Tan50°/ cot40° Cảm ơn m.n đã giúp

Ta có:

$50^o+40^o=90^o$

$\Leftrightarrow 40^o=90^o-50^o$

Vậy:

$\dfrac{\tan 50^o}{\cot 40^o}$

$=\dfrac{\tan 50^o}{\tan 50^o}$

$=1$