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

\(A=sin\left(\frac{\pi}{4}+x\right)-sin\left(\frac{\pi}{2}-\frac{\pi}{4}+x\right)=sin\left(\frac{\pi}{4}+x\right)-sin\left(\frac{\pi}{4}+x\right)=0\)

a) \(A=sin\left(\dfrac{\pi}{4}+x\right)-cos\left(\dfrac{\pi}{4}-x\right)\)

\(\Leftrightarrow A=sin\dfrac{\pi}{4}.cosx+cos\dfrac{\pi}{4}.sinx-\left(cos\dfrac{\pi}{4}.cosx+sin\dfrac{\pi}{4}.sinx\right)\)

\(\Leftrightarrow A=sin\dfrac{\pi}{4}.cosx+cos\dfrac{\pi}{4}.sinx-cos\dfrac{\pi}{4}.cosx-sin\dfrac{\pi}{4}.sinx\)

\(\Leftrightarrow A=\dfrac{\sqrt{2}}{2}.cosx+\dfrac{\sqrt{2}}{2}.sinx-\dfrac{\sqrt{2}}{2}.cosx-\dfrac{\sqrt{2}}{2}.sinx\)

\(\Leftrightarrow A=0\)

b) \(B=cos\left(\dfrac{\pi}{6}-x\right)-sin\left(\dfrac{\pi}{3}+x\right)\)

\(\Leftrightarrow B=cos\dfrac{\pi}{6}.cosx+sin\dfrac{\pi}{6}.sinx-\left(sin\dfrac{\pi}{3}.cosx+cos\dfrac{\pi}{3}.sinx\right)\)

\(\Leftrightarrow B=cos\dfrac{\pi}{6}.cosx+sin\dfrac{\pi}{6}.sinx-sin\dfrac{\pi}{3}.cosx-cos\dfrac{\pi}{3}.sinx\)

\(\Leftrightarrow B=\dfrac{\sqrt{3}}{2}.cosx+\dfrac{1}{2}.sinx-\dfrac{\sqrt{3}}{2}.cosx-\dfrac{1}{2}.sinx\)

\(\Leftrightarrow B=0\)

c) \(C=sin^2x+cos\left(\dfrac{\pi}{3}-x\right).cos\left(\dfrac{\pi}{3}+x\right)\)

\(\Leftrightarrow C=sin^2x+\left(cos\dfrac{\pi}{3}.cosx+sin\dfrac{\pi}{3}.sinx\right).\left(cos\dfrac{\pi}{3}.cosx-sin\dfrac{\pi}{3}.sinx\right)\)

\(\Leftrightarrow C=sin^2x+\left(\dfrac{1}{2}.cosx+\dfrac{\sqrt{3}}{2}.sinx\right).\left(\dfrac{1}{2}.cosx-\dfrac{\sqrt{3}}{2}.sinx\right)\)

\(\Leftrightarrow C=sin^2x+\dfrac{1}{4}.cos^2x-\dfrac{3}{4}.sin^2x\)

\(\Leftrightarrow C=\dfrac{1}{4}.sin^2x+\dfrac{1}{4}.cos^2x\)

\(\Leftrightarrow C=\dfrac{1}{4}\left(sin^2x+cos^2x\right)\)

\(\Leftrightarrow C=\dfrac{1}{4}\)

d) \(D=\dfrac{1-cos2x+sin2x}{1+cos2x+sin2x}.cotx\)

\(\Leftrightarrow D=\dfrac{1-\left(1-2sin^2x\right)+2sinx.cosx}{1+2cos^2a-1+2sinx.cosx}.cotx\)

\(\Leftrightarrow D=\dfrac{2sin^2x+2sinx.cosx}{2cos^2x+2sinx.cosx}.cotx\)

\(\Leftrightarrow D=\dfrac{2sinx\left(sinx+cosx\right)}{2cosx\left(cosx+sinx\right)}.cotx\)

\(\Leftrightarrow D=\dfrac{sinx}{cosx}.cotx\)

\(\Leftrightarrow D=tanx.cotx\)

\(\Leftrightarrow D=1\)

a1)\(\dfrac{sin110}{cos110}+\dfrac{cos20}{sin20}\)

\(=\dfrac{sin\left(180-70\right)}{cos\left(180-70\right)}+\dfrac{cos\left(90-70\right)}{sin\left(90-70\right)}\)

\(=\dfrac{sin70}{-cos70}+\dfrac{sin70}{cos70}=0\)

a2) \(sin^2x+sin^2\left(\dfrac{\pi}{3}-x\right)+sinx.sin\left(\dfrac{\pi}{3}-x\right)\)

\(=\dfrac{1}{2}\left(1-cos2x\right)+\dfrac{1}{2}\left[1-cos\left(\dfrac{2\pi}{3}-2x\right)\right]+\dfrac{1}{2}\left[cos\left(2x-\dfrac{\pi}{3}\right)-cos\left(\dfrac{\pi}{3}\right)\right]\)

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

\(=\dfrac{3}{4}-\dfrac{1}{2}\left[cos2x+cos\left(\dfrac{2\pi}{3}-2x\right)-cos\left(2x-\dfrac{\pi}{3}\right)\right]\)

\(=\dfrac{3}{4}-\dfrac{1}{2}\left[cos2x-2.sin\dfrac{\pi}{6}.sin\left(\dfrac{\pi-4x}{2}\right)\right]\)

\(=\dfrac{3}{4}-\dfrac{1}{2}\left(cos2x-cos2x\right)\)

