Introduction to Trigonometry: CBSE Class 10 Maths

Understanding Trigonometric Ratios

Sine (sin)

In a right-angled triangle, the sine of an angle (θ) is the ratio of the length of the side opposite the angle to the length of the hypotenuse. Formula: sin θ = Opposite / Hypotenuse.

Cosine (cos)

The cosine of an angle (θ) is the ratio of the length of the side adjacent to the angle to the length of the hypotenuse. Formula: cos θ = Adjacent / Hypotenuse.

Tangent (tan)

The tangent of an angle (θ) is the ratio of the length of the side opposite the angle to the length of the side adjacent to the angle. Formula: tan θ = Opposite / Adjacent. Also, tan θ = sin θ / cos θ.

Cosecant (cosec)

The reciprocal of sine. The ratio of the length of the hypotenuse to the length of the side opposite the angle (θ). Formula: cosec θ = 1 / sin θ = Hypotenuse / Opposite.

Secant (sec)

The reciprocal of cosine. The ratio of the length of the hypotenuse to the length of the side adjacent to the angle (θ). Formula: sec θ = 1 / cos θ = Hypotenuse / Adjacent.

Cotangent (cot)

The reciprocal of tangent. The ratio of the length of the side adjacent to the angle to the length of the side opposite the angle (θ). Formula: cot θ = 1 / tan θ = Adjacent / Opposite.

How to Calculate Trigonometric Ratios: A Step-by-Step Guide

1. Step 1: Identify the Triangle and Angle — Ensure you are working with a right-angled triangle. Identify the specific angle (often denoted by θ, A, B, etc.) for which you need to find the ratios.
2. Step 2: Label the Sides — Relative to your chosen angle (θ), label the three sides: - Hypotenuse (H): The longest side, always opposite the right angle. - Opposite (O): The side directly facing your angle θ. - Adjacent (A): The side next to your angle θ that is not the hypotenuse.
3. Step 3: Use Pythagoras Theorem (if needed) — If the length of one side is unknown, use the Pythagoras theorem (Opposite² + Adjacent² = Hypotenuse²) to find it.
4. Step 4: Apply the Ratio Formulas — Use the definitions of the ratios to calculate their values. A helpful mnemonic to remember the primary ratios is SOH CAH TOA: - Sin = Opposite / Hypotenuse - Cos = Adjacent / Hypotenuse - Tan = Opposite / Adjacent

Worked Example: Finding Ratios

• Question: In a triangle ABC, right-angled at B, if AB = 12 cm and BC = 5 cm, find the value of sin A and tan C. Solution: Step 1: Find the length of the hypotenuse AC. Using Pythagoras theorem in ΔABC: AC² = AB² + BC² AC² = 12² + 5² AC² = 144 + 25 AC² = 169 AC = √169 = 13 cm. Step 2: Find sin A. For angle A: - Opposite side = BC = 5 cm - Adjacent side = AB = 12 cm - Hypotenuse = AC = 13 cm Using the formula sin A = Opposite / Hypotenuse: sin A = 5 / 13. Step 3: Find tan C. For angle C: - Opposite side = AB = 12 cm - Adjacent side = BC = 5 cm - Hypotenuse = AC = 13 cm Using the formula tan C = Opposite / Adjacent: tan C = 12 / 5. Final Answers: sin A = 5/13 and tan C = 12/5.

The Power of Trigonometric Identities

A trigonometric identity is an equation involving trigonometric ratios that is true for all values of the angles for which the ratios are defined. These identities are the fundamental tools for simplifying expressions and proving more complex trigonometric statements. The most important identity, often called the Pythagorean Identity, is sin²θ + cos²θ = 1. This can be easily derived from a right-angled triangle with sides a, b, and hypotenuse c. By Pythagoras theorem, a² + b² = c². Dividing the entire equation by c² gives (a/c)² + (b/c)² = 1. Since sin θ = a/c and cos θ = b/c, we get sin²θ + cos²θ = 1. From this single identity, we can derive two others:

1. Dividing by cos²θ gives: 1 + tan²θ = sec²θ
2. Dividing by sin²θ gives: cot²θ + 1 = cosec²θ

Mastering these three identities is non-negotiable for your board exams. They allow you to transform and manipulate trigonometric expressions to solve problems that would otherwise be very difficult.

