\(\frac{sinx+cosx-1}{sinx-cosx+1}=\frac{\left(sinx+cosx-1\right)\left(sinx-\left(cosx-1\right)\right)}{\left(sinx-cosx+1\right)^2}\)

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

\(=\frac{1-cos^2x-cos^2x+2cosx-1}{2\left(1-cosx\right)\left(1+sinx\right)}=\frac{cosx\left(1-cosx\right)}{\left(1-cosx\right)\left(1+sinx\right)}=\frac{cosx}{1+sinx}\)

\(=\frac{1+cosx}{sinx}\left(1-\frac{1+cos^2x-2cosx}{sin^2x}\right)=\frac{1+cosx}{sinx}\left(\frac{sin^2x-1-cos^2x+2cosx}{sin^2x}\right)\)

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

\(=\frac{2cosx\left(1-cos^2x\right)}{sinx.sin^2x}=\frac{2cosx.sin^2x}{sinx.sin^2x}=\frac{2cosx}{sinx}=2cotx\)

Chọn C.

Ta có:

\(B=\dfrac{1-4\sin^2x\cdot\cos^2x}{\sin^2x+2\sin x\cdot\cos x+\cos^2}+2\sin x\cdot\cos x\\ B=\dfrac{1-4\sin^2x\cdot\cos^2x}{2\sin x\cdot\cos x}+2\sin x\cdot\cos x\\ B=\dfrac{1-4\sin^2x\cdot\cos^2x+4\sin^2x\cdot\cos^2x}{2\sin x\cdot\cos x}=\dfrac{1}{2\sin x\cdot\cos x}\)

\(A=cosa\left(sinb.cosc-cosb.sinc\right)+cosb\left(sinc.cosa-cosc.sina\right)+cosc\left(sinacosb-cosasinb\right)\)

\(A=cosasinbcosc-cosacosbsinc+cosacosbsinc-sinacosbcosc+sinacosbcosc-cosasinbcosc\)

\(A=0\)

\(B=sin^2x+\frac{1}{2}\left(cos\frac{2\pi}{3}+cos2x\right)\)

\(B=\frac{1}{2}-\frac{1}{2}cos2x-\frac{1}{4}+\frac{1}{2}cos2x\)

\(B=\frac{1}{4}\)

\(C=\frac{1}{2}-\frac{1}{2}cos2x+\frac{1}{2}-\frac{1}{2}cos\left(\frac{4\pi}{3}+2x\right)+\frac{1}{2}-\frac{1}{2}cos\left(\frac{4\pi}{3}-2x\right)\)

\(C=\frac{3}{2}-\frac{1}{2}cos2x-\frac{1}{2}\left(cos\left(\frac{4\pi}{3}+2x\right)+cos\left(\frac{4\pi}{3}-2x\right)\right)\)

\(C=\frac{3}{2}-\frac{1}{2}cos2x-cos\frac{4\pi}{3}.cos2x\)

\(C=\frac{3}{2}-\frac{1}{2}cos2x+\frac{1}{2}cos2x\)

\(C=\frac{3}{2}\)

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

\(D=\frac{1}{2}\left(sinx+cosx\right)^2-sin^2x-sinx\left(sinx-cosx\right)\)

\(D=\frac{1}{2}\left(1+2sinx.cosx\right)-sin^2x-sin^2x+sinx.cosx\)

\(D=\frac{1}{2}+sinxcosx+sinxcosx=\frac{1}{2}+sin2x\)

Góc độ cao của thang dựa vào tường là 60º và chân thang cách tường 4,6 m. Chiều dài của thang là

\(A = \frac{{ \sin 2x }}{{1+ \cos 2x }} = \frac{{2.\sin x.\cos x }}{{1+(2\cos ^2x-1)}} =  \frac{{2.\sin x.\cos x }}{{2\cos ^2x}} = \frac{{\sin x}}{{\cos x}}= tanx\)

\(A=4sinx.cosx.sin\left(-3x\right)+cosx\)

\(=-2sin2x.sin3x+cosx\)

\(=cos5x-cosx+cosx\)

\(=cos5x\)

\(cos\left(2x+\frac{\pi}{6}\right)cos\left(2x-\frac{\pi}{6}\right)=\frac{1}{2}\left(cos4x+cos\frac{\pi}{3}\right)=\frac{1}{2}\left(cos4x+\frac{1}{2}\right)\)

\(sin\left(x+\frac{\pi}{6}\right)sin\left(x-\frac{\pi}{6}\right)=\frac{1}{2}\left(cos\frac{\pi}{3}-cos2x\right)=\frac{1}{2}\left(\frac{1}{2}-cos2x\right)\)

\(\Rightarrow C=\frac{1}{2}sinx.cos4x+\frac{1}{4}sinx+\frac{1}{4}sin3x-\frac{1}{2}sin3x.cos2x\)

\(=\frac{1}{4}sin5x-\frac{1}{4}sin3x+\frac{1}{4}sinx+\frac{1}{4}sin3x-\frac{1}{4}sin5x+\frac{1}{4}sinx\)

\(=\frac{1}{2}sinx\)