Arctan

f définie sur :
f(x)=(racine(x²+1)+x)Arctan(racine(x²+1)-x)
1)calculer les limites en -infini et +infini.
je trouve:lim-infini=0 et lim+infini=1
2)On pose: a=Arctan(x), Montrer que: f(x)=(Pi-2a)/(4tan((pi/4)-(a/2))) j'ai essayé de simplifier:
4tan((pi/4)-(a/2))=(1-tan(a/2))/(1+tan(a/2)) mais en vain.
Merci d'avance.

Bonsoir,
On note: t=tan(a/2) , F(x)=(Pi-2a)/(4tan(pi/4-a/2)).
On doit montrer que f(x)=F(x).
Et on rappelle que:
cos(a)=(1-t²)/(1+t²) , tan(a)=2t/(1-t²)
On a:
f(x)=(racine(x²+1)+x).Arctan(racine(x²+1)-x)
<==>f(x)=Arctan(racine(x²+1)-x)/(racine(x²+1)-x)
<==>f(x)=Arctan(racine(tan(a)²+1)-tan(a))/(racine(tan(a)²+1)-tan(a))
<==>f(x)=Arctan((1/cos(a))-tan(a))/((1/cos(a))-tan(a))
<==>f(x)=[Arctan((1-t)²/(1-t²))]/[((1-t)²/(1-t²))]
Or:
F(x)=(Pi-2a)/(4tan(pi/4-a/2))
<==>F(x)=(Pi/4-a/2)/(tan(Pi/4)-a/2)
<==>F(x)=[(Pi/4-a/2)]/[(1-t)/(1+t)]
Donc:
f(x)=F(x)
<==>[Arctan((1-t)²/(1-t²))]/[((1-t)²/(1-t²))]=[(Pi/4-a/2)]/[(1-t)/(1+t)]
<==>[Arctan((1-t)²/(1-t²))]=[(Pi/4-a/2)]
<==>(1-t)²/(1-t²)=tan(Pi/4-a/2)
<==>(1-t)²/(1-t²)=(1-t)/(1+t)
Ce qui est vrai...
Ainsi:
f(x)=F(x)
C.Q.F.D.