a)Ta có:
\(A=sin^2x.tan^2x+4sin^2x-tan^2x+3cos^2x\)
\(=tan^2x\left(sin^2x-1\right)+3+sin^2x\)
\(=-tan^2x.cos^2x+3+sin^2x\)
\(=-sin^2x+3+sin^2x=3\), không phụ thuộc vào x
b) Ta có:
\(B=\dfrac{\left(1-tan^2x\right)^2}{4tan^2x}-\dfrac{1}{4sin^2x.cos^2x}\)
\(=\dfrac{\left(1-\dfrac{sin^2x}{cos^2x}\right)^2}{4\dfrac{sin^2x}{cos^2x}}-\dfrac{1}{4sin^2x.cos^2x}\)
\(=\dfrac{\left(cos^2x-sin^2x\right)^2-1}{4sin^2x.cos^2x}\)
\(=\dfrac{cos^22x-1}{sin^22x}=\dfrac{-sin^22x}{sin^22x}=-1\), không phụ thuộc vào x
c)Ta có:
\(C=sin^6x+cos^6x+3sin^2x.cos^2x\)
\(=\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)+3sin^2x.cos^2x\)
\(=1-3sin^2x.cos^2x+3sin^2x.cos^2x=1\), không phụ thuộc vào x
d)Ta có:
\(D=\dfrac{cos^4x+sin^2x-cos^2x}{sin^4x+cos^2x-sin^2x}\)
\(=\dfrac{-cos^2x\left(1-cos^2x\right)+sin^2x}{-sin^2x\left(1-sin^2x\right)+cos^2x}\)
\(=\dfrac{-cos^2x.sin^2x+sin^2x}{-sin^2x.cos^2x+cos^2x}\)
\(=\dfrac{sin^4x}{cos^4x}=tan^4x\), vẫn phụ thuộc vào x
