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NCERT Class 10 Maths Chapter 8, Introduction to Trigonometry, explained simply
Introduction to Trigonometry is Chapter 8 of the NCERT Class 10 Mathematics textbook (Reprint 2026-27), printed pages 113 to 132. It teaches six ratios between the sides of a right triangle, their values for a few special angles, and three identities that connect them. This guide explains each idea in plain words, shows how the CBSE 2026-27 sample paper asks the chapter, walks through one item from each exercise, and ends with a self-check and a way to test an AI's answer.
Keep the textbook open next to it (chapter PDF). Our explanations and examples are our own; the textbook is the authority.
How much it counts in the board exam
CBSE's Class 10 Mathematics curriculum for 2026-27 gives Unit V, Trigonometry, 12 of the 80 theory marks (curriculum, PDF page 3). The unit has three parts (PDF page 6):
- Introduction to trigonometry: the ratios for an acute angle in a right triangle, which of them are defined at 0° and 90°, the values for 30°, 45° and 60°, and how the ratios relate to each other.
- Trigonometric identities: the proof and use of sin²A + cos²A = 1, with "only simple identities".
- Heights and distances: angles of elevation and depression, with no more than two right triangles and only 30°, 45° and 60°. That part is the next NCERT chapter, Some Applications of Trigonometry.
The 2026-27 sample paper (Mathematics Standard) asks trigonometry in five questions, which add up to exactly the unit's 12 marks:
| Question | Marks | What it asks | Chapter |
|---|---|---|---|
| Q8 | 1 | tan (A + B) = √3 and tan (A − B) = 1/√3: find A and B | this one |
| Q17 | 1 | A kite string of 52 m with tan θ = 12/5: find the height of the kite | this one (ratios) |
| Q23 | 2 | Given tan θ = 1/√5, find (cosec²θ − sec²θ)/(cosec²θ + sec²θ) | this one (identities) |
| Q31 | 3 | Find the error in a student's simplification, or prove an identity (internal choice) | this one (identities) |
| Q35 | 5 | A kite seen from two places, or two ships seen from a lighthouse (internal choice) | heights and distances |
So this chapter alone carries 7 marks of that paper, and the ratios you learn here are the tools for the other 5.
1. The six trigonometric ratios (section 8.2)
Take a triangle ABC with the right angle at B, and look at angle A. The side opposite A is BC, the side adjacent to A is AB, and the hypotenuse is AC. The six ratios of angle A are:
| Ratio | Definition | The reciprocal of |
|---|---|---|
| sin A | opposite / hypotenuse = BC/AC | cosec A |
| cos A | adjacent / hypotenuse = AB/AC | sec A |
| tan A | opposite / adjacent = BC/AB | cot A |
| cosec A | hypotenuse / opposite = AC/BC | sin A |
| sec A | hypotenuse / adjacent = AC/AB | cos A |
| cot A | adjacent / opposite = AB/BC | tan A |
Also, tan A = sin A / cos A and cot A = cos A / sin A.
Worked, our own. In triangle ABC, right-angled at B, AB = 8 cm and BC = 6 cm. By Pythagoras, AC = √(64 + 36) = 10 cm. So sin A = 6/10 = 3/5, cos A = 8/10 = 4/5, tan A = 6/8 = 3/4, and their reciprocals are cosec A = 5/3, sec A = 5/4 and cot A = 4/3.
Three things the book makes a point of (pages 116-117):
- The ratios depend only on the angle, not on the size of the triangle. Triangles with the same angles are similar, so their sides are in proportion and the ratios come out equal.
- sin A is one symbol, not "sin times A". sin² A means (sin A)². But sin⁻¹ A is not 1/sin A; it means something else, which comes in later classes.
- Where the word comes from. The book traces sine to Aryabhata's ardha-jya (half-chord) in the Aryabhatiyam, shortened to jya or jiva, which became sinus in Latin (page 116).
If one ratio is given, find the rest
Draw the triangle and give the two known sides a common multiplier k. Our own example: if sin A = 5/13, take the opposite side 5k and the hypotenuse 13k. The third side is √(169k² − 25k²) = 12k, so cos A = 12/13 and tan A = 5/12. Because the hypotenuse is the longest side, sin A and cos A can never be more than 1 (page 118).
The Q17 type. The sample paper's kite string is 52 m long and tan θ = 12/5. Take the height as 12k and the ground distance as 5k; the string is then 13k, so sin θ = 12/13. Height = 52 × 12/13 = 48 m, option (C) in the marking scheme.
