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📑 In This Chapter Guide (Table of Contents)
1. Trigonometric Ratios (T-Ratios) in a Right Triangle
For a right-angled triangle ΔABC with right angle at B and acute angle θ = ∠A:
sin θ = Opposite Side (Perpendicular) / Hypotenuse = P / Hcos θ = Adjacent Side (Base) / Hypotenuse = B / Htan θ = Opposite / Adjacent = P / B = sin θ / cos θcosec θ = 1 / sin θ = H / Psec θ = 1 / cos θ = H / Bcot θ = 1 / tan θ = B / P = cos θ / sin θ
2. Standard Trigonometric Values Table (0° to 90°)
| Ratio | 0° | 30° | 45° | 60° | 90° |
|---|---|---|---|---|---|
| sin θ | 0 | 1/2 | 1/√2 | √3/2 | 1 |
| cos θ | 1 | √3/2 | 1/√2 | 1/2 | 0 |
| tan θ | 0 | 1/√3 | 1 | √3 | Not Defined |
3. Fundamental Identities & Top Proving Strategies
The 3 Fundamental Identities:
sin²θ + cos²θ = 1 ⟹ sin²θ = 1 - cos²θ ⟹ cos²θ = 1 - sin²θ1 + tan²θ = sec²θ ⟹ sec²θ - tan²θ = 1 ⟹ (sec θ - tan θ)(sec θ + tan θ) = 11 + cot²θ = cosec²θ ⟹ cosec²θ - cot²θ = 1 ⟹ (cosec θ - cot θ)(cosec θ + cot θ) = 1
4 Rules of Thumb for Identity Proof Questions:
- Convert everything into terms of sin θ and cos θ when you are stuck.
- Take the Lowest Common Denominator (LCM) of fractional terms.
- Rationalize the numerator or denominator when terms like
(1 ± sin θ)or(1 ± cos θ)appear inside square roots. - Apply algebraic identities:
a² - b² = (a - b)(a + b),a³ + b³ = (a + b)(a² - ab + b²).
💡 Frequently Asked Questions (FAQ)
❓ Prove that (sin θ - 2sin³θ) / (2cos³θ - cos θ) = tan θ.
LHS = [sin θ(1 - 2sin²θ)] / [cos θ(2cos²θ - 1)]. Since 1 - 2sin²θ = (cos²θ + sin²θ) - 2sin²θ = cos²θ - sin²θ, and 2cos²θ - 1 = 2cos²θ - (cos²θ + sin²θ) = cos²θ - sin²θ. Both bracket terms cancel out completely! LHS = sin θ / cos θ = tan θ = RHS. (Hence Proved)
❓ If tan A = 4/3, find all other trigonometric ratios of angle A.
Given tan A = P/B = 4/3. Let P = 4k, B = 3k. Hypotenuse H = √(P² + B²) = √(16k² + 9k²) = 5k. Therefore: sin A = 4/5, cos A = 3/5, cosec A = 5/4, sec A = 5/3, cot A = 3/4.
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