Question

Помогите с алгеброй, очень прошу, дам 50б.
Упростить выражения:

Answer 1

$frac{Sin( alpha - beta )}{Cos alpha Cos beta } +tg beta = frac{Sin alpha Cos beta -Sin beta Cos alpha }{Cos alpha Cos beta }+tg beta = frac{Sin alpha Cos beta }{Cos alpha Cos beta }- frac{Sin beta Cos alpha }{Cos alpha Cos beta }+$$+tg beta =tg alpha -tg beta +tg beta =tg alpha$

$frac{Cos( alpha + beta )+Cos( alpha - beta )}{Cos alpha Cos beta }= frac{Cos alpha Cos beta -Sin alpha Sin beta +Cos alpha Cos beta +Sin alpha Sin beta }{Cos alpha Cos beta }=$$= frac{2Cos alpha Cos beta }{Cos alpha Cos beta } =2$

$frac{tg alpha +Ctg beta }{Cos( alpha - beta )}*Sin alpha Sin beta = frac{ (frac{Sin alpha }{Cos alpha } + frac{Cos beta }{Sin beta })*Sin alpha Sin beta }{Cos( alpha - beta )}=$$= frac{ frac{Sin alpha Sin beta +Cos alpha Cos beta }{Cos alpha Sin beta }*Sin alpha Sin beta }{Cos( alpha - beta )}= frac{Cos( alpha - beta )*Sin alpha }{Cos alpha *Cos( alpha - beta )}=tg alpha$

$Cos( frac{ pi }{3}- alpha )+Cos( frac{ pi }{3}+ alpha )=Cos frac{ pi }{3}Cos alpha +Sin frac{ pi }{3}Sin alpha +Cos frac{ pi }{3}Cos alpha -$$-Sin frac{ pi }{3}Sin alpha =2Cos frac{ pi }{3}Cos alpha =2* frac{1}{2}*Cos alpha=Cos alpha$

$frac{2Cos alpha Sin beta +Sin( alpha - beta )}{2Cos alpha Cos beta -Cos( alpha - beta )} = frac{2Cos alpha Sin beta +Sin alpha Cos beta -Sin beta Cos alpha }{2Cos alpha Cos beta -Cos alpha Cos beta -Sin alpha Sin beta } =$$= frac{Cos alpha Sin beta +Sin alpha Cos beta }{Cos alpha Cos beta -Sin alpha Sin beta }= frac{Sin( alpha + beta )}{Cos( alpha + beta )}=tg( alpha + beta )$