\(VP=\frac{2\sin^2x-1}{\sin^4x}=\frac{\sin^2x+\sin^2x-1}{\sin^4x}=\frac{\sin^2x-\cos^2x}{\sin^4x}\)

\(=\frac{\left(\sin^2x-\cos^2x\right).1}{\sin^4x}=\frac{\left(\sin^2x-\cos^2x\right)\left(\sin^2x+\cos^2x\right)}{\sin^4x}=\frac{\sin^4x-\cos^4x}{\sin^4x}\)

\(=1-\cot^4x\)=VT

\(\sin^6x+\cos^6x\\ =\left(\sin^2x\right)^3+\left(\cos^2x\right)^3\\ =\left(\sin^2x+\cos^2x\right)^3-3\sin^2x\cos^2x\left(\sin^2x+\cos^2x\right)\\ =1-3\sin^2x\cos^2x\left(đpcm\right)\)

\(sin^6x+cos^6x\)

=\(\left(sin^2x+cos^2x\right)\left(sin^4x-sin^2x.cos^2x+cos^4x\right)\)

=\(sin^4x-sin^2x.cos^2x+cos^4x\)

=\(\left(1-2sin^2x.cos^2x\right)-sin^2x.cos^2x\)

=\(1-3sin^2x.cos^2x\)(đpcm)

➞\(sin^6x+cos^6x\)=\(1-3sin^2x.cos^2x\)

\(\cos^4x-\sin^4x=\left(\cos^2x-\sin^2x\right)\left(\cos^2x+\sin^2x\right)\)

\(=\cos^2x-\sin^2x=\cos^2x-\left(1-\cos^2x\right)=2\cos^2x-1\)

(đpcm)

Có \(\sin^2x+\cos^2x=1\Rightarrow\sin^2x-\cos^2x=1-2\cos^2x\)

\(\Rightarrow VT=\frac{\sin^2x-\cos^2x}{\sin^2x.\cos^2x}=\frac{\sin^4x-\cos^4x}{\sin^2x.\cos^2x}=\frac{\sin^2x}{\cos^2x}-\frac{\cos^2x}{\sin^2x}=\tan^2x-\cot^2x=VP\)

Có \(\sin^2x+\cos^2x=1\Rightarrow2\sin^2x=1-\cos^2x+\sin^2x\)

\(\Rightarrow1+\sin^2x=2\sin^2x+\cos^2x\)

\(\Rightarrow VT=\frac{2\sin^2x+\cos^2x}{\cos^2x}=2\tan^2x+1\)

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.

Giả sử có \(\Delta ABC\) có \(A=90^o;AH\) là đường cao

Có \(\sin\widehat{B}=\frac{AC}{BC};\cos\widehat{B}=\frac{AB}{BC};\tan\widehat{B}=\frac{AC}{AB};\cot\widehat{B}=\frac{AB}{AC}\)

\(\frac{\cot^2\widehat{B}-\cos^2\widehat{B}}{\cot^2\widehat{B}}+\frac{\sin\widehat{B}.\cos\widehat{B}}{\cot\widehat{B}}=\frac{\frac{AB^2}{AC^2}-\frac{AB^2}{BC^2}}{\frac{AB^2}{AC^2}}+\frac{\frac{AC}{BC}.\frac{AB}{BC}}{\frac{AB}{AC}}\)

\(=\frac{\frac{AB^2}{AC^2}}{\frac{AB^2}{AC^2}}-\frac{\frac{AB^2}{BC^2}}{\frac{AB^2}{AC^2}}+\frac{\frac{AC.AB}{BC^2}}{\frac{AB}{AC}}=1-\frac{AC^2}{BC^2}+\frac{AC^2}{BC^2}=1\)

Mấy bài nè vận dụng hệ thức sin cos tan cot

Tớ gắn vào tam giấc bạn vẽ hình là biết liền ấy mà