ta có:

 . \(\hept{\begin{cases}tan\alpha=\frac{sin\alpha}{cos\alpha}\\cot\alpha=\frac{cos\alpha}{sin\alpha}\\tan\alpha\times cot\alpha=1\end{cases}}\)

Gọi tam giác vuông đề bài cho là ΔABC vuông tại A, có \(\hat{B}=\alpha\)

Xét ΔABC vuông tại A có

\(\sin\alpha=\sin B=\frac{AC}{BC};cos\alpha=cosB=\frac{AB}{BC}\)

\(\tan\alpha=\tan B=\frac{AC}{AB};\cot\alpha=\cot B=\frac{AB}{AC}\)

\(\frac{\sin\alpha}{cos\alpha}=\frac{AC}{BC}:\frac{AB}{BC}=\frac{AC}{AB}=tan\alpha\)

\(\frac{cosa}{\sin a}=\frac{cosB}{\sin B}=\frac{AB}{BC}:\frac{AC}{BC}=\frac{AB}{AC}=\cot\alpha\)

\(\tan a\cdot\cot\alpha=\frac{\sin\alpha}{cos\alpha}\cdot\frac{cos\alpha}{\sin\alpha}=1\)

\(\sin^2\alpha+cos^2\alpha=\left(\frac{AB}{BC}\right)^2+\left(\frac{AC}{BC}\right)^2=\frac{AB^2+AC^2}{BC^2}=\frac{BC^2}{BC^2}=1\)

b: \(\cot\alpha=\dfrac{\cos\alpha}{\sin\alpha}\)

Ta có : \(\tan\left(2a+30\right)=\cot a=\dfrac{1}{\tan a}\)

\(\Rightarrow\tan a.\tan\left(2a+30\right)=1\)

\(\Rightarrow\cot\left(90-a\right).\tan\left(2a+30\right)=1\)

\(\Rightarrow90-a=2a+30\)

=> a = 20

Vậy ..

\(tan^3a+cot^3a=\frac{sin^3a}{cos^3a}+\frac{cos^3a}{sin^3a}\ge2\sqrt{\frac{sin^3a}{cos^3a}.\frac{cos^3a}{sin^3a}}=2\)

A

Chọn A

a) sin anpha = 2/3 => góc anpha = 42^o 

cos 42^o = 0,743

tan 42^o =  0,9

cot  42^o = 1/tan 42^o = 1/0,9 = 1,111

b) tan anpha + cot anpha = 3

<=> tan anpha + 1/tan anpha = 3

<=> tan²  anpha = 2

<=> tan anpha = \(\sqrt{2}\)

=> góc anpha =  55^o 

Ta có: a = sin 55^o . cos 55^o

<=> a = 0,469

A B C H M

a) Do AM là trung tuyến nên BM = MC

Ta có :  \(HC-HB-2HM\)

\(=HM+MC-HB-HM-HM\)

\(=MC-HB-HM\)

\(=MC-\left(HB+HM\right)\)

\(=MC-MB=0\)

\(\Rightarrow HC-HB=2MC\left(đpcm\right)\)

b) Xét  \(\Delta AHM\)có  \(\tan a=\frac{HM}{AH}\)

Xét  \(\Delta AHC\)có  \(\cot C=\frac{HC}{AH}\)

Xét  \(\Delta AHB\)có  \(\cot B=\frac{HB}{AH}\)

Ta có :  \(\frac{\cot C-\cot B}{2}=\left(\frac{HC}{AH}-\frac{HB}{AH}\right)\div2=\frac{HC-HB}{AH}\div2\)

Mà  \(HC-HB=2HM\)( câu a )

\(\Rightarrow\frac{\cot C-\cot B}{2}=\frac{2HM}{AH}\div2=\frac{HM}{AH}=\tan a\left(đpcm\right)\)

Vậy ...