Biểu thức này chỉ rút gọn được khi mẫu là \(1-2sin^210^0\)

\(tan40+tan50=\dfrac{sin40}{cos40}+\dfrac{sin50}{cos50}=\dfrac{sin40.cos50+cos50.sin40}{cos40.cos50}\)

\(=\dfrac{sin\left(40+50\right)}{\dfrac{1}{2}\left(cos90+cos10\right)}=\dfrac{2}{cos10}\)

\(\Rightarrow tan30+tan60+tan40+tan50=\dfrac{\sqrt{3}}{3}+\sqrt{3}+\dfrac{2}{cos10}\)

\(=\dfrac{4\sqrt{3}}{3}+\dfrac{2}{cos10}=\dfrac{4\sqrt{3}cos10+6}{3.cos10}=\dfrac{4\sqrt{3}\left(cos10+\dfrac{\sqrt{3}}{2}\right)}{3.cos10}\)

\(=\dfrac{4\sqrt{3}\left(cos10+cos30\right)}{3cos10}=\dfrac{8\sqrt{3}cos20.cos10}{3cos10}=\dfrac{8\sqrt{3}}{3}cos20\)

\(\Rightarrow G=\dfrac{\dfrac{8\sqrt{3}}{3}cos20}{1-2sin^210}=\dfrac{\dfrac{8\sqrt{3}}{3}cos20}{cos20}=\dfrac{8\sqrt{3}}{3}\)

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 \(A< 0,\forall20^0< \alpha< 90^0\)

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

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

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

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

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

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

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

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

H/tan 60+H/tan30=80

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

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

H=46,7653718 

Ta có \(\sin x=\cos\left(90^0-x\right)\)

\(\Rightarrow M=\left(\sin^242^0+\sin^248^0\right)+\left(\sin^243^0+\sin^247^0\right)+\left(\sin^244^0+\sin^246^0\right)+\sin^245^0\)

\(=\left(\sin^242^0+\cos^242^0\right)+\left(\sin^243^0+\cos^243^0\right)+\left(\sin^244^0+\cos^244^0\right)+\sin^245^0\)

\(=1+1+1+\left(\frac{\sqrt{2}}{2}\right)^2=3+\frac{1}{2}=\frac{7}{2}\)

Gọi H,K lần lượt là các tiếp điểm của các tiếp tuyến cắt nhau tại M của (O;r)

=>OH=OK và OH\(\perp\)MB tại H và OK\(\perp\)MD tại K

Xét (O,R) có

OH,OK lần lượt là khoảng cách từ O xuống các dây AB,CD

OH=OK

Do đó: \(sđ\stackrel\frown{AB}=sđ\stackrel\frown{CD}\)