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By Shailendra Singh
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Updated on 12 Nov 2025, 14:39 IST
Trigonometry is a fundamental branch of mathematics that studies the relationships between the angles and sides of triangles. The word originates from three Greek roots: 'tri' (three), 'gonia' (angle), and 'metron' (measure). Essentially, trigonometry explores how angles and side lengths interact within triangular shapes.
Trigonometry has widespread applications across multiple disciplines:
Even with today's advanced technology, engineering and architectural calculations depend heavily on trigonometric concepts.
In a right-angled triangle (a triangle with one 90° angle), we identify three sides relative to a given acute angle:
In triangle ABC with ∠B = 90°:
For angle A:
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For angle C:
Point: The hypotenuse remains constant, but the base and perpendicular change depending on which angle you're considering.
Trigonometric ratios express relationships between the sides of a right triangle relative to an acute angle. There are six fundamental trigonometric ratios.
For an acute angle θ in a right triangle:
| Ratio | Formula | Description |
| Sine (sin θ) | Perpendicular / Hypotenuse | Ratio of opposite side to hypotenuse |
| Cosine (cos θ) | Base / Hypotenuse | Ratio of adjacent side to hypotenuse |
| Tangent (tan θ) | Perpendicular / Base | Ratio of opposite to adjacent side |
| Cosecant (cosec θ) | Hypotenuse / Perpendicular | Reciprocal of sine |
| Secant (sec θ) | Hypotenuse / Base | Reciprocal of cosine |
| Cotangent (cot θ) | Base / Perpendicular | Reciprocal of tangent |
A popular mnemonic for remembering the primary three ratios:
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Three fundamental reciprocal relationships connect the trigonometric ratios:
| Relationship | Formula |
| Sine and Cosecant | sin θ × cosec θ = 1 |
| Cosine and Secant | cos θ × sec θ = 1 |
| Tangent and Cotangent | tan θ × cot θ = 1 |
These relationships express tangent and cotangent in terms of sine and cosine:
| Ratio | Formula |
| Tangent | tan θ = sin θ / cos θ |
| Cotangent | cot θ = cos θ / sin θ |
For angle θ in right triangle:
Therefore: tan θ = (P/H) / (B/H) = P/B = sin θ / cos θ
The trigonometric table provides exact values for commonly used angles. Memorizing these values is essential for quick problem-solving.
| Angle | 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 |
| cot θ | Not defined | √3 | 1 | 1/√3 | 0 |
| sec θ | 1 | 2/√3 | √2 | 2 | Not defined |
| cosec θ | Not defined | 2 | √2 | 2/√3 | 1 |
For 45°: Consider an isosceles right triangle with equal sides = a
For 30° and 60°: Consider an equilateral triangle with side 2a, divided in half
Trigonometric identities are equations involving trigonometric functions that are true for all values of the variables (within the domain).
| Identity | Valid Range |
| sin²θ + cos²θ = 1 | 0° ≤ θ ≤ 90° |
| 1 + tan²θ = sec²θ | 0° ≤ θ < 90° |
| 1 + cot²θ = cosec²θ | 0° < θ ≤ 90° |
In right triangle ABC (right-angled at B):
By Pythagoras theorem: AB² + BC² = AC²
Dividing throughout by AC²:
(AB/AC)² + (BC/AC)² = 1
cos²θ + sin²θ = 1
From the fundamental identities, we can derive:
| Derived Form | From Identity |
| sin²θ = 1 - cos²θ | sin²θ + cos²θ = 1 |
| cos²θ = 1 - sin²θ | sin²θ + cos²θ = 1 |
| tan²θ = sec²θ - 1 | 1 + tan²θ = sec²θ |
| cot²θ = cosec²θ - 1 | 1 + cot²θ = cosec²θ |
Two angles are complementary if their sum equals 90°. Trigonometric ratios of complementary angles have special relationships.
