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Aidez moi svp c important merci d avance
1+tan2y+1/sin2y = 1/sin2y×cos2y


Aidez Moi Svp C Important Merci D Avance 1tan2y1sin2y 1sin2ycos2y class=

Répondre :

[tex]1+tan^2y+ \frac{1}{sin^2y} = \frac{1}{sin^2y*cos^2y} \\ \\(\frac{sin(y)}{cos(y)})^2 + \frac{1}{sin^2y} = \frac{1}{sin^2y*cos^2y} -1=0 \\\\ \frac{sin(y)^2}{cos(y)^2} + \frac{1}{sin^2y} - \frac{1}{sin^2y*cos^2y} = -1 \\\\ \frac{sin(y)^2*sin(y)^2}{sin(y)^2*cos(y)^2} + \frac{cos(y)^2*1}{cos(y)^2*sin(y)^2}- \frac{1}{sin^2y*cos^2y}=-1 \\\\ \frac{sin(y)^4}{sin(y)^2*cos(y)^2} + \frac{cos(y)^2}{sin(y)^2*cos(y)^2} - \frac{1}{sin^2y*cos^2y} =-1\\\\ [tex]\frac{sin(y)^4-(1-cos(y)^2}{sin(y)^2*cos(y)^2} =-1\\\\ \frac{sin(y)^4-sin(y)^2}{sin(y)^2*cos(y)^2}=-1\\\\ \frac{(sin(y)^2-1)*sin(y)^2}{sin(y)^2*cos(y)^2}=-1\\\\ \frac{-(-1sin(y)^2)*sin(y)^2}{sin(y)^2*cos(y)^2}=-1\\\\ \frac{-cos(y)^2*sin(y)^2}{sin(y)^2*cos(y)^2}=-1\\\\ -1=-1 [/tex]

y ∈ R \ {[tex] \frac{k\pi}{2} ,[/tex]K ∈ Z}