\(=\dfrac{3}{4}\)

a3) \(sin^2x+cos\left(\dfrac{\pi}{3}-x\right).cos\left(\dfrac{\pi}{3}+x\right)\)

\(=\dfrac{1-cos2x}{2}+\dfrac{1}{2}\left[cos\left(-2x\right)+cos\left(\dfrac{2\pi}{3}\right)\right]\)

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

\(=\dfrac{1}{2}-\dfrac{1}{4}\)

\(=\dfrac{1}{4}\)

\(=cos\left(x-\frac{\pi}{3}\right)cos\left(x+\frac{\pi}{4}\right)+sin\left(\frac{\pi}{2}-x-\frac{\pi}{6}\right)sin\left(\frac{\pi}{2}-x-\frac{3\pi}{4}\right)\)

\(=cos\left(x-\frac{\pi}{3}\right)cos\left(x+\frac{\pi}{4}\right)+sin\left(\frac{\pi}{3}-x\right)sin\left(-x-\frac{\pi}{4}\right)\)

\(=cos\left(x-\frac{\pi}{3}\right)cos\left(x+\frac{\pi}{4}\right)+sin\left(x-\frac{\pi}{3}\right)sin\left(x+\frac{\pi}{4}\right)\)

\(=cos\left(x-\frac{\pi}{3}-x-\frac{\pi}{4}\right)=cos\left(-\frac{7\pi}{12}\right)=cos\frac{7\pi}{12}=\frac{\sqrt{2}-\sqrt{6}}{4}\)

\(D=\frac{1-cos2x+sin2x}{1+cos2x+sin2x}.cotx\)

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

\(=\frac{2sinx\left(cosx+sinx\right)}{2cosx\left(sinx+cosx\right)}.cotx=tanx.cotx=1\)

Câu thứ 2 bạn ấn vào gõ công thức trực quan nhập lại đề nha, khó hiểu quá

\(A=\frac{\sqrt{3}sinx.\left(cosx.cos\frac{\pi}{6}-sinx.sin\frac{\pi}{6}\right)+cosx\left(sin\frac{\pi}{3}cosx-cos\frac{\pi}{6}.sinx\right)}{sin\left(2x+\frac{\pi}{3}\right)}\)

\(A=\frac{\frac{3}{2}sinx.cosx-\frac{\sqrt{3}}{2}sin^2x+\frac{\sqrt{3}}{2}cos^2x-\frac{1}{2}sinx.cosx}{sin\left(2x+\frac{\pi}{3}\right)}\)

\(A=\frac{sinx.cosx+\frac{\sqrt{3}}{2}\left(cos^2x-sin^2x\right)}{sin\left(2x+\frac{\pi}{3}\right)}\)

\(A=\frac{\frac{1}{2}sin2x+\frac{\sqrt{3}}{2}cos2x}{sin\left(2x+\frac{\pi}{3}\right)}=\frac{sin2x.cos\frac{\pi}{3}+cos2x.sin\frac{\pi}{3}}{sin\left(2x+\frac{\pi}{3}\right)}\)

\(A=\frac{sin\left(2x+\frac{\pi}{3}\right)}{sin\left(2x+\frac{\pi}{3}\right)}=1\)

a: pi/2<x<pi

=>cosx<0

=>\(cosx=-\sqrt{1-\left(\dfrac{1}{5}\right)^2}=-\dfrac{2\sqrt{6}}{5}\)

\(sin2x=2\cdot sinx\cdot cosx=2\cdot\dfrac{1}{5}\cdot\dfrac{-2\sqrt{6}}{5}=\dfrac{-4\sqrt{6}}{25}\)

\(cos2x=2\cdot cos^2x-1=2\cdot\dfrac{24}{25}-1=\dfrac{48}{25}-1=\dfrac{23}{25}\)

\(tan2x=-\dfrac{4\sqrt{6}}{25}:\dfrac{23}{25}=-\dfrac{4\sqrt{6}}{23}\)

\(cot2x=1:\dfrac{-4\sqrt{6}}{23}=\dfrac{-23}{4\sqrt{6}}\)

b: \(sin\left(x-\dfrac{pi}{6}\right)=sinx\cdot cos\left(\dfrac{pi}{6}\right)-cosx\cdot sin\left(\dfrac{pi}{6}\right)\)

\(=sinx\cdot\dfrac{\sqrt{3}}{2}-cosx\cdot\dfrac{1}{2}\)

\(=\dfrac{1}{5}\cdot\dfrac{\sqrt{3}}{2}-\dfrac{-2\sqrt{6}}{5}\cdot\dfrac{1}{2}=\dfrac{\sqrt{3}+2\sqrt{6}}{10}\)

c: \(cos\left(x-\dfrac{pi}{3}\right)=cosx\cdot cos\left(\dfrac{pi}{3}\right)+sinx\cdot sin\left(\dfrac{pi}{3}\right)\)

\(=-\dfrac{2\sqrt{6}}{5}\cdot\dfrac{1}{2}+\dfrac{1}{5}\cdot\dfrac{1}{2}=\dfrac{-2\sqrt{6}+1}{10}\)

d: \(tan\left(x-\dfrac{pi}{4}\right)=\dfrac{tanx-tan\left(\dfrac{pi}{4}\right)}{1+tanx\cdot tan\left(\dfrac{pi}{4}\right)}\)

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

\(=\dfrac{\dfrac{1}{-2\sqrt{6}}-1}{1+\dfrac{1}{-2\sqrt{6}}}=\dfrac{-25-4\sqrt{6}}{23}\)