Question

x=cosθ-cos2θ y=sinθ-sin2θ θ ∈ R- {Kπ | K ∈ Z}, cosθ ≠ 1/4,Find dy/dx :(Wherever y is defined as a function of x and dx/dt or dx/dθ≠0)

Answer 1

x = cosθ - cos2θ

differentiate with respect to θ,

dx/dθ = d(cosθ - cos2θ)/dθ

= dcosθ/dθ - d(cos2θ)/dθ

= -sinθ -(-sin2θ) × d(2θ)/dθ

= -sinθ + 2sinθ ..........(1)

y = sinθ - sin2θ

differentiate with respect to θ,

dy/dθ = d(sinθ - sin2θ)/dθ

= dsinθ/dθ - d(sin2θ)/dθ

= cosθ - 2cos2θ..........(2)

now, dy/dx = {dy/dθ}/{dx/dθ}

from equations (1) and (2),

dy/dx = (cosθ - 2cos2θ)/(sinθ - 2sin2θ)

Answer 2

HELLO DEAR,



GIVEN:-
x = cosθ - cos2θ,

differentiating x w.r.t θ

dx/dθ = d(cosθ - cos2θ)/dθ

=> dx/dθ = -sinθ - 2(-sin2θ)

=> dx/dθ = 2sin2θ - sinθ-------( 1 )


y = sinθ - sin2θ

differentiating y w.r.t θ

dy/dθ = d(sinθ - sin2θ)/dθ

=> dy/dθ = cosθ - 2cos2θ-------( 2 )


NOW, divide---( 1 ) by ---( 2 )

dy/dx = {dy/dθ}/{dx/dθ}

=> dy/dx = (cosθ - 2cos2θ)/(2sinθ - sinθ)


I HOPE ITS HELP YOU DEAR,
THANKS