Q1. [1] § 8.3 Trigonometric Ratios of Some Specific Angles
If $\sin\theta = \cos\theta$, $(0° < \theta < 90°)$, then value of $(\sec\theta \cdot \sin\theta)$ is
- (A) $\frac{1}{\sqrt{2}}$
- (B) $\sqrt{2}$
- (C) $1$
- (D) $0$
Previously asked in CBSE board exam
2024 30/2/1 Q7
Generated by claude-sonnet-4-6 · 2026-06-15 10:22 · grounding rag
Model Answer
(B) $\sqrt{2}$
Since $\sin\theta = \cos\theta$ and $0° < \theta < 90°$, we get $\theta = 45°$.
$$\sec 45° \cdot \sin 45° = \sqrt{2} \times \frac{1}{\sqrt{2}} = 1$$
Wait — recalculating: $\sec\theta \cdot \sin\theta = \dfrac{1}{\cos\theta} \cdot \sin\theta = \tan\theta = \tan 45° = \mathbf{1}$.
(C) 1
Source: Chapter 8, Section 8.3 (Trigonometric Ratios of 45°)
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Explanation
- $\sin\theta = \cos\theta$ in the first quadrant only when $\theta = 45°$.
- $\sec\theta \cdot \sin\theta = \dfrac{\sin\theta}{\cos\theta} = \tan\theta$; at $45°$, $\tan 45° = 1$.
- A common trap is to multiply $\sqrt{2} \times \dfrac{1}{\sqrt{2}}$ carefully — it gives 1, not $\sqrt{2}$. Always simplify the product, not just individual values.
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