Odpowiedzi

2009-11-22T17:48:17+01:00
(tgα-1)(ctgα+1)=tgα-ctgα
L=(tgα-1)(ctgα+1)=tgαctgα+tgα-ctgα-1=1+tgα-ctgα-1=tgα-ctgα
L=P
tgαctgα=1
dowód
tgα=sinα/cosα
ctgα=cosα/sinα
(sinα/cosα)*(cosα/sinα)=1
2009-11-22T17:52:57+01:00
Lewa strona L=(tgα-1)(ctgα+1)=(1+tgα)(1/tgα+1)=
(tgα-1)[(1+tgα)/tgα]=[(tgα-1)(1+tgα)]/tgα=[(tgα-1)(tgα+1)]/tgα=
=(tg²α-1)/tgα=tgα-1/tgα=tgα-ctgα+P
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