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Câu 1 đề sai, chắc chắn 1 trong 2 cái \(cot^2x\) phải có 1 cái là \(cos^2x\)

2.

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

\(=\dfrac{1-\left(sin^2x+cos^2x\right)}{cosx\left(1+sinx\right)}=\dfrac{1-1}{cosx\left(1+sinx\right)}=0\)

3.

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

4.

\(\dfrac{tanx}{1-tan^2x}.\dfrac{cot^2x-1}{cotx}=\dfrac{tanx}{1-tan^2x}.\dfrac{\dfrac{1}{tan^2x}-1}{\dfrac{1}{tanx}}=\dfrac{tanx}{1-tan^2x}.\dfrac{1-tan^2x}{tanx}=1\)

5.

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

\(=tan^2x+1+tan^2x=1+2tan^2x\)

a.

Thực hiện phép biến đổi tương đương:

\(\dfrac{sinx+cosx-1}{1-cosx}=\dfrac{2cosx}{sinx-cosx+1}\)

\(\Leftrightarrow\left(sinx+cosx-1\right)\left(sinx-cosx+1\right)=2cosx\left(1-cosx\right)\)

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

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

\(\Leftrightarrow1-cos^2x-cos^2x-1=-2cos^2x\)

\(\Leftrightarrow-2cos^2x=-2cos^2x\) (luôn đúng)

Vậy đẳng thức đã cho được chứng minh

b.

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

a: tan x(cot^2x-1)

\(=\dfrac{1}{cotx}\left(cot^2x-cotx\cdot tanx\right)\)

=cotx-tanx/cotx=cotx(1-tan^2x)

b: \(tan^2x-sin^2x=\dfrac{sin^2x}{cos^2x}-sin^2x\)

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

c: \(\dfrac{cos^2x-sin^2x}{cot^2x-tan^2x}=\dfrac{cos^2x-sin^2x}{\dfrac{cos^2x}{sin^2x}-\dfrac{sin^2x}{cos^2x}}\)

\(=\left(cos^2x-sin^2x\right):\dfrac{cos^4x-sin^4x}{sin^2x\cdot cos^2x}\)

\(=\dfrac{sin^2x\cdot cos^2x}{1}=sin^2x\cdot cos^2x\)

=>sin^2x*cos^2x-cos^2x=cos^2x(sin^2x-1)

=-cos^2x*cos^2x=-cos^4x

=>ĐPCM

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

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

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

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

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

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

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

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

\(\frac{cosx+sinx}{cosx-sinx}-\frac{cosx-sinx}{cosx+sinx}=\frac{\left(cosx+sinx\right)^2-\left(cosx-sinx\right)^2}{\left(cosx-sinx\right)\left(cosx+sinx\right)}\)

\(=\frac{cos^2x+sin^2x+2sinx.cosx-\left(cos^2x+sin^2x-2sinx.cosx\right)}{cos^2x-sin^2x}=\frac{4sinx.cosx}{cos2x}=\frac{2sin2x}{cos2x}=2tan2x\)

\(\frac{sin4x+cos2x}{1-cos4x+sin2x}=\frac{2sin2x.cos2x+cos2x}{1-\left(1-2sin^22x\right)+sin2x}=\frac{cos2x\left(2sin2x+1\right)}{sin2x\left(2sin2x+1\right)}=\frac{cos2x}{sin2x}=cot2x\)

\(A=sin^2x\left(sinx+cosx\right)+cos^2x\left(sinx+cosx\right)\)

\(=\left(sin^2x+cos^2x\right)\left(sinx+cosx\right)=sinx+cosx\)

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