Question

Demonstrati ca ctg (a+b) = $\frac{1}{tg(a+b)}$

Answer 1

$ctg(a+b)=\frac{cos(a+b)}{sin(a+b)}$
$\frac{1}{ctg(a+b)}=\frac{sin(a+b)}{cos(a+b)}=tg(a+b)$
Deci
$ctg(a+b)=\frac{1}{tg(a+b)}$
Chestia asta e prin definitie.
tg(x)=1/ctg(x)