Q.If \(0^\circ < \theta < 90^\circ\), then \(\sin\theta < \sin^2\theta\).

The statement is **false**.
If \(0^\circ < \theta < 90^\circ\), then \(\sin\theta < 1\) and \(\sin\theta \ne 0\) \[ \therefore \sin\theta \times \sin\theta < 1 \times \sin\theta \quad [Multiplying both sides by \(\sin\theta\)] \Rightarrow \sin^2\theta < \sin\theta \Rightarrow \sin\theta > \sin^2\theta \]
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