1) Ta có:

$sin\widehat{B}= \dfrac{AH}{AB}$

$\Rightarrow AB = \dfrac{AH}{sin\widehat{B}} = \dfrac{6}{sin60^o} = 4\sqrt{3} \, cm$

$cos\widehat{B} = \dfrac{AB}{BC}$

$\Rightarrow BC = \dfrac{AB}{cos\widehat{B}} = \dfrac{4\sqrt{3}}{cos60^o} = 8\sqrt{3} \, cm$

$tan\widehat{B} =\dfrac{AC}{AB}$

$\Rightarrow AC = AB.tan\widehat{B} = 4\sqrt{3}.tan60^o = 12 \, cm$

2) Xét $∆ABC$ vuông tại $A$ có $\widehat{C} = \alpha$

a) $tan\alpha = tan\widehat{C} = \dfrac{AB}{AC}$

$= \dfrac{AB}{BC}.\dfrac{BC}{AC}$

$=\dfrac{AB}{BC}.\dfrac{1}{\dfrac{AC}{BC}}$

$= sin\widehat{C}.\dfrac{1}{cos\widehat{C}}$

$=\dfrac{sin\alpha}{cos\alpha}$

b) $cot\alpha = cot\widehat{C} = \dfrac{AC}{AB}$

$= \dfrac{AC}{BC}.\dfrac{BC}{AB}$
$=\dfrac{AC}{BC}.\dfrac{1}{\dfrac{AB}{BC}}$
$= cos\widehat{C}.\dfrac{1}{sin\widehat{C}}$
$=\dfrac{cos\alpha}{sin\alpha}$

c) Áp dụng kết quả đã chứng minh ở câu a và b, ta có:

$tan\alpha = \dfrac{sin\alpha}{cos\alpha}$

$cot\alpha = \dfrac{cos\alpha}{sin\alpha}$

$\Rightarrow tan\alpha.cot\alpha = \dfrac{sin\alpha}{cos\alpha}\cdot \dfrac{cos\alpha}{sin\alpha} = 1$