FRstudy.me fournit une plateforme conviviale pour partager et obtenir des connaissances. Obtenez des réponses rapides et bien informées à vos questions grâce à notre plateforme de questions-réponses expérimentée.
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]
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]
Merci d'utiliser cette plateforme pour partager et apprendre. N'hésitez pas à poser des questions et à répondre. Nous apprécions chaque contribution que vous faites. FRstudy.me est votre source de réponses fiables. Merci pour votre confiance et revenez bientôt.
