📚 CBSE Grade-10 Study Guide
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Mathematics — Standard (041) — CBSE Class 10 board question

Q1. [2] § 8.3 Trigonometric Ratios of Some Specific Angles
Evaluate the following : $\dfrac{3\sin 30° - 4\sin^3 30°}{2\sin^2 50° + 2\cos^2 50°}$
Previously asked in CBSE board exam
2025 30/4/1 Q24(B)
Generated by claude-sonnet-4-6 · 2026-06-15 10:25 · grounding rag
Model Answer

Numerator: $3\sin 30° - 4\sin^3 30° = 3\times\dfrac{1}{2} - 4\times\left(\dfrac{1}{2}\right)^3 = \dfrac{3}{2} - 4\times\dfrac{1}{8} = \dfrac{3}{2} - \dfrac{1}{2} = 1$

Denominator: $2\sin^2 50° + 2\cos^2 50° = 2(\sin^2 50° + \cos^2 50°) = 2\times 1 = 2$

$$\therefore \quad \frac{3\sin 30° - 4\sin^3 30°}{2\sin^2 50° + 2\cos^2 50°} = \frac{1}{2}$$

Source: Chapter 8, Section 8.3 (Trigonometric Ratios of Specific Angles)

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Explanation
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