+\frac{BH^2}{CH^2}\right)=\frac{1}{2}\left(2.\frac{AH^2}{CH^2}\right)=\frac{AH^2}{CH^2}\)
Ta thấy VT=VP\(\Rightarrow\)giả sử đúng
Vậy ........
\(A=\left(\sin\alpha+\cos\alpha+\sin\alpha-\cos\alpha\right)^2-2\left(\sin\alpha+\cos\alpha\right)\left(\sin\alpha-\cos\alpha\right)\)
\(=4\sin^2\alpha-2\sin^2\alpha+2\cos^2\alpha=2\left(\sin^2\alpha+\cos^2\alpha\right)=2\)
\(B=\sin^4\alpha+\cos^4\alpha+2\sin^2\alpha.\cos^2\alpha\left(\sin^2\alpha+\cos^2\alpha\right)=\sin^4\alpha+\cos^4\alpha+2\sin^2\alpha.\cos^2\alpha\)
\(=\left(\sin^2\alpha+\cos^2\alpha\right)^2-1=0\)
\(C=3\left(\sin^4\alpha+\cos^4\alpha\right)-2\sin^2\alpha.\cos^2\alpha\left(\sin^2\alpha+\cos^2\alpha\right)=3\left(\sin^4\alpha+\cos^4\alpha\right)-2\sin^2\alpha.\cos^2\alpha\)
\(=3\left(\sin^2\alpha+\cos^2\alpha-\frac{1}{9}\right)^2-\frac{1}{9}=\frac{61}{27}\)
a/\(sin^4\alpha+cos^4\alpha+2sin^2\alpha.cos^2\alpha=\left(sin^2\alpha+cos^2\alpha\right)^2=1\)
b/ \(tan^2\alpha-sin^2\alpha.tan^2\alpha=tan^2\alpha\left(1-sin^2\alpha\right)=\frac{sin^2\alpha}{cos^2\alpha}.cos^2\alpha=sin^2\alpha\)
c/ \(cos^2\alpha+tan^2\alpha.cos^2\alpha=cos^2\alpha\left(1+tan^2\alpha\right)\)
\(=cos^2\alpha.\left(1+\frac{sin^2\alpha}{cos^2\alpha}\right)=cos^2\alpha.\left(\frac{sin^2\alpha+cos^2\alpha}{cos^2\alpha}\right)\)
\(=cos^2.\frac{1}{cos^2\alpha}=1\)