Question

dy/dx{(Cos^2theta)/(sin^2theta-cos2theta)}

Answer 1

Answer:

D / Dix {(Cell ^ 2 Theta) / (Sin ^ 2 Theta-Cell 2 Theta).

Step-by-step explanation:

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Answer 2

Answer:

Correct option is

A

tan

2

3θ

Given, x=2cosθ−cos2θ and y=2sinθ−sin2θ

dθ

dx

=−2sinθ+2sin2θ

and

dθ

dy

=2cosθ−2cos2θ

∴

dx

dy

=

−2sinθ+2sin2θ

2cosθ−2cos2θ

=

sin2θ−sinθ

cosθ−cos2θ

=

2cos(

2

θ+2θ

)sin(

2

2θ−θ

)

2sin(

2

θ+2θ

)sin(

2

2θ−θ

)

=tan

2

3θ