\(A=2(\sin ^6x+\cos ^6x)-3(\sin ^4x+\cos ^4x)\)

\(=2(\sin ^2x+\cos ^2x)(\sin ^4x-\sin ^2x\cos ^2x+\cos ^4x)-3(\sin ^4x+\cos ^4x)\)

\(=2(\sin ^4x-\sin ^2x\cos ^2x+\cos ^4x)-3(\sin ^4x+\cos ^4x)\)

\(=-(\sin ^4x+2\sin ^2x\cos ^2x+\cos ^4x)=-(\sin ^2x+\cos ^2x)^2=-1^2=-1\)

là giá trị không phụ thuộc vào biến (đpcm)

-----------------------

\(B=\sin ^6x+\cos ^6x-2\sin ^4x-\cos ^4x+\sin ^2x\)

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

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

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

\(=-\sin ^2x+\sin ^2x=0\)

là giá trị không phụ thuộc vào biến (đpcm)

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

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

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

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

là giá trị không phụ thuộc vào biến $x$

--------------------

\(D=\frac{1}{\cos ^6x}-\tan ^6x-\frac{\tan ^2x}{\cos ^2x}=\frac{1}{\cos ^6x}-\frac{\sin ^6x}{\cos ^6x}-\frac{\sin ^2x}{\cos ^4x}\)

\(=\frac{1-\sin ^6x-\sin ^2x\cos ^2x}{\cos ^6x}=\frac{(\sin ^2x+\cos ^2x)^3-\sin ^6x-\sin ^2x\cos ^2x}{\cos ^6x}\)

\(=\frac{\sin ^6x+\cos ^6x+3\sin ^2x\cos ^2x(\sin ^2x+\cos ^2x)-\sin ^6x-\sin ^2x\cos ^2x}{\cos ^6x}\)

\(=\frac{\cos ^6x+3\sin ^2x\cos ^2x-\sin ^2x\cos ^2x}{\cos ^6x}=\frac{\cos ^4x+2\sin ^2x}{\cos ^4x}\)

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

Giá trị biểu thức này vẫn phụ thuộc vào $x$. Bạn xem lại đề.

\(\frac{1+sin^4x-cos^4x}{1-sin^6x-cos^6x}=\frac{1+\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)}{1-\left(sin^2x+cos^2x\right)^2+3sin^2x.cos^2x\left(sin^2x+cos^2x\right)}\)

\(=\frac{1+sin^2x-cos^2x}{1-1+3sin^2x.cos^2x}=\frac{\left(1-cos^2x\right)+sin^2x}{3sin^2x.cos^2x}=\frac{2sin^2x}{3sin^2x.cos^2x}=\frac{2}{3cos^2x}\)

\(VT=\frac{\frac{\sin^2x}{\cos^2x}-\sin^2x}{\frac{\cos^2x}{\sin^2x}-\cos^2x}=\frac{\frac{\sin^2x-\sin^2x.\cos^2x}{\cos^2x}}{\frac{\cos^2x-\cos^2x.\sin^2x}{\sin^2x}}\)

\(=\frac{\sin^2x}{\cos^2x}.\frac{\sin^2x-\sin^2x.\cos^2x}{\cos^2-\cos^2x.\sin^2x}\)

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

\(1+\tan^2x=\frac{1}{\cos^2x}\Rightarrow\frac{\tan^2x}{\cos^2x}=\tan^2x\left(1+\tan^2x\right)\)

\(\Rightarrow VT=\tan^2x.\tan^4x=\tan^6x=VP\)

\(\frac{tan^2x-sin^2x}{cot^2x-cos^2x}=\frac{sin^2x.cos^2x\left(tan^2x-sin^2x\right)}{sin^2x.cos^2x\left(cot^2x-cos^2x\right)}=\frac{sin^4x\left(1-cos^2x\right)}{cos^4x\left(1-sin^2x\right)}=\frac{sin^6x}{cos^6x}=tan^6x\)

Lời giải:

* $x$ là biến chứ không phải tham số bạn nhé*
\(A=2[(\cos ^2x)^3+(\sin ^2x)^3]-3(\cos ^4x+\sin ^4x)\)

\(=2(\cos ^2x+\sin ^2x)(\cos ^4x-\cos ^2x\sin ^2x+\sin ^4x)-3(\cos ^4x+\sin ^4x)\)

\(=2(\cos ^4x-\cos ^2x\sin ^2x+\sin ^4x)-3(\cos ^4x+\sin ^4x)\)

\(=-(\cos ^4x+2\cos ^2x\sin ^2x+\sin ^4x)=-(\cos ^2x+\sin ^2x)^2=-1^2=-1\)

là giá trị không phụ thuộc vào biến (đpcm)

--------------------------

\(B=\frac{\tan ^2x}{\sin ^2x\cos ^2x}-(1+\tan ^2x)^2=\frac{\sin ^2x}{\cos ^2x.\sin ^2x\cos ^2x}-(1+\frac{\sin ^2x}{\cos ^2x})^2\)

\(=\frac{1}{\cos ^4x}-(\frac{\cos ^2x+\sin ^2x}{\cos ^2x})^2=\frac{1}{\cos ^4x}-(\frac{1}{\cos ^2x})^2=\frac{1}{\cos ^4x}-\frac{1}{\cos ^4x}=0\)

là giá trị không phụ thuộc vào biến $x$ (đpcm)

Ta có: \(\sin^6x+\cos^6x=\left(\sin^2x\right)^3+\left(\cos^2x\right)^3\)

\(=\left(\sin^2x+\cos^2x\right)^3-3\left(\cos^2x+\sin^2x\right)\cos^2x.\sin^2x\)

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

Giả sử tất cả các biểu thức đều xác định

a/

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

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

b/

\(tanx+cotx=\frac{sinx}{cosx}+\frac{cosx}{sinx}=\frac{sin^2x+cos^2x}{sinx.cosx}=\frac{1}{sinx.cosx}\)

c/

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

d/

\(\frac{1}{1+tanx}+\frac{1}{1+cotx}=\frac{1}{1+tanx}+\frac{1}{1+\frac{1}{tanx}}=\frac{1}{1+tanx}+\frac{tanx}{1+tanx}=\frac{1+tanx}{1+tanx}=1\)

e/

\(\left(1-\frac{1}{cosx}\right)\left(1+\frac{1}{cosx}\right)+tan^2x=1-\frac{1}{cos^2x}+tan^2x\)

\(=\frac{cos^2x-1}{cos^2x}+tan^2x=\frac{-sin^2x}{cos^2x}+tan^2x=-tan^2x+tan^2x=0\)