What is the value of (dy/dx)2 + 1 if x = a sin2θ(1 + cos2θ) and y = a cos2θ(1 – cos2θ)?

What is the value of (dy/dx)2 + 1 if x = a sin2θ(1 + cos2θ) and y = a cos2θ(1 – cos2θ)? Correct Answer Sec2θ

Here, x = a sin2θ(1 + cos2θ) and y = a cos2θ(1 – cos2θ) => x = 2a cos2θ*sin2θ and y = 2a sin2θ*cos2θ Now differentiating x and y with respect to θ we get, dx/dθ = 2a = 4a cosθ (cosθ cos2θ – sinθ sin2θ ) = 4a cosθ cos(θ + 2θ) = 4a cosθ cos3θ dy/dθ = 2a = 4a sinθ (cosθ cos2θ – sinθ sin2θ ) = 4a sinθ cos(θ + 2θ) = 4a sinθ cos3θ Thus, dy/dx = (dy/dθ)/(dx/dθ) = (= 4a cosθ cos3θ)/( 4a sinθ cos3θ) = tanθ So, (dy/dx)2 + 1 = 1 + tan2θ = sec2θ