👤

Découvrez de nouvelles perspectives et obtenez des réponses sur FRstudy.me. Posez vos questions et obtenez des réponses rapides et bien informées de la part de notre réseau de professionnels expérimentés.

Bonjour J'ai Un Dm De math aider moi Svp
Montrè Que:

1+tan²y=1/cos²y
cos² y-sin² y/sin² y+sin y*cosy=1-tan y/tan y
1+cosy/siny=siny/1-cosy
(tany-1/cosy)=1-siny/1+siny


Sagot :

Bonjour houda0638937882

Exercice 1

[tex]1+\tan^2y=1+\dfrac{\sin^2y}{\cos^2y}\\\\1+\tan^2y=\dfrac{\cos^2y}{\cos^2y}+\dfrac{\sin^2y}{\cos^2y}[/tex]

[tex]1+\tan^2y=\dfrac{\cos^2y+\sin^2y}{\cos^2y}\\\\\boxed{1+\tan^2y=\dfrac{1}{\cos^2y}}[/tex]

Exercice 2

[tex]\dfrac{\cos^2 y-\sin^2 y}{\sin^2 y+\sin y\times \cos y}=\dfrac{(\cos y-\sin y)(\cos y+\sin y)}{\sin y(\sin y+\cos y)}[/tex]

[tex]\dfrac{\cos^2 y-\sin^2 y}{\sin^2 y+\sin y\times \cos y}=\dfrac{\cos y-\sin y}{\sin y}[/tex]

[tex]\dfrac{\cos^2 y-\sin^2 y}{\sin^2 y+\sin y\times \cos y}=\dfrac{\dfrac{\cos y}{\cos y}-\dfrac{\sin y}{\cos y}}{\dfrac{\sin y}{\cos y}}[/tex]

[tex]\boxed{\dfrac{\cos^2 y-\sin^2 y}{\sin^2 y+\sin y\times \cos y}=\dfrac{1-\tan y}{\tan y}}[/tex]

Exercice 3

[tex]\dfrac{1+\cos y}{\sin y}=\dfrac{(1+\cos y)(1-\cos y)}{\sin y(1-\cos y)}[/tex]

[tex]\dfrac{1+\cos y}{\sin y}=\dfrac{1^2-\cos^2 y}{\sin y(1-\cos y)}[/tex]

[tex]\dfrac{1+\cos y}{\sin y}=\dfrac{1-\cos^2 y}{\sin y(1-\cos y)}[/tex]

[tex]\dfrac{1+\cos y}{\sin y}=\dfrac{\sin^2 y}{\sin y(1-\cos y)}[/tex]

[tex]\dfrac{1+\cos y}{\sin y}=\dfrac{\sin y\times \sin y}{\sin y(1-\cos y)}[/tex]

[tex]\boxed{\dfrac{1+\cos y}{\sin y}=\dfrac{\sin y}{1-\cos y}}[/tex]

Exercice 4


.[tex](\tan y-\dfrac{1}{\cos y})^2=(\dfrac{\sin y}{\cos y}-\dfrac{1}{\cos y})^2[/tex]

[tex](\tan y-\dfrac{1}{\cos y})^2=(\dfrac{\sin y-1}{\cos y})^2[/tex]

[tex](\tan y-\dfrac{1}{\cos y})^2=\dfrac{(\sin y-1)^2}{\cos^2 y}[/tex]

[tex](\tan y-\dfrac{1}{\cos y})^2=\dfrac{(1-\sin y)^2}{\cos^2 y}[/tex]

[tex](\tan y-\dfrac{1}{\cos y})^2=\dfrac{(1-\sin y)^2}{1-\sin^2 y}[/tex]

[tex](\tan y-\dfrac{1}{\cos y})^2=\dfrac{(1-\sin y)^2}{(1+\sin y)(1-\sin y)}[/tex]

[tex](\tan y-\dfrac{1}{\cos y})^2=\dfrac{(1-\sin y)(1-\sin y)}{(1+\sin y)(1-\sin y)}[/tex]

[tex]\boxed{(\tan y-\dfrac{1}{\cos y})^2=\dfrac{1-\sin y}{1+\sin y}}[/tex]