a) Ta có: \(A=2\cdot\cot37^0\cdot\cot53^0+\sin^228^0+\sin^262^0-\dfrac{3\cdot\tan54^0}{\cot36^0}\)

\(=2\cdot\tan53^0\cdot\cot53^0+\sin^228^0+\cos^228^0-\dfrac{3\cdot\tan54^0}{\tan54^0}\)

\(=2+1-3\)

=0

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

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

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

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

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

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

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

cos 50=sin 40(2 góc phụ nhau)

50>40=>sin 50> sin 40=> sin 50> cos 50 (1)

sin 50<1 (2)

tan 50 =sin50/cos 50=sin50 / sin40 > 1(tử lớn hơn mẫu)=>tan 50>1 (3)

(1)(2)(3)=> tan50>sin50>cos50

cos50 = sin40

<=> cos50 < sin50

tan50=cot40

:v.... sao k thấy lq j hết

\(\cot65^0=\tan25^0< \cot60^0=\tan30^0< \tan50^0< \tan70^0\)

\(cos\left(2x-18^o\right).tan50^0+sin\left(2x-18^o\right)=\dfrac{1}{2cos130^0}\)

⇔\(cos\left(2x-18^o\right).sin50^0+sin\left(2x-18^o\right).cos50^0=\dfrac{cos50^0}{2cos130^0}\)

(Nhân cả 2 vế với cos50⁰)

⇔ sin (50⁰ + 2x - 18⁰) = \(-\dfrac{1}{2}\)

⇔ \(\left[{}\begin{matrix}2x+32^0=-30^0+k.360^0\\2x+32^0=210^0+k.360^0\end{matrix}\right.\) với k là số nguyên

\(=\left(1+tan^220\right).cos^220-tan40.cot\left(90-50\right)\)

\(=\left(1+\frac{sin^220}{cos^220}\right).cos^220-tan40.cot40\)

\(=cos^220+sin^220-1\)

\(=1-1=0\)

a) Ta có: \(sin\alpha=cos\left(90-\alpha\right)\Rightarrow sin42=cos48\)

\(\Rightarrow sin42-cos48=0\)

b) Ta có: \(sin\alpha=cos\left(90-\alpha\right)\Rightarrow sin61=cos29\Rightarrow sin^261=cos^229\)

\(\Rightarrow sin^261+sin^229=sin^229+cos^229=1\)

c) Ta có: \(tan\alpha=\dfrac{1}{tan\left(90-\alpha\right)}\Rightarrow tan40=\dfrac{1}{tan50}\)

\(\Rightarrow tan40.tan50=1\) mà \(tan45=1\Rightarrow tan40.tan45.tan50=1\)

\(sin42^0-cos48^0=sin42^0-sin\left(90^0-48^0\right)=sin42^0-sin42^0=0\)

\(sin^261^0+sin^229^0=sin^261^0+cos^2\left(90^0-29^0\right)=sin^261^0+cos^261^0=1\)

\(tan40^0.tan50^0.tan45^0=tan40^0.cot\left(90^0-50^0\right).1=tan40^0.cot40^0=1\)

Sử dụng các công thức:

\(cosa=sin\left(90^0-a\right)\) ; \(sina=cos\left(90^0-a\right)\) ; \(tana=cot\left(90^0-a\right)\) ; \(tana.cota=1\)

ta có : \(5tan40.tan50-cos^247-3-cos^243\)

\(=5tan40.tan\left(90-40\right)-cos^247-cos^2\left(90-47\right)-3\)

\(=5.tan40.cot40-cos^247-sin^247-3=5-1-3=1\)

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

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

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

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

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

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

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

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