\(A=\frac{\left(1+cos2x\right)}{cos2x}.tanx=\frac{\left(1+2cos^2x-1\right)}{cos2x}.\frac{sinx}{cosx}=\frac{2cos^2x.sinx}{cos2x.cosx}=\frac{2sinx.cosx}{cos2x}=\frac{sin2x}{cos2x}=tan2x\)

\(B=\frac{1+2sin2a.cos2a-1+2sin^22a}{1+2sin2a.cos2a+2cos^22a-1}=\frac{2sin2a\left(sin2a+cos2a\right)}{2cos2a\left(sin2a+cos2a\right)}=\frac{sin2a}{cos2a}=tan2a\)

\(C=\frac{2sina.cosa+sina}{1+2cos^2a-1+cosa}=\frac{sina\left(2cosa+1\right)}{cosa\left(2cosa+1\right)}=\frac{sina}{cosa}=tana\)

Ta có ( sin a   +   cos a ) 2  = 1 + sin2a ⇒ 1/2 = 1 + sin2a. Vậy sin2a = (-1)/2

Đáp án: C

Chia cả tử và mẫu cho \(cosa\)

\(D=\dfrac{\dfrac{cosa}{cosa}+\dfrac{sina}{cosa}}{\dfrac{cosa}{cosa}-\dfrac{sina}{cosa}}=\dfrac{1+tana}{1-tana}=\dfrac{1+\dfrac{1}{2}}{1-\dfrac{1}{2}}=3\)

có A=\(\dfrac{1-cosa+2cos^2a-1}{2sina.cosa-sina}=\dfrac{cosa\left(2cosa-1\right)}{sina\left(2cosa-1\right)}=\dfrac{cosa}{sina}=cota\)

3/

\(cos4A+cos4B+cos4C=2cos\left(2A+2B\right).cos\left(2A-2B\right)+2cos^22C-1\)

\(=2cos2C.cos\left(2A-2B\right)+2cos^22C-1\)

\(=2cos2C\left(cos\left(2A-2B\right)+cos2C\right)-1\)

\(=2cos2C\left(cos\left(2A-2B\right)+cos\left(2A+2B\right)\right)-1\)

\(=4cos2A.cos2B.cos2C-1\Rightarrow\left\{{}\begin{matrix}a=-1\\b=4\end{matrix}\right.\)

4/

\(cos^2A+cos^2B+cos^2C=\frac{1}{2}+\frac{1}{2}cos2A+\frac{1}{2}+\frac{1}{2}cos2B+\frac{1}{2}+\frac{1}{2}cos2C\)

\(=\frac{3}{2}+\frac{1}{2}\left(cos2A+cos2B+cos2C\right)\)

\(=\frac{3}{2}+\frac{1}{2}\left[2cos\left(A+B\right).cos\left(A-B\right)+2cos^2C-1\right]\)

\(=1+\frac{1}{2}\left(-2cosC.cos\left(A-B\right)+2cos^2C\right)\)

\(=1-cosC\left(cos\left(A-B\right)-cosC\right)\)

\(=1-cosC\left(cos\left(A-B\right)+cos\left(A+B\right)\right)\)

\(=1-2cosA.cosB.cosC\) \(\Rightarrow\left\{{}\begin{matrix}a=1\\b=-2\end{matrix}\right.\)

1/ \(sinA+sinB+sin2\frac{C}{2}=2sin\frac{A+B}{2}cos\frac{A-B}{2}+2sin\frac{C}{2}cos\frac{C}{2}\)

\(=2cos\frac{C}{2}.cos\frac{A-B}{2}+2cos\frac{A+B}{2}.cos\frac{C}{2}=2cos\frac{C}{2}\left(cos\frac{A-B}{2}+cos\frac{A+B}{2}\right)\)

\(=4cos\frac{A}{2}cos\frac{B}{2}cos\frac{C}{2}\Rightarrow\left\{{}\begin{matrix}a=0\\b=4\end{matrix}\right.\)

2/ \(sin4A+sin4B+sin4C=2sin\left(2A+2B\right)cos\left(2A-2B\right)+2sin2C.cos2C\)

\(=-2sin2C.cos\left(2A-2B\right)+2sin2C.cos2C\)

\(\)\(=2sin2C\left(cos2C-cos\left(2A-2B\right)\right)\)

\(=-4sin2C.sin\left(C+A-B\right)sin\left(C-A+B\right)\)

\(=-4sin2A.sin2B.sin2C\Rightarrow\left\{{}\begin{matrix}a=0\\b=-4\end{matrix}\right.\)

\(A=2sin2x.cos2x.cos4x=sin4x.cos4x=\frac{1}{2}sin8x\)

\(B=sin^4x+cos^6x-6sin^2x.cos^2x\)

\(=\left(sin^2x+cos^2x\right)^2-8sin^2x.cos^2x\)

\(=1-2\left(2sinx.cosx\right)^2=1-2sin^22x=cos4x\)

\(C=\frac{cos2a+1-2cos^22a}{2sin2a.cos2a+sin2a}=\frac{\left(1-cos2a\right)\left(2cos2a+1\right)}{sin2a\left(2cos2a+1\right)}=\frac{1-cos2a}{sin2a}\)

\(=\frac{1-\left(1-2sin^2a\right)}{2sina.cosa}=\frac{2sin^2a}{2sina.cosa}=\frac{sina}{cosa}=tana\)

\(D=\frac{2cos3a.cos2a+cos3a}{2sin3a.cos2a+sin3a}=\frac{cos3a\left(2cos2a+1\right)}{sin3a\left(2cos2a+1\right)}=\frac{cos3a}{sin3a}=cot3a\)

\(E=\frac{1}{2}-\frac{1}{2}cos\left(\frac{\pi}{4}+x\right)-\frac{1}{2}+\frac{1}{2}cos\left(\frac{\pi}{4}+x\right)\)

\(=\frac{1}{2}\left[cos\left(\frac{\pi}{4}+x\right)-cos\left(\frac{\pi}{4}-x\right)\right]=-sin\frac{\pi}{4}.sinx=-\frac{\sqrt{2}}{2}sinx\)

\(A=cos^2a+cos^2b+2cosa.cosb+sin^2a+sin^2b+2sina.sinb\)

\(=cos^2a+sin^2a+cos^2b+sin^2b+2\left(cosa.cosb+sina.sinb\right)\)

\(=2+2cos\left(a-b\right)=2+2cos\frac{\pi}{3}=3\)

\(\left(cosa+sina\right)^2=\frac{36}{25}\Leftrightarrow1+2sina.cosa=\frac{36}{25}\)

\(\Rightarrow sin2a=\frac{36}{25}-1=\frac{11}{25}\)

\(cos2a=cos^2a-sin^2a=\left(cosa-sina\right)\left(cosa+sina\right)>0\)

\(\Rightarrow cos2a=\sqrt{1-sin^22a}=\frac{6\sqrt{14}}{25}\)