CBSE Class 10 Mathematics: NCERT Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry (2025–2026)
Mastering CBSE Class 10 Mathematics begins with a solid grasp of NCERT Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry. For students preparing for the 2025–2026 board exam 10, Chapter 8 serves as the foundation for higher-level calculus, physics vectors, and coordinate geometry. This comprehensive guide breaks down trigonometric ratios, standard angle values, and fundamental identities with step-by-step derivations and official-pattern practice solutions aligned with the latest CBSE syllabus.
Key Concepts: Foundations of Trigonometry
The word trigonometry is derived from the Greek words tri (meaning three), gon (meaning sides), and metron (meaning measure). In CBSE Mathematics, trigonometry investigates the precise geometric and algebraic relationships between the side lengths and acute angles of a right-angled triangle.
1. The Six Trigonometric Ratios
Consider a right-angled triangle ΔABC, right-angled at ∠B. With respect to the acute angle ∠A (or reference angle θ):
- Opposite Side (Perpendicular, P): The side opposite to ∠A (BC).
- Adjacent Side (Base, B): The side adjacent to ∠A (AB).
- Hypotenuse (H): The longest side opposite to the 90° angle (AC).
The six fundamental trigonometric ratios are defined as follows:
| Primary Ratio | Formula | Reciprocal Ratio | Formula |
|---|---|---|---|
| Sine (∠A) = sin A | Opposite / Hypotenuse = BC / AC | Cosecant (∠A) = cosec A | Hypotenuse / Opposite = 1 / sin A = AC / BC |
| Cosine (∠A) = cos A | Adjacent / Hypotenuse = AB / AC | Secant (∠A) = sec A | Hypotenuse / Adjacent = 1 / cos A = AC / AB |
| Tangent (∠A) = tan A | Opposite / Adjacent = BC / AB | Cotangent (∠A) = cot A | Adjacent / Opposite = 1 / tan A = AB / BC |
Crucial Quotient Relations:
- tan A = sin A / cos A
- cot A = cos A / sin A
2. Trigonometric Ratios of Specific Standard Angles
For the CBSE Class 10 board examination, committing standard values (0°, 30°, 45°, 60°, and 90°) to memory is essential for fast and accurate computation:
| ∠A / Ratio | 0° | 30° | 45° | 60° | 90° |
|---|---|---|---|---|---|
| sin A | 0 | 1/2 | 1/√2 | √3/2 | 1 |
| cos A | 1 | √3/2 | 1/√2 | 1/2 | 0 |
| tan A | 0 | 1/√3 | 1 | √3 | Not defined |
| cosec A | Not defined | 2 | √2 | 2/√3 | 1 |
| sec A | 1 | 2/√3 | √2 | 2 | Not defined |
| cot A | Not defined | √3 | 1 | 1/√3 | 0 |
3. Fundamental Trigonometric Identities
An equation involving trigonometric ratios of an angle is called a trigonometric identity if it is true for all values of the angles involved. For all acute angles θ (0° ≤ θ ≤ 90°):
- sin²θ + cos²θ = 1 ⇒ sin²θ = 1 − cos²θ and cos²θ = 1 − sin²θ
- 1 + tan²θ = sec²θ ⇒ sec²θ − tan²θ = 1 (valid for 0° ≤ θ < 90°)
- 1 + cot²θ = cosec²θ ⇒ cosec²θ − cot²θ = 1 (valid for 0° < θ ≤ 90°)
Important CBSE Questions with Complete Step-by-Step Solutions
Question 1 (Short Answer Type): Ratios from Right Triangle Relations
Problem: Given 15 cot A = 8, find the values of sin A and sec A.
Solution:
- We are given 15 cot A = 8 ⇒ cot A = 8 / 15.
- In right triangle ΔABC right-angled at B, cot A = Adjacent Side / Opposite Side = AB / BC = 8 / 15.
- Let AB = 8k and BC = 15k, where k is a positive real number.
- By Pythagoras Theorem in ΔABC:
AC² = AB² + BC²
AC² = (8k)² + (15k)² = 64k² + 225k² = 289k²
AC = √(289k²) = 17k - Now, determine the required ratios:
sin A = BC / AC = 15k / 17k = 15/17
sec A = AC / AB = 17k / 8k = 17/8
Question 2 (Evaluation Type): Standard Angle Substitutions
Problem: Evaluate the value of:
(5 cos² 60° + 4 sec² 30° − tan² 45°) / (sin² 30° + cos² 30°)
Solution:
- Substitute the standard values into the expression:
- cos 60° = 1/2
- sec 30° = 2/√3
- tan 45° = 1
- sin 30° = 1/2
- cos 30° = √3/2
- Evaluate the Numerator:
Numerator = 5(1/2)² + 4(2/√3)² − (1)²
= 5(1/4) + 4(4/3) − 1
= 5/4 + 16/3 − 1
= (15 + 64 − 12) / 12 = 67/12 - Evaluate the Denominator using identity sin²θ + cos²θ = 1:
Denominator = (1/2)² + (√3/2)² = 1/4 + 3/4 = 4/4 = 1 - Calculate final answer:
Result = (67/12) / 1 = 67/12.
