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x=sin^3p/cos^2p
y=cos^3p/sin^2p
x+y= (sin^3p*sin^2p+cos^2p*cos^3p)/sin^2p*cos^2p
by expanding this u get
{ [ (sinp+cosp)(sin^2p+cos^2p-sinp*cosp)]-sin^2p*cos^2p(sinp+cosp)}/sin^2pcos^2p

{ [1/4] - [9/128] } *[64/9]
u get the answer as 23/18

if there is any wrong in my answer plz tell me
i hope u understand my solution manu

3ycosA=2xsinA---be the first equ
4x-3ycosecA=2---be the second equ
u get thevalue of x =cosA/2cosA-1
y=2sinA/3(2cosA-1)
sub in the equ 4x^2+9y^2=4
u get A=0
sub in the 2cosA-1
u get the required equ

The question is challenging and good . It is really a good question

The second u have sent is really in confusion . Send it in a proper way

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my id is dimpy@yahoo.com