$\dfrac{\text{cos θ. cos (90° - θ)}}{\text{tan (90° - θ)}}$ is equivalent to :

1. cos² θ - 1

2. sin² θ

3. sin² θ - cos² θ

4. sin² θ - 1

Solving,

$\Rightarrow \dfrac{\text{cos θ. cos (90° - θ)}}{\text{tan (90° - θ)}} \\[1em] \Rightarrow \dfrac{\text{cos θ . sin θ}}{\text{cot θ}} \\[1em] \Rightarrow \dfrac{\text{cos θ . sin θ}}{\dfrac{\text{cos θ}}{\text{sin θ}}} \\[1em] \Rightarrow \dfrac{\text{cos θ . sin θ . sin θ}}{\text{cos θ}} \\[1em] \Rightarrow \text{sin}^2 θ.$

Hence, Option 2 is the correct option.