\(A=3\left[\left(sin^2x+cos^2x\right)^2-2\cdot sin^2x\cdot cos^2x\right]-2\left[\left(sin^2x+cos^2x\right)^3-3\cdot sin^2x\cdot cos^2x\left(sin^2x+cos^2x\right)\right]\)
 

\(=3\left[1-2\cdot sin^2x\cdot cos^2x\right]-2\left[1-3\cdot sin^2x\cdot cos^2x\right]\)

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

=1

Ta có:

Vậy giá trị của biểu thức  không phụ thuộc vào x.

1/ \(3-4\sin^2=4\cos^2x-1\Leftrightarrow4\left(\sin^2x+\cos^2x\right)-4=0\Leftrightarrow4.1-4=0\left(ld\right)\Rightarrow dpcm\)

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

3/ \(\sin^4x+\cos^4x=\left(\sin^2x+\cos^2x\right)^2-2\sin^2x.\cos^2x=1-2\sin^2x.\cos^2x\)

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

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

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

\(=1-2sin^2xcos^2x\)

b) \(\dfrac{1+cotx}{1-cotx}=\dfrac{tanx.cotx+cotx}{tanx.cotx-cotx}\)

\(=\dfrac{cotx.\left(tanx+1\right)}{cotx.\left(tanx-1\right)}\)

\(=\dfrac{tanx+1}{tanx-1}\)

c) \(\dfrac{cosx+sinx}{cos^3x}=\dfrac{1}{cos^2x}+\dfrac{tanx}{cos^2x}\)

\(=1+tan^2x+tanx.\dfrac{1}{cos^2x}\)

\(=1+tan^2x+tanx.\left(1+tan^2x\right)\)

\(=1+tan^2x+tanx+tan^3x\)

\(=tan^3x+tan^2x+tanx+1\)

Lời giải:

a.

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

$=1-2\sin ^2x\cos ^2x$

b.

$\frac{1+\cot x}{1-\cot x}=\frac{1+\frac{\cos x}{\sin x}}{1-\frac{\cos x}{\sin x}}=\frac{\cos x+\sin x}{\sin x-\cos x}(1)$

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

Từ $(1); (2)$ ta có đpcm

c.

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

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

$=(1+\tan x)(\tan ^2x+1)=\tan ^3x+\tan ^2x+\tan x+1$

Ta có đpcm.

\(\frac{sin4x-sin2x}{1-cos2x+cos4x}=\frac{2sin2x.cos2x-sin2x}{1-cos2x+2cos^22x-1}=\frac{sin2x\left(2cos2x-1\right)}{cos2x\left(2cos2x-1\right)}=\frac{sin2x}{cos2x}=tan2x\)

\(\Rightarrow\) đề sai

b/

\(\frac{1-cos4x}{sin4x}=\frac{1-\left(1-2sin^22x\right)}{2sin2x.cos2x}=\frac{2sin^22x}{2sin2x.cos2x}=\frac{sin2x}{cos2x}=tan2x\)

Đề sai tiếp lần 2

a, Ta có:  sin 4 x + cos 4 x = sin 2 x + cos 2 x 2 - 2 sin 2 x . cos 2 x = 1 - 2 sin 2 x . cos 2 x

b, Ta có:  sin 6 x + cos 6 x = sin 2 x + cos 2 x 3 - 3 sin 2 x cos 2 x sin 2 x + cos 2 x =  1 - 3 sin 2 x cos 2 x

\(\frac{sin2x-sin4x}{1-cos2x+cos4x}=\frac{sin2x-2sin2x.cos2x}{1-cos2x+2cos^22x-1}=\frac{sin2x\left(1-2cos2x\right)}{-cos2x\left(1-2cos2x\right)}=\frac{-sin2x}{cos2x}=-tan2x\)

\(\frac{sin4x-sin2x}{1-cos2x+cos4x}=-\left(\frac{sin2x-sin4x}{1-cos2x+cos4x}\right)=-\left(-tan2x\right)=tan2x\) lấy luôn kết quả câu trên cho lẹ, biến đổi thì làm y hệt