| Original | Complementary |
| sin(90° - θ) = cos θ | cos(90° - θ) = sin θ |
| tan(90° - θ) = cot θ | cot(90° - θ) = tan θ |
| sec(90° - θ) = cosec θ | cosec(90° - θ) = sec θ |
In right triangle ABC with ∠B = 90° and ∠A = θ:
For angle A: sin A = BC/AC
For angle C: cos C = BC/AC
Therefore: sin A = cos(90° - A)
These relationships allow us to:
Example: Evaluate sin 36° / cos 54°
Solution: sin 36° / cos 54° = sin 36° / cos(90° - 36°) = sin 36° / sin 36° = 1
| Category | Formula | Notes |
| Basic Ratios | sin θ = P/H, cos θ = B/H, tan θ = P/B | P=Perpendicular, B=Base, H=Hypotenuse |
| Reciprocal | cosec θ = 1/sin θ, sec θ = 1/cos θ, cot θ = 1/tan θ | Reciprocal relationships |
| Quotient | tan θ = sin θ/cos θ, cot θ = cos θ/sin θ | Express in terms of sin and cos |
| Pythagorean #1 | sin²θ + cos²θ = 1 | Most fundamental identity |
| Pythagorean #2 | 1 + tan²θ = sec²θ | Valid for 0° ≤ θ < 90° |
| Pythagorean #3 | 1 + cot²θ = cosec²θ | Valid for 0° < θ ≤ 90° |
| Complementary | sin(90°-θ) = cos θ, tan(90°-θ) = cot θ | Co-function identities |
Transformation formulas:
Modern calculators have dedicated trigonometric function buttons. Here's how to use them effectively:
Example: Find sin 37°
| Mode | Use When |
| DEG (Degrees) | Working with angles in degrees (most CBSE problems) |
| RAD (Radians) | Advanced mathematics |
| GRAD (Gradians) | Rarely used in school mathematics |
Step 1: Draw and Label
Step 2: Identify the Appropriate Ratio
Step 3: Set Up the Equation
Step 4: Solve
Step 5: Verify
Problem 1: In triangle ABC, right-angled at B, if tan A = 3/4, find all other trigonometric ratios of angle A.
Solution:
Given: tan A = 3/4 = Perpendicular/Base
Let BC = 3k, AB = 4k (where k > 0)
By Pythagoras theorem: AC² = AB² + BC² = (4k)² + (3k)² = 16k² + 9k² = 25k² AC = 5k
Now calculate:
Problem 2: If sin θ = 1/2, verify that 2 sin θ cos θ = 2 tan θ/(1 + tan²θ)
Solution:
Given: sin θ = 1/2
Let BC = k, AC = 2k
By Pythagoras: AB = √(4k² - k²) = √(3k²) = k√3
Therefore:
LHS: 2 sin θ cos θ = 2 × (1/2) × (√3/2) = √3/2
RHS: 2 tan θ/(1 + tan²θ) = 2(1/√3) / (1 + 1/3) = (2/√3) / (4/3) = (2/√3) × (3/4) = √3/2
LHS = RHS (Verified)
Problem 3: In triangle PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Find sin P, cos P, and tan P.
Solution:
Given: PR + QR = 25, PQ = 5
Let PR = x, then QR = 25 - x
By Pythagoras theorem: PR² = PQ² + QR² x² = 25 + (25-x)² x² = 25 + 625 - 50x + x² 50x = 650 x = 13
Therefore: PR = 13 cm, QR = 12 cm
Trigonometric ratios:
Application: Determining distances and heights without direct measurement
Example: Finding the height of a building
Practical Use:
Application: Structural design and force analysis
Uses in Engineering:
Example: A bridge support forms a 60° angle with the ground. If the horizontal distance is 10m, the support length = 10/cos 60° = 20m
Application: Determining position and direction
Marine Navigation:
Aviation:
Application: Measuring cosmic distances
Uses:
Application: Understanding periodic phenomena
Topics:
For any triangle ABC (not necessarily right-angled):
a/sin A = b/sin B = c/sin C = 2R
Where:
When to Use: When you know:
Example: In triangle ABC, A = 78°, C = 18°, AB = 26 cm. Find BC.
Solution: B = 180° - 78° - 18° = 84°
Using sine rule: BC/sin A = AB/sin C BC/sin 78° = 26/sin 18° BC = 26 × sin 78° / sin 18° ≈ 82.3 cm
For any triangle ABC:
a² = b² + c² - 2bc cos A
Similarly:
When to Use: When you know:
Example: In triangle ABC, AB = 4 cm, BC = 4√2 cm, ∠B = 45°. Find AC.
Solution: Using cosine rule: AC² = AB² + BC² - 2(AB)(BC)cos B AC² = 16 + 32 - 2(4)(4√2)cos 45° AC² = 48 - 32(1/√2) = 48 - 32/√2 = 16 AC = 4 cm
Formula: Area = (1/2) ab sin C
Where a and b are two sides, and C is the included angle.
Alternative: Area = abc/(4R), where R is circumradius
Bearing represents direction measured clockwise from North.
Example: If the bearing of point P from Q is 65°, it means P is 65° clockwise from North when standing at Q.
