Question

\[\frac{\cos \theta \cdot \cos(90^\circ - \theta)}{\tan(90^\circ - \theta)}\] is equivalent to ______.

Options

• cos² θ − 1

• sin² θ

• sin² θ - cos² θ

• sin² θ − 1

Answer 1

\[\frac{\cos \theta \cdot \cos(90^\circ - \theta)}{\tan(90^\circ - \theta)}\] is equivalent to sin² θ.

Explanation: 

Solving,

\[\Rightarrow \frac{\cos \theta \cdot \cos(90^\circ - \theta)}{\tan(90^\circ - \theta)}\]

\[\Rightarrow \frac{\cos \theta \cdot \sin \theta}{\cot \theta}\]

\[\Rightarrow \frac{\cos \theta \cdot \sin \theta}{\frac{\cos \theta}{\sin \theta}}\]

\[\Rightarrow \frac{\cos \theta \cdot \sin \theta \cdot \sin \theta}{\cos \theta}\]

\[\Rightarrow \sin^2 \theta.\]