Chọn C.

Ta có

C = [ ( sin²x + cos²x) – sin²cos²x]² - [ ( sin⁴x + cos⁴x) ² - 2sin⁴x.cos⁴x]

= 2[ 1-sin²x.cos²x]² - [ ( sin²x + cos²x) ² - 2sin²x.cos²x]² + 2sin⁴x.cos⁴x

= 2[ 1-sin²x.cos²x]² - [1-sin²x.cos²x]² + 2sin⁴x.cos⁴x

= 2( 1 - 2sin²x.cos²x + sin⁴x.cos⁴x)- ( 1 - 4sin²xcos²x + 4sin⁴x.cos⁴x) + 2sin⁴x.cos⁴x

= 1.

Chọn C.

Ta có: C = 2( sin⁴x + cos⁴x + sin²x.cos²x) ² - ( sin⁸x + cos⁸x)

= 2 [ (sin²x + cos²x) ² - sin²x.cos²x]² - [ (sin⁴x + cos⁴x)² - 2sin⁴x.cos⁴x]

= 2[ 1 - sin²x.cos²x]² - [ (sin²x+ cos²x) ² - 2sin²x.cos²x]² + 2sin⁴x.cos⁴x

= 2[ 1- sin²x.cos²x]² - [ 1 - 2sin²x.cos²x]²  + 2sin⁴x.cos⁴x

= 2( 1 - 2sin²xcos²x+ sin⁴x.cos⁴x) –( 1- 4sin²xcos²x+ 4sin⁴xcos⁴x) + 2sin⁴x.cos⁴x

=  1.

\(=\dfrac{-cos4x\left(2sin3x+1\right)}{sin4x\left(2sin3x+1\right)}=-cot4x\)

\(D=\frac{1+sin2x+cos2x}{1+sin2x-cos2x}=\frac{1+2sinxcosx+2cos^2x-1}{1+2sinxcosx-1+2sin^2x}\)

\(D=\frac{cosx\left(sinx+cosx\right)}{sinx\left(sinx+cosx\right)}=cotx\)

\(F=\frac{sinx+sin4x+sin7x}{cosx+cos4x+cos7x}\)

\(F=\frac{2sin4xcos3x+sin4x}{2cos4xcos3x+cos4x}\)

\(F=\frac{2sin4x\left(cos3x+1\right)}{2cos4x\left(cos3x+1\right)}=tan4x\)

\(G=\frac{cos2x-sin4x-cos6x}{cos2x+sin4x-cos6x}=\frac{-2sin4xsin2x-sin4x}{-2sin4xsin2x+sin4x}\)

\(G=\frac{-sin4x\left(2sin2x+1\right)}{-sin4x\left(2sin2x-1\right)}=\frac{2sin2x+1}{2sin2x-1}\)

\(P=\dfrac{-2sin5x.sinx-sinx}{2sin5x.cosx+cosx}=\dfrac{-sinx\left(2sin5x+1\right)}{cosx\left(2sin5x+1\right)}=-tanx\)

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.

Chọn A

y = cos6 x+ sin2xcos2x(sin2x + cos2x) + sin4x - sin2x

= cos6x + sin2x(1 - sin2x) + sin4x - sin2x = cos6x

Do đó : y' = -6cos5xsinx.