Question 3 (Algebraic System Type): Finding Acute Angle Measures
Problem: If tan (A + B) = √3 and tan (A − B) = 1/√3; 0° < A + B ≤ 90°; A > B, find the values of ∠A and ∠B.
Solution:
- From the first given equation:
tan (A + B) = √3
Since tan 60° = √3, we have:
A + B = 60° — (Equation 1) - From the second given equation:
tan (A − B) = 1/√3
Since tan 30° = 1/√3, we have:
A − B = 30° — (Equation 2) - Add Equation 1 and Equation 2:
(A + B) + (A − B) = 60° + 30°
2A = 90° ⇒ A = 45° - Substitute A = 45° into Equation 1:
45° + B = 60° ⇒ B = 15° - Final Answer: ∠A = 45° and ∠B = 15° (Satisfies 0° < 60° ≤ 90° and 45° > 15°).
Question 4 (Long Answer Proof Type): Proving Trigonometric Identity
Problem: Prove that: (sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A
Solution:
- Expand the Left Hand Side (LHS) using the algebraic identity (a + b)² = a² + 2ab + b²:
LHS = (sin² A + 2 sin A cosec A + cosec² A) + (cos² A + 2 cos A sec A + sec² A) - Apply reciprocal relations (sin A · cosec A = 1 and cos A · sec A = 1):
LHS = sin² A + 2(1) + cosec² A + cos² A + 2(1) + sec² A
LHS = (sin² A + cos² A) + 4 + cosec² A + sec² A - Substitute standard identities (sin² A + cos² A = 1, cosec² A = 1 + cot² A, and sec² A = 1 + tan² A):
LHS = 1 + 4 + (1 + cot² A) + (1 + tan² A)
LHS = 7 + tan² A + cot² A - LHS = RHS. Hence Proved.
Question 5 (Standard CBSE Identity Proof): Factoring with Identity Substitution
Problem: Prove that: (cos A − sin A + 1) / (cos A + sin A − 1) = cosec A + cot A using the identity cosec² A = 1 + cot² A.
Solution:
- Divide both numerator and denominator of the LHS by sin A:
LHS = [(cos A / sin A) − (sin A / sin A) + (1 / sin A)] / [(cos A / sin A) + (sin A / sin A) − (1 / sin A)]
= (cot A − 1 + cosec A) / (cot A + 1 − cosec A)
= [(cosec A + cot A) − 1] / [cot A − cosec A + 1] - Replace the numeral '1' in the numerator with (cosec² A − cot² A):
Numerator = (cosec A + cot A) − (cosec² A − cot² A)
= (cosec A + cot A) − [(cosec A − cot A)(cosec A + cot A)] - Factor out the common term
(cosec A + cot A)from the numerator:
= (cosec A + cot A) [1 − (cosec A − cot A)]
= (cosec A + cot A) (1 − cosec A + cot A) - Divide by the denominator
(1 − cosec A + cot A):
LHS = [(cosec A + cot A) (1 − cosec A + cot A)] / (1 − cosec A + cot A)
= cosec A + cot A - LHS = RHS. Hence Proved.
How to Prepare for This Topic for CBSE Class 10 Board Exams
To score full marks in Chapter 8 in your CBSE Class 10 examination, adopt these subject-specific strategies:
- Master Reference Angles: In right triangles, always verify which angle is the reference angle. The side opposite to ∠A is BC, whereas the side opposite to ∠C is AB. Confusing base and perpendicular is the #1 source of lost marks.
- Convert to Sine and Cosine in Proofs: When stuck on identity proof questions involving tan, cot, sec, or cosec, rewrite all ratios in terms of
sin θandcos θand simplify algebraically. - Memorize the Hand Trick / Ratio Table: Ensure you can reconstruct the 0° to 90° value chart on your scratch sheet within 60 seconds at the start of the examination.
- Write Clear Steps and State Identities: CBSE marking schemes award step-wise marks. Always mention the exact formula used (e.g., "Using identity: sin²θ + cos²θ = 1") alongside your working.
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