20 1.) which of the following polar equations is equivalent to the parametric equations below? x=t^2 y=2t a.) r=4cot(theta)csc(theta) b.) r=4tan(theta)sec(theta) c.) r=tan(theta)sec(theta)/4 d.) r=16cot(theta)csc(theta) 2.) which polar equation is equivalent to the parametric equations below? x=sin(theta)cos(theta)+cos(theta) y=sin^2(theta)+sin(theta) a.) r=cos(theta)+1 b.) r=sin(theta)+1 c.) r=sin^2(theta)+1 d.) r=1-cos(theta)

Part (1)
x = t²   ,   y = 2t
we should know that ⇒ x = r cos θ   ,  y = r sin θ
By equating the values of x ⇒ ∴ t² = r cos θ   →(1)
By equating the values of y ⇒ ∴ 2t = r sin θ ⇒ ∴ 4t² = r² sin² θ  → (2)

dividing equation (2) over (1)
4t²/t² = (r² sin² θ)/(r cos θ)
∴ 4 = r (sin² θ/cos θ) 
∴ r = 4 cos θ / sin² θ = 4 ( cos θ / sin θ ) ( 1 / sin θ)
∴ r = 4 (cot θ) (csc θ)

The correct answer is option A.) r=4cot(theta) csc(theta)

Part (2)
x = sin(θ) cos(θ) + cos(θ) ,   y = sin²(θ) + sin(θ)
we should know that ⇒ x = r cos θ   ,  y = r sin θ

By equating the values of x ⇒
∴ r cos θ = sin(θ) cos(θ) + cos(θ) ⇒⇒ take cos θ as common from right side

∴ r cos θ = cos θ ( 1 + sin θ)  ⇒⇒⇒ divide both sides over cos θ

∴ r = 1 + sin θ  → (1)


By equating the values of y ⇒ 
∴ r sin θ = sin²(θ) + sin(θ) ⇒⇒ take sin θ as common from right side

∴ r sin θ = sin θ ( 1 + sin θ) ⇒⇒⇒ divide both sides over sinθ

∴ r = 1 + sin θ  → (2)


From (1) and (2)
∴ The correct answer is option B.) r = sin θ + 1

2/1 or just 2. sorry if i get this wrong