Rozwiązane

Udowodnij tożsamości trygonometryczne:
a) cos²α =ctgα/tgα+ctgα

b)tgα *ctgα/1-sin²α=tg²α+1



Odpowiedź :

a)
(cos x)^2 =ctg x/(tg x +ctg x)
P =[cos x/sin x]/ [ sinx/ cos x + cos x/ sin x] =

= [cos x/ sin x]/{ [ (sin x)^2 + (cos x)^2 ]/cos x*sin x} =
=( cos x/sin x ) *[ cos x *sin x/ 1] =(cos x)^2 = L
cbdo.
b)
[tg x * ctg x]/ [ 1 - (sin x)^2 =(tg x)^2 + 1
L = [ ( sin x/cos x)*(cos x/ sin x)] / (1 - (sin x)^2 ) =
= 1/ [ 1 - ( sin x)^2]
P = [ sin x/cos x]^2 + 1 = (sin x)^2 /(cos x)^2 + 1 =
= [(sin x)^2 +(cos x)^2 ] / ( cos x)^2 = 1/(cos x)^2 =
= 1 /[ 1 - (sin x)^2]
L = P
cbdo.