\(A=\frac{sinx}{cosx}+\frac{cosx}{sinx}+\frac{sin3x}{cos3x}+\frac{cos3x}{sin3x}\)

\(=\frac{sin^2x+cos^2x}{sinx.cosx}+\frac{sin^23x+cos^23x}{sin3x.cos3x}=\frac{2}{2sinx.cosx}+\frac{2}{2sin3x.cos3x}\)

\(=\frac{2}{sin2x}+\frac{2}{sin6x}=\frac{2\left(sin2x+sin6x\right)}{sin2x.sin6x}=\frac{4sin4x.cos2x}{sin2x.sin6x}\)

\(=\frac{8sin2x.cos^22x}{sin2x.sin6x}=\frac{8cos^22x}{sin6x}\)

\(B=\frac{sin30}{cos30}+\frac{sin60}{cos60}+\frac{sin40}{cos40}+\frac{sin50}{cos50}=\frac{sin30.cos60+cos30.sin60}{cos30.cos60}+\frac{sin40.cos50+sin50.cos40}{cos40.cos50}\)

\(=\frac{sin90}{cos30.cos60}+\frac{sin90}{cos40.cos50}=\frac{1}{\frac{1}{2}.\frac{\sqrt{3}}{2}}+\frac{1}{\frac{1}{2}cos90+\frac{1}{2}cos10}\)

\(=\frac{4\sqrt{3}}{3}+\frac{2}{cos10}=\frac{4\sqrt{3}\left(cos10+\frac{\sqrt{3}}{2}\right)}{3cos10}=\frac{4\sqrt{3}\left(cos10+cos30\right)}{3cos10}\)

\(=\frac{8\sqrt{3}cos20.cos10}{3cos10}=\frac{8\sqrt{3}}{3}cos20\)

\(\frac{3cot60^o}{2cos^230^o-1}=\frac{3.\frac{\sqrt{3}}{3}}{2.\left(\frac{\sqrt{3}}{2}\right)^2-1}=\frac{\sqrt{3}}{\frac{3}{2}-1}=2\sqrt{3}\)

Sửa: \(A=\dfrac{\cos70^0-\sin\alpha}{\tan60^0-\cot70^0}\)

Vì \(\sin\alpha>\sin20^0\Leftrightarrow\cos70^0-\sin\alpha< \sin20^0-\sin20^0=0\)

Mà \(\tan60^0-\cot70^0=\tan60^0-\tan20^0>0\)

Do đó \(A< 0,\forall20^0< \alpha< 90^0\)

\(a,C=\dfrac{2x^2-x-x-1+2-x^2}{x-1}\left(x\ne1\right)\\ C=\dfrac{x^2-2x+1}{x-1}=\dfrac{\left(x-1\right)^2}{x-1}=x-1\\ b,D=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\left(a>0;a\ne1\right)\\ D=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)

Có 

H/tan 60+H/tan30=80

h.(1/tan60+1/tan30)=80

h=80:(1/tan60+1/tan30)

H=46,7653718 

\(M=cot28^o\cdot cot62^o+tan60^o\)

\(M=\dfrac{1}{tan28^o}\cdot\dfrac{1}{tan62^o}+tan60^o\)

\(M=\dfrac{1}{tan28^o\cdot tan62^o}+tan60^o\)

\(M=\dfrac{1}{1}+tan60^o\)

\(M=1+\sqrt{3}\)

\(sin58^0=cos32^0\)

Đáp án D đúng