A=\(\frac{\left(cos7x+cos10x\right)-\left(cos8x+cos9x\right)}{\left(sin7x+sin10x\right)-\left(sin8x+sin9x\right)}\) =\(\frac{2cos\frac{17x}{2}.cos\frac{3x}{2}-2cos\frac{17x}{2}.cos\frac{x}{2}}{2sin\frac{17x}{2}.cos\frac{3x}{2}-2sin\frac{17x}{2}.cos\frac{x}{2}}\)

=\(\frac{2cos\frac{17x}{2}\left(cos\frac{3x}{2}-cos\frac{x}{2}\right)}{2sin\frac{17x}{2}\left(cos\frac{3x}{2}-cos\frac{x}{2}\right)}\)=\(\frac{cos\frac{17x}{2}}{sin\frac{17x}{2}}\)=cotg\(\frac{17x}{2}\)

\(A=\frac{cos7x-cos8x-cos9x+cos10x}{sin7x-sin8x-sin9x+sin10x}=\frac{(cos10x+cos7x)-\left(cos9x+cos8x\right)}{\left(sin10x+sin7x\right)-\left(sin9x+sin8x\right)}.\) 

     \(=\frac{2cos\frac{17x}{2}cos\frac{3x}{2}-2cos\frac{17x}{2}cos\frac{x}{2}}{2sin\frac{17x}{2}cos\frac{3x}{2}-2sin\frac{17x}{2}cos\frac{x}{2}}=\frac{2cos\frac{17x}{2}\left(cos\frac{3x}{2}-cos\frac{x}{2}\right)}{2sin\frac{17x}{2}\left(cos\frac{3x}{2}-cos\frac{x}{2}\right)}=cotan\frac{17x}{2}.\)  

\(D=\frac{sin4x+sin5x+sin6x}{cos4x+cos5x+cos6x}\)

\(=\frac{\left(sin4x+sin6x\right)+sin5x}{\left(cos4x+cos6x\right)+cos5x}\)

\(=\frac{2sin\frac{4x+6x}{2}.cos\frac{4x-6x}{2}+sin5x}{2cos\frac{4x+6x}{2}.cos\frac{4x-6x}{2}+cos5x}\)

\(=\frac{2sin5x.cos\left(-x\right)+sin5x}{2cos5x.cos\left(-x\right)+cos5x}=\frac{sin5x\left(2.cos\left(-x\right)+1\right)}{cos5x\left(2.cos\left(-x\right)+1\right)}=\frac{sin5x}{cos5x}=tan5x\)

\(A=\frac{cos3x+cos9x+cos5x+cos7x}{sin3x+sin9x+sin5x+sin7x}=\frac{2cos6x.cos3x+2cos6x.cosx}{2sin6x.cos3x+2sin6x.cosx}\)

\(=\frac{2cos6x\left(cos3x+cosx\right)}{2sin6x\left(cos3x+cosx\right)}=tan6x\)

\(A=1\Rightarrow tan6x=1\Rightarrow x=\frac{\pi}{24}+\frac{k\pi}{6}\)

a/

\(\Leftrightarrow2sin4x.cos3x=2sin7x.cos3x\)

\(\Leftrightarrow\left[{}\begin{matrix}cos3x=0\\sin7x=sin4x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=\frac{\pi}{2}+k\pi\\7x=4x+k2\pi\\7x=\pi-4x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+\frac{k\pi}{3}\\x=\frac{k2\pi}{3}\\x=\frac{\pi}{11}+\frac{k2\pi}{11}\end{matrix}\right.\)

b.

\(\Leftrightarrow2cos4x.cosx=2cos8x.cosx\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cos8x=cos4x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\8x=4x+k2\pi\\8x=-4x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\x=\frac{k\pi}{2}\\x=\frac{k\pi}{6}\end{matrix}\right.\) \(\Leftrightarrow x=\frac{k\pi}{6}\)

M = 3(sin^8x-cos^8x) + 4(cos^6x-2sin^6x)+6sin^4x
Ta có:
sin^8(x) - cos^8(x) = [sin^4(x) ]² - [cos^4(x)]²
= (sin²x + cos²x)(sin²x -cos²x).[ sin^4(x) + cos^4(x) ]
= (sin²x -cos²x)[ sin^4(x) + cos^4(x) ]
= sin^6(x) - cos^6(x) + sin²x.cos^4(x) -cos²x.sin^4(x)
Lúc đó M viết lại là:
M = 3.[sin^6(x) - cos^6(x) + sin²x.cos^4(x) -cos²x.sin^4(x) ] + 4.[ cos^6(x) -2sin^6(x) ] + 6sin^4(x)
M = -5sin^6(x) + cos^6(x) -3sin^4(x).cos²x + 3sin²x.cos^4(x) +6sin^4(x)
M = -3sin^(6)x - 3cos²x.sin^4(x) + cos^4(x).sin²x + cos^6(x) - 2sin^6(x) + 2sin²x.cos^4(x) + 6sin^4(x)
M = -3sin^4(x).(sin²x + cos²x ) + cos^4(x).[sin²x + cos²x ] -2sin²x.[sin^4(x) - cos^4(x) ] + 6sin^4(x)
M = 3sin^4(x) + cos^4(x) -2sin²x.[sin²x - cos²x]
M = 3sin^4(x) + cos^4(x) -2sin^4(x) + 2sin²x.cos²x
M = sin^4(x) + 2sin²x.cos²x + cos^4(x)
M = [sin²x + cos²x ]² = 1

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.