Question

Simplifique:
F= tanβ+cotβ+secβ
---------------------------
csc β+1
Plis necesito procedimiento

Answer 1

resolviendo :

$F= \frac{tan \beta +cot \beta +sec \beta }{csc \beta +1} \\ \\ F= \frac{ \frac{sen \beta }{cos \beta }+ \frac{cos \beta }{sen \beta } + \frac{1}{cos \beta } }{ \frac{1}{sen \beta } +1} \\ \\ F= \frac{ \frac{sen \beta +1}{cos \beta }+ \frac{cos \beta }{sen \beta } }{ \frac{1+sen \beta }{sen \beta } } \\ \\ F= \frac{ \frac{sen \beta (sen \beta +1)+cos \beta (cos \beta )}{sen \beta .cos \beta } }{ \frac{1+sen \beta }{sen \beta } }$
$F= \frac{ \frac{sen \beta (sen \beta +1)+cos \beta (cos \beta )}{sen \beta .cos \beta } }{ \frac{1+sen \beta }{sen \beta } } \\ \\ F= \frac{sen \beta (sen \beta +1)+cos \beta (cos \beta )}{cos \beta (1+sen \beta )} \\ \\ F= \frac{sen \beta }{cos \beta } + \frac{cos \beta }{1+sen \beta } \\ \\ F= \frac{sen \beta (1+sen \beta )+cos ^{2} \beta }{cos \beta (1+sen \beta) } \\ \\ F= \frac{sen \beta +sen ^{2 \beta }+cos^{2} \beta }{cos \beta (1+sen \beta )}$
$F= \frac{sen \beta +sen ^{2 \beta }+cos^{2} \beta }{cos \beta (1+sen \beta )} \\ \\ F= \frac{(1+sen \beta )}{cos \beta (1+sen \beta )} \\ \\ F= \frac{1}{cos \beta } \\ \\ F=sec \beta$

recordar :

sen²x+cos²x=1 

senx/cosx= tanx 
cosx/senx=cotx
senx.cscx=1
cosx.secx=1 
tanx.cotx=1 

Saludos Isabela