📚 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
If $4\cot^2 45° - \sec^2 60° + \sin^2 60° + p = \frac{3}{4}$, then find the value of $p$.
Previously asked in CBSE board exam
2023 30/2/1 Q22(a) (OR-1)
Generated by claude-sonnet-4-6 · 2026-06-15 10:23 · grounding rag
Model Answer

Known values: $\cot 45° = 1$, $\sec 60° = 2$, $\sin 60° = \dfrac{\sqrt{3}}{2}$

Substituting in the equation:

$$4(1)^2 - (2)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 + p = \frac{3}{4}$$

$$4 - 4 + \frac{3}{4} + p = \frac{3}{4}$$

$$\frac{3}{4} + p = \frac{3}{4}$$

$$\boxed{p = 0}$$

Source: Introduction to Trigonometry, Section 8.3

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Explanation
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