Question

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

(a) Tan^2θ

(b) Cosec^2θ

(c) Cot^2θ

(d) Sec^2θ

The question was asked in an internship interview.

The query is from First Order Derivative in chapter Limits and Derivatives of Mathematics – Class 11

Answer 1

Correct choice is (d) Sec^2θ

Explanation: Here, x = a sin2θ(1 + cos2θ) and y = a cos2θ(1 – cos2θ)

=> x = 2a cos^2θ*sin2θ and y = 2a sin^2θ*cos2θ

Now differentiating x and y with respect to θ we get,

dx/dθ = 2a[cos^2θ*2cos2θ + sin2θ*2cos2θ cos2θ]

= 4a cosθ (cosθ cos2θ – sinθ sin2θ )

= 4a cosθ cos(θ + 2θ)

= 4a cosθ cos3θ

dy/dθ = 2a[cos2θ*2cosθ sinθ + sin^2θ (-2sin2θ)]

= 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 + tan^2θ = sec^2θ