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/

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]