Ta có $\tan A+\tan B+\tan C=\tan A\tan B\tan C$
Thật vậy:
$VT=\tan A+\tan B+\tan C$
$=\dfrac{\sin A}{\cos A}+\dfrac{\sin B}{\cos B}+\dfrac{\sin C}{\cos C}$
$=\dfrac{\sin A\cos B+\cos A\sin B}{\cos A\cos B}+\dfrac{\sin C}{\cos C}$
$=\dfrac{\sin(A+B)}{\cos A\cos B}+\dfrac{\sin C}{\cos C}$
$=\dfrac{\sin C}{\cos A\cos B}+\dfrac{\sin C}{\cos C}$
$=\dfrac{\sin C\cos C+\cos A\cos B\sin C}{\cos A\cos B\cos C}$
$=\tan C.\dfrac{\cos C+\cos A\cos B}{\cos A\cos B}$
$=\tan C.\dfrac{ \cos(180^o-(A+B))+\cos A\cos B}{\cos A\cos B}$
$=\tan C.\dfrac{\cos A\cos B-\cos(A+B)}{\cos A\cos B}$
$=\tan C.\dfrac{\cos A\cos B-\cos A\cos B+\sin A\sin B}{\cos A\cos B}$
$=\tan A\tan B\tan C$
$=VP$ (đpcm)
Do đó:
$\cot A\cot B+\cot B\cot C+\cot A\cot C$
$=\dfrac{1}{\tan A\tan B}+\dfrac{1}{\tan B\tan C}+\dfrac{1}{\tan A\tan C}$
$=\dfrac{1}{ \dfrac{\tan A+\tan B+\tan C}{\tan C} } +\dfrac{1}{ \dfrac{\tan A+\tan B+\tan C}{\tan A} }+\dfrac{1}{ \dfrac{\tan A+\tan B+\tan C}{\tan B} }$
$=\dfrac{\tan A+\tan B+\tan C}{\tan A+\tan B+\tan C}$
$=1$ (đpcm)
