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Question:
X= sectheta - costheta, Y= sec^2theta- cos^theta then yy'
Answer:

Given, y = sec2 θ - cos2 θ

Now, dy/dθ = 2 * sec θ * (sec θ * tan θ) - 2 * cos θ *(-sin θ)

=> dy/dθ = 2 * sec2 θ * tan θ + 2 * cos θ * sin θ

=> dy/dθ = 2 * sec2 θ * (sin θ/cos θ) + 2 * cos θ * sin θ

=> dy/dθ = 2 * sec3 θ * sin θ + 2 * cos θ * sin θ

Again x = sec θ - cos θ

dx/dθ = sec θ * tan θ - (-sin θ)

=> dx/dθ = sec θ * tan θ + sin θ

=> dx/dθ = sec θ * (sin θ/cos θ) + sin θ

=> dy/dθ = sec2 θ * sin θ +  sin θ

Now, dy/dx = (dy/dθ)/(dx/dθ)

                   = (2 * sec3 θ * sin θ + 2 * cos θ * sin θ)/(sec2 θ * sin θ + sin θ)

                   = (2 * sec3 θ + 2 * cos θ)/(sec2 θ + 1)

Now, y*(dy/dx) = (sec2 θ - cos2 θ) * {(2 * sec3 θ + 2 * cos θ)/(sec2 θ + 1)}     

=> y*(dy/dx) = {(sec2 θ - cos2 θ) * (2 * sec3 θ + 2 * cos θ)}/(sec2 θ + 1)}

=> y*(dy/dx) = {(1/cos2 θ - cos2 θ) * (2 * 1/cos3 θ + 2 * cos θ)}/(1/cos2 θ + 1)}

=> y*(dy/dx) = {(1 - cos4 θ)/cos2 θ * (2 + 2 * cos4 θ)/cos3 θ}/(1 + cos2 θ)/cos2 θ}

=> y*(dy/dx) = 2{(1 - cos4 θ)/cos2 θ * (1 + cos4 θ)/cos3 θ}/(1 + cos2 θ)/cos2 θ}

=> y*(dy/dx) = 2(1 - cos8 θ)/{cos3 θ *(1 + cos2 θ)}

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