Odpowiedzi

2010-03-07T20:14:16+01:00
A) tgα+ctgα=1/sinα cosα
sina/cosa+cosa/sina=1/sinα cosα /*sina
sin²a/cosa+cosa=1/cosa /*cosa
sin²a+cos²a=1
1=1 ( z jedynki tryg.)

b) (tgα-1)(ctgα+1)=tgα-ctgα
(sina/cosa-1)(cosa/sina+1=tgα-ctgα
(1+sina/cosa-cosa/sina-1)=tgα-ctgα
tga-ctga=tgα-ctgα
2010-03-07T20:14:57+01:00
Tgx+ctgx=1/sinxcosx
L= sinx/cosx + cosx/sinx= sin2x/sinxcosx+ cos2x/ sinxcosx= sin2x+cos2x/sinxcosx = 1/ sinxcosx=P
b) (tgx-1) (ctgx+1)= tgx- ctgx
L= tgxctgx + tgx - ctgx -1= 1-1 + tgx-ctgx = tgx-ctgx= P