Exam Tips & Common Mistakes

Trigonometry can be a high-scoring topic if you avoid these common errors:

• Confusing Sides: Always re-identify the Opposite and Adjacent sides when the reference angle changes within the same problem (like finding ratios for angle A and then angle C). What is Opposite for A is Adjacent for C.
• Incorrect Notation: Remember that sin A is not sin multiplied by A. It is a function 'sine' of angle 'A'. So, sin(A+B) is NOT equal to sin A + sin B.
• Squaring Errors: The notation sin²A means (sin A)². Don't mistake it for sin(A²). Always find the value of the ratio first, then square it.
• Forgetting Identities: The three Pythagorean identities (sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ, 1 + cot²θ = cosec²θ) are the backbone of 'Prove That' questions. Write them down and memorize them perfectly.
• Exam Tip: Create a table of trigonometric ratios for standard angles (0°, 30°, 45°, 60°, 90°) and revise it daily. Many questions in the exam are based on directly substituting these values.

Practice Questions with Solutions

• Q: If 15 cot A = 8, find the values of sin A and sec A. A: Step 1: From the given information, we have cot A = 8/15. We know that cot A = Adjacent / Opposite. So, let Adjacent side = 8k and Opposite side = 15k, where k is a positive constant. Step 2: Find the Hypotenuse using Pythagoras theorem. Hypotenuse² = Opposite² + Adjacent² = (15k)² + (8k)² = 225k² + 64k² = 289k². So, Hypotenuse = √(289k²) = 17k. Step 3: Calculate sin A and sec A. sin A = Opposite / Hypotenuse = 15k / 17k = 15/17. sec A = Hypotenuse / Adjacent = 17k / 8k = 17/8. Final answer: sin A = 15/17 and sec A = 17/8.
• Q: Evaluate: 2 tan²45° + cos²30° – sin²60° A: Step 1: Recall the standard values of the trigonometric ratios. We know that tan 45° = 1, cos 30° = √3/2, and sin 60° = √3/2. Step 2: Substitute these values into the given expression. Expression = 2 (tan 45°)² + (cos 30°)² – (sin 60°)² = 2(1)² + (√3/2)² – (√3/2)² Step 3: Simplify the expression. = 2(1) + 3/4 – 3/4 = 2 + 0 = 2. Final answer: 2
• Q: Prove the identity: (cosec θ - cot θ)² = (1 - cos θ) / (1 + cos θ) A: Step 1: Start with the Left Hand Side (LHS) and express cosec θ and cot θ in terms of sin θ and cos θ. LHS = (cosec θ - cot θ)² = (1/sin θ - cos θ/sin θ)² Step 2: Simplify the expression inside the bracket. LHS = ((1 - cos θ) / sin θ)² = (1 - cos θ)² / sin²θ Step 3: Use the identity sin²θ = 1 - cos²θ in the denominator. LHS = (1 - cos θ)² / (1 - cos²θ) Step 4: Factor the denominator using the algebraic identity a² - b² = (a - b)(a + b). LHS = (1 - cos θ)(1 - cos θ) / ((1 - cos θ)(1 + cos θ)) Step 5: Cancel the common term (1 - cos θ) from the numerator and denominator. LHS = (1 - cos θ) / (1 + cos θ) = RHS. Final answer: Hence, the identity is proved.
• Q: If tan (A + B) = √3 and tan (A – B) = 1/√3; 0° < A + B ≤ 90°; A > B, find A and B. A: Step 1: Use the given equations and standard angle values to find expressions for (A + B) and (A - B). We have tan (A + B) = √3. Since tan 60° = √3, we get A + B = 60° ---(1) We have tan (A – B) = 1/√3. Since tan 30° = 1/√3, we get A – B = 30° ---(2) Step 2: We now have a pair of linear equations in two variables, A and B. Add equation (1) and (2). (A + B) + (A - B) = 60° + 30° 2A = 90° A = 45° Step 3: Substitute the value of A in equation (1). 45° + B = 60° B = 60° - 45° B = 15° Final answer: A = 45° and B = 15°.

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