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CBSE Class 12-science Answered

find
Asked by | 24 May, 2009, 08:58: PM
Expert Answer

x=a sin2θ(1+cos2θ)

dx/dθ=2acos2θ(1+cos2θ)-2asin22θ=2acos 2θ+2acos2 4θ

Similarily:

y=a cos2θ(1-cos2θ)

dy/dθ=-2asin2θ(1-cos2θ)+2asin2θcos2θ=2asin2θ

Dividing:

dy/dx=2acos 2θ+2acos2 4θ/2asin2θ=tan

x=a sin2θ(1+cos2θ)

dx/dθ=2acos2θ(1+cos2θ)-2asin22θ=2acos 2θ+2acos2 4θ

Similarily:

y=a cos2θ(1-cos2θ)

dy/dθ=-2asin2θ(1-cos2θ)+2asin2θcos2θ=2asin2θ

Dividing:

dy/dx=2acos 2θ+2acos2 4θ/2asin2θ
             =(cos2θ+cos2 4θ)/sin2θ

 

 

Answered by | 15 Jun, 2009, 12:10: AM
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