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x=sincube t /√cos2t. ,. Y=coscube t/√cos2t. …

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x=sincube t /√cos2t. ,. Y=coscube t/√cos2t. Find dy/dx

Posted by Muskan K 7 years, 3 months ago

CBSE > Class 12 > Mathematics

1 answers

Arpajit Adhikary 7 years, 3 months ago

dx/dt= [(3sin^2(t)cos(t)(√cos(2t))) + {(sin^3(t)(sin(2t)))/(√cos(2t))}]/cos(2t) dy/dt= [(-3cos^2(t)sin(t)(√cos(2t))) + {(cos^3(t)(sin(2t)))/(√cos(2t))}]/cos(2t) dy/dx= [{-3cos^2(t)cos(2t) + sin^2(t)sin(2t)}/{3sin(t)cos(t)cos(2t) + sin(2t)sin^2(t)}]

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