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Giả sử có \(\Delta ABC\) có \(A=90^o;AH\) là đường cao
Có \(\sin\widehat{B}=\frac{AC}{BC};\cos\widehat{B}=\frac{AB}{BC};\tan\widehat{B}=\frac{AC}{AB};\cot\widehat{B}=\frac{AB}{AC}\)
\(\frac{\cot^2\widehat{B}-\cos^2\widehat{B}}{\cot^2\widehat{B}}+\frac{\sin\widehat{B}.\cos\widehat{B}}{\cot\widehat{B}}=\frac{\frac{AB^2}{AC^2}-\frac{AB^2}{BC^2}}{\frac{AB^2}{AC^2}}+\frac{\frac{AC}{BC}.\frac{AB}{BC}}{\frac{AB}{AC}}\)
\(=\frac{\frac{AB^2}{AC^2}}{\frac{AB^2}{AC^2}}-\frac{\frac{AB^2}{BC^2}}{\frac{AB^2}{AC^2}}+\frac{\frac{AC.AB}{BC^2}}{\frac{AB}{AC}}=1-\frac{AC^2}{BC^2}+\frac{AC^2}{BC^2}=1\)
\(VP=\frac{2\sin^2x-1}{\sin^4x}=\frac{\sin^2x+\sin^2x-1}{\sin^4x}=\frac{\sin^2x-\cos^2x}{\sin^4x}\)
\(=\frac{\left(\sin^2x-\cos^2x\right).1}{\sin^4x}=\frac{\left(\sin^2x-\cos^2x\right)\left(\sin^2x+\cos^2x\right)}{\sin^4x}=\frac{\sin^4x-\cos^4x}{\sin^4x}\)
\(=1-\cot^4x\)=VT
\(1-\frac{sin^3x}{sinx+cosx}-\frac{cos^3x}{sinx+cosx}=1-\frac{sin^3x+cos^3x}{sinx+cosx}\)
\(=1-\frac{\left(sinx+cosx\right)\left(sin^2x+cos^2x-sinx.cosx\right)}{sinx+cosx}=1-\left(1-sinxcosx\right)\)
\(=sinx.cosx\)
Có \(\sin^2x+\cos^2x=1\Rightarrow\sin^2x-\cos^2x=1-2\cos^2x\)
\(\Rightarrow VT=\frac{\sin^2x-\cos^2x}{\sin^2x.\cos^2x}=\frac{\sin^4x-\cos^4x}{\sin^2x.\cos^2x}=\frac{\sin^2x}{\cos^2x}-\frac{\cos^2x}{\sin^2x}=\tan^2x-\cot^2x=VP\)
\(=cot^2x\left(cos^2x-1\right)+cos^2x+4\left(sin^2x+cos^2x\right)\)
\(=\frac{cos^2x}{sin^2x}\left(-sin^2x\right)+cos^2x+4\)
\(=-cos^2x+cos^2x+4=4\)
Khỏi tick
\(2a+3b+\frac{6}{a}+\frac{10}{b}=\left(\frac{3}{2}a+\frac{6}{a}\right)+\left(\frac{5}{2}b+\frac{10}{b}\right)+\frac{1}{2}\left(a+b\right)\)
\(\ge2\sqrt{\frac{3}{2}a.\frac{6}{a}}+2\sqrt{\frac{5}{2}b.\frac{10}{b}}+\frac{1}{2}.4=18\)
Lời giải:
Ta có:
\(\frac{\cot ^2a-\cos ^2}{\cot ^2a}+\frac{\sin a\cos a}{\cot a}=1-\frac{\cos ^2a}{\cot ^2a}+\frac{\sin a\cos a}{\cot a}\)
\(=1-\frac{\cos ^2a}{\frac{\cos ^2a}{\sin ^2a}}+\frac{\sin a\cos a}{\frac{\cos a}{\sin a}}=1-\sin ^2a+\sin ^2a=1\)
Ta có đpcm.