\(\frac{sin^22x+4sin^2x-4}{1-8sin^2x-cos4x}=\frac{4sin^2x.cos^2x-4\left(1-sin^2x\right)}{1-8sin^2x-\left(1-2sin^22x\right)}=\frac{4sin^2x.cos^2x-4cos^2x}{2sin^22x-8sin^2x}\)

\(=\frac{-4cos^2x\left(1-sin^2x\right)}{8sin^2x.cos^2x-8sin^2x}=\frac{-4cos^2x.cos^2x}{-8sin^2x\left(1-cos^2x\right)}=\frac{cos^4x}{2sin^4x}=\frac{1}{2}cot^4x\)

\(\frac{cos2x}{cot^2x-tan^2x}=\frac{cos2x.sin^2x.cos^2x}{cos^4x-sin^4x}=\frac{\left(cos^2x-sin^2x\right).\left(2sinx.cosx\right)^2}{4\left(cos^2x-sin^2x\right)\left(cos^2x+sin^2x\right)}=\frac{1}{4}sin^22x\)

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. ĐKXĐ: ...

\(\frac{sinx}{cosx}+\frac{sin2x}{cos2x}+\frac{sin3x}{cos3x}=0\)

\(\Leftrightarrow\frac{sin2x.cosx+cos2x.sinx}{cosx.cos2x}+\frac{sin3x}{cos3x}=0\)

\(\Leftrightarrow\frac{sin3x}{cosx.cos2x}+\frac{sin3x}{cos3x}=0\)

\(\Leftrightarrow sin3x\left(\frac{cosx.cos2x+cos3x}{cosx.cos2x.cos3x}\right)=0\)

\(\Leftrightarrow sin3x\left(\frac{cosx\left(2cos^2x-1\right)+4cos^3x-3cosx}{cosx.cos2x.cos3x}\right)=0\)

\(\Leftrightarrow sin3x\left(\frac{6cos^2x-4}{cos2x.cos3x}\right)=0\)

\(\Leftrightarrow sin3x\left(\frac{3cos2x-1}{cos2x.cos3x}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin3x=0\\cos2x=\frac{1}{3}\end{matrix}\right.\)

b.

\(cos2x\left(2cos^22x-1\right)=\frac{1}{2}\)

\(\Leftrightarrow4cos^32x-2cos2x-1=0\)

Pt bậc 3 này ko giải được, chắc bạn ghi nhầm đề

c. ĐKXĐ: ...

\(\frac{cosx}{sinx}-\frac{sinx}{cosx}=cosx-sinx\)

\(\Leftrightarrow\frac{\left(cosx-sinx\right)\left(cosx+sinx\right)}{sinx.cosx}=cosx-sinx\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx-sinx=0\Rightarrow x=...\\\frac{cosx+sinx}{sinx.cosx}=1\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow cosx+sinx=sinx.cosx\)

Đặt \(sinx+cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sinx.cosx=\frac{t^2-1}{2}\end{matrix}\right.\)

\(\Rightarrow t=\frac{t^2-1}{2}\Rightarrow t^2-2t-1=0\Rightarrow\left[{}\begin{matrix}t=1+\sqrt{2}\left(l\right)\\t=1-\sqrt{2}\end{matrix}\right.\)

\(\Rightarrow\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=1-\sqrt{2}\Rightarrow sin\left(x+\frac{\pi}{4}\right)=\frac{1-\sqrt{2}}{\sqrt{2}}\Rightarrow x=...\)

d.

\(\Leftrightarrow2\left(cosx-cos3x\right)sin4x=sin3x\)

\(\Leftrightarrow2sin4x.cosx-2sin4x.cos3x=sin3x\)

\(\Leftrightarrow sin5x+sin3x-sin7x-sinx=sin3x\)

\(\Leftrightarrow sin5x-sin7x-sinx=0\)

\(\Leftrightarrow-2cos6x.sinx-sinx=0\)

\(\Leftrightarrow sinx\left(2cos6x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cos6x=-\frac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow...\)

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

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.

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