x=A sin theta, y= B tan theta prove that A^2/x^2-B^2/y^2=1
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A^2/x^2-B^2/y^2
(A^2/A^2sin^2€)- (B^2/B^2tan^2€)
=(1/sin^2€)- (1/tan^2€)
=(tan^2€-sin^2€)/)tan^2€*sin^2€)
=(sin^2€/cos^2€-sin^2€)/[sin^2€/cos^2€)*sin^2€)】
=[sin^2€(1-cos^2€)/cos^2€]/[(sin^2€*sin^2€)/cos^2€】
=[(1-cos^2€)/cos^2€]/sin^2€)/cos^2€】
=(1-cos^2€)/sin^2€
=sin^2€/sin^2€
=1
(A^2/A^2sin^2€)- (B^2/B^2tan^2€)
=(1/sin^2€)- (1/tan^2€)
=(tan^2€-sin^2€)/)tan^2€*sin^2€)
=(sin^2€/cos^2€-sin^2€)/[sin^2€/cos^2€)*sin^2€)】
=[sin^2€(1-cos^2€)/cos^2€]/[(sin^2€*sin^2€)/cos^2€】
=[(1-cos^2€)/cos^2€]/sin^2€)/cos^2€】
=(1-cos^2€)/sin^2€
=sin^2€/sin^2€
=1
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