If bearing of A from B = θ, then bearing of B from A = θ + 180° (or θ - 180°)
Problem 1: If cos A = 7/25, find sin A and tan A.
Solution:
Given: cos A = 7/25 = Base/Hypotenuse
Let Base = 7k, Hypotenuse = 25k
Perpendicular = √(625k² - 49k²) = √(576k²) = 24k
sin A = 24k/25k = 24/25 tan A = 24k/7k = 24/7
Problem 2: Evaluate: (sin 30° + cos 60°)²
Solution:
sin 30° = 1/2, cos 60° = 1/2
(sin 30° + cos 60°)² = (1/2 + 1/2)² = 1² = 1
Problem 3: Prove that: (1 + tan²θ)/(1 + cot²θ) = tan²θ
Solution:
LHS = (1 + tan²θ)/(1 + cot²θ) = sec²θ / cosec²θ = (1/cos²θ) / (1/sin²θ) = sin²θ/cos²θ = tan²θ = RHS
Problem 4: If sec θ + tan θ = p, prove that sin θ = (p² - 1)/(p² + 1)
Solution:
Given: sec θ + tan θ = p ... (1)
We know: sec²θ - tan²θ = 1 (sec θ - tan θ)(sec θ + tan θ) = 1 sec θ - tan θ = 1/p ... (2)
Adding (1) and (2): 2 sec θ = p + 1/p = (p² + 1)/p sec θ = (p² + 1)/(2p)
Subtracting (2) from (1): 2 tan θ = p - 1/p = (p² - 1)/p tan θ = (p² - 1)/(2p)
sin θ = tan θ/sec θ = [(p² - 1)/(2p)] / [(p² + 1)/(2p)] = (p² - 1)/(p² + 1)
Problem 5: A ladder 15m long reaches a window which is 9m above the ground on one side of the street. Keeping its foot at the same point, the ladder is turned to the other side of the street to reach a window 12m high. Find the width of the street.
Solution:
First position: Let distance from wall = x₁ x₁² + 9² = 15² x₁² = 225 - 81 = 144 x₁ = 12m
Second position: Let distance from wall = x₂ x₂² + 12² = 15² x₂² = 225 - 144 = 81 x₂ = 9m
Width of street = x₁ + x₂ = 12 + 9 = 21m
Problem 6: From the top of a 7m high building, the angle of elevation of the top of a tower is 60° and the angle of depression of its foot is 45°. Find the height of the tower.
Solution:
Let height of tower = H Distance between building and tower = d
From angle of depression (45°): tan 45° = 7/d 1 = 7/d d = 7m
From angle of elevation (60°): tan 60° = (H - 7)/d √3 = (H - 7)/7 H - 7 = 7√3 H = 7(1 + √3) = 7 + 7√3 ≈ 7 + 12.12 = 19.12m
Trigonometry forms a critical foundation for higher mathematics and numerous real-world applications. By mastering the fundamental ratios, memorizing key values, understanding identities, and practicing diverse problems, students can excel in this topic.
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Start with the more complex side, use fundamental identities to simplify, and work toward the simpler side. Never assume what you're trying to prove.
tan θ = sin θ/cos θ. At 90°, cos 90° = 0, making tan 90° = 1/0, which is undefined (division by zero).
Memorize the standard table (0°, 30°, 45°, 60°, 90°). Use complementary angle relationships for values like sin 75° = cos 15°.
There's no difference both notations mean the same thing. sin²θ is just shorthand.
For Class 10 CBSE, always use degrees unless specifically stated otherwise. Radians are introduced in Class 11.
Trigonometry is the branch of mathematics that deals with the relationships between the sides and angles of triangles. The word comes from Greek roots meaning "triangle measurement." It's crucial for many real-world applications including astronomy, navigation, engineering, surveying, and geography. For instance, astronomers historically used trigonometry to calculate distances to stars and planets, and engineers still rely on trigonometric concepts for advanced technology and construction projects.
The six fundamental trigonometric ratios are:
These ratios depend only on the angle, not on the size of the triangle.
Here are the key values you should memorize:
For 0°: sin 0° = 0, cos 0° = 1, tan 0° = 0
For 30°: sin 30° = 1/2, cos 30° = √3/2, tan 30° = 1/√3
For 45°: sin 45° = 1/√2, cos 45° = 1/√2, tan 45° = 1
For 60°: sin 60° = √3/2, cos 60° = 1/2, tan 60° = √3
For 90°: sin 90° = 1, cos 90° = 0, tan 90° = undefined
For complementary angles, these important relationships hold:
These formulas help simplify many trigonometric problems, especially when angles are expressed as complements.