Question : Which of the following relations is correct for $0< \theta < 90^{\circ}$?
Option 1: $\sin \theta = \sin^{2}\theta$
Option 2: $\sin \theta < \sin^{2} \theta$
Option 3: $\sin \theta > \sin^{2}\theta$
Option 4: $\sin \theta > \operatorname{cosec}^{2}\theta$

Question

Question : The expression $\frac{\cos ^4 \theta-\sin ^4 \theta+2 \sin ^2 \theta+3}{(\operatorname{cosec} \theta+\cot \theta+1)(\operatorname{cosec} \theta-\cot \theta+1)-2}, 0^{\circ}<\theta<90^{\circ}$, is equal to:
Option 1: $\frac{1}{2} \sin \theta$
Option 2: $2 \sin \theta$
Option 3: $\sec \theta$
Option 4: $2\operatorname{cosec} \theta$

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Answer 1

Correct Answer: $\sin \theta > \sin^{2}\theta$

Solution : For $(0 < \theta < 90^{\circ})$, the correct relation is:
$\sin\theta > \sin^{2}\theta$
Here's why:
⇒ $\sin^{2}\theta$ is the square of $\sin\theta$ and since $\sin\theta$ is less than 1 for $0 < \theta < 90^0$, $\sin^{2}\theta$ will be less than $\sin\theta$.
⇒ The other relations are not correct because:
$\sin\theta =\sin^{2}\theta$ and $\sin\theta < \sin^{2}\theta$ are not true for the same reason mentioned above.
For $\sin\theta > \operatorname{cosec}^{2}\theta$, remember that $\operatorname{cosec}\theta = \frac{1}{\sin\theta}$. So, $\operatorname{cosec}^{2}\theta = \frac{1}{(\sin^{2}\theta)}$, which is always greater than $\sin\theta$ for $(0 < \theta < 90^{\circ})$ . So, this relation is also not correct.
Hence, the correct answer is $\sin\theta > \sin^{2}\theta$.