Introduction to Trigonometry

Board: CBSE | Class: Class 10

Comprehensive study notes for Introduction to Trigonometry by Ajay Yadav (Math King of Katargam). Master every concept with clear explanations, solved examples, and practice problems.

Key Concepts

Trigonometric Ratios

In a right triangle ABC with ∠B = 90°: sinθ = opposite/hypotenuse, cosθ = adjacent/hypotenuse, tanθ = opposite/adjacent. Also: cosecθ = 1/sinθ, secθ = 1/cosθ, cotθ = 1/tanθ.

Trigonometric Table

Common angles: 0°, 30°, 45°, 60°, 90°. Mnemonic: sin values are √0/2, √1/2, √2/2, √3/2, √4/2 (0, 1/2, 1/√2, √3/2, 1). cos is sin in reverse order.

Complementary Angles

sin(90° – θ) = cosθ, cos(90° – θ) = sinθ, tan(90° – θ) = cotθ, cot(90° – θ) = tanθ, sec(90° – θ) = cosecθ, cosec(90° – θ) = secθ.

Trigonometric Identities

Identity 1: sin²θ + cos²θ = 1. Identity 2: 1 + tan²θ = sec²θ. Identity 3: 1 + cot²θ = cosec²θ.

Signs in Quadrants

Quadrant I (0-90°): All positive. Quadrant II (90-180°): sin & cosec positive. Quadrant III (180-270°): tan & cot positive. Quadrant IV (270-360°): cos & sec positive.

Important Formulas

sinθ	opposite/hypotenuse
cosθ	adjacent/hypotenuse
tanθ	opposite/adjacent = sinθ/cosθ
Identity 1	sin²θ + cos²θ = 1
Identity 2	1 + tan²θ = sec²θ
Identity 3	1 + cot²θ = cosec²θ

Solved Examples

Example 1: Given sinθ = 3/5, find cosθ and tanθ.

Solution: cosθ = √1 – 9/25 = √16/25 = 4/5. tanθ = sinθ/cosθ = (3/5)/(4/5) = 3/4.

Example 2: Evaluate: sin60°cos30° + cos60°sin30°

Solution: (√3/2)(√3/2) + (1/2)(1/2) = 3/4 + 1/4 = 1.

Example 3: Prove: (1 – sinθ)(1 + sinθ) = cos²θ

Solution: LHS = 1 – sin²θ = cos²θ = RHS. Hence proved.

Practice Questions

1. Find sinθ and tanθ if cosθ = 12/13.
2. Evaluate: tan45°cos30° – cot45°sin60°
3. Prove: sec²θ + cosec²θ = sec²θcosec²θ
4. If tanA = 3/4, find all other trigonometric ratios.
5. Show that (sinA + cosA)² = 1 + 2sinAcosA.

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