2. The values for 0°, 30°, 45°, 60° and 90° (section 8.3)
These are the values from the book's Table 8.1 (page 125):
| 0° | 30° | 45° | 60° | 90° | |
|---|---|---|---|---|---|
| sin | 0 | 1/2 | 1/√2 | √3/2 | 1 |
| cos | 1 | √3/2 | 1/√2 | 1/2 | 0 |
| tan | 0 | 1/√3 | 1 | √3 | not defined |
| cosec | not defined | 2 | √2 | 2/√3 | 1 |
| sec | 1 | 2/√3 | √2 | 2 | not defined |
| cot | not defined | √3 | 1 | 1/√3 | 0 |
A way to remember the sine row (our own trick): write √0/2, √1/2, √2/2, √3/2, √4/2 for 0° to 90°. That gives 0, 1/2, 1/√2, √3/2 and 1. The cosine row is the same list backwards, and tan is sin ÷ cos. As the angle grows from 0° to 90°, sin goes up from 0 to 1 and cos comes down from 1 to 0.
Worked, our own (like the book's Example 6). In triangle ABC, right-angled at B, angle C = 60° and BC = 4 cm. Then tan C = AB/BC, so AB = 4 × tan 60° = 4√3 cm, and cos C = BC/AC, so AC = 4 ÷ cos 60° = 4 ÷ (1/2) = 8 cm.
The Q8 type. If sin (A + B) = 1 and cos (A − B) = √3/2, with A > B, then A + B = 90° and A − B = 30°. Adding and halving: A = 60° and B = 30°. The sample paper's Q8 works the same way with tan: A + B = 60°, A − B = 30°, so A = 45° and B = 15°.
3. Trigonometric identities (section 8.4)
An identity is true for every angle where both sides are defined. The book proves three, all from Pythagoras: in triangle ABC right-angled at B, AB² + BC² = AC², and you divide through by a different side each time (pages 128-129).
- Divide by AC²: sin²A + cos²A = 1, for 0° ≤ A ≤ 90°.
- Divide by AB²: 1 + tan²A = sec²A, for 0° ≤ A < 90°.
- Divide by BC²: cot²A + 1 = cosec²A, for 0° < A ≤ 90°.
The Q23 type, our own numbers. If tan θ = 1/√3, find (cosec²θ − sec²θ)/(cosec²θ + sec²θ). Then cot θ = √3, so cosec²θ = 1 + 3 = 4 and sec²θ = 1 + 1/3 = 4/3. The value is (4 − 4/3)/(4 + 4/3) = (8/3)/(16/3) = 1/2. In the marking scheme for the sample paper's Q23 (where the answer is 2/3), each squared ratio earns half a mark and the final value 1 mark.
How to prove an identity
- Work on one side (usually the more complicated one) until it becomes the other side. Do not move terms across the equals sign as if solving an equation.
- If you are stuck, write everything in sin and cos.
- Look for sin²A + cos²A, sec²A − tan²A or cosec²A − cot²A, each of which equals 1.
Our own: prove tan A + cot A = sec A cosec A. LHS = sin A/cos A + cos A/sin A = (sin²A + cos²A)/(sin A cos A) = 1/(sin A cos A) = sec A cosec A = RHS.
Our own: prove (1 − cos²A)(1 + cot²A) = 1. LHS = sin²A × cosec²A = sin²A × (1/sin²A) = 1 = RHS.
The Q31 type: find the error
The sample paper shows a student's working for cos θ/(1 − sin θ) + (1 − sin θ)/cos θ and asks where it goes wrong. The error is in step 2: the student replaced (1 − sin θ)² by cos²θ, as if (1 − sin θ)² were 1 − sin²θ. It is not: (1 − sin θ)² = 1 − 2 sin θ + sin²θ. Done correctly, the numerator becomes cos²θ + 1 − 2 sin θ + sin²θ = 2 − 2 sin θ = 2(1 − sin θ), which cancels to give 2/cos θ = 2 sec θ. The marking scheme gives 1 mark for naming the wrong step, 1 for the correct expansion, and half a mark for each of the last two lines.
4. Exercises 8.1, 8.2 and 8.3: what each asks, with one worked item
| Exercise | Printed page (PDF page) | What it asks |
|---|---|---|
| 8.1 | 121 (9) | 11 questions: find ratios from sides, find all ratios from one, and true-or-false about the ratios |
| 8.2 | 127 (15) | Evaluate expressions with the special angles, choose the right option, find A and B, and true-or-false |
| 8.3 | 131-132 (19-20) | Write ratios in terms of one ratio, choose the right option, and prove ten identities |
Exercise 8.1, Q1. In triangle ABC right-angled at B, AB = 24 cm and BC = 7 cm. The hypotenuse is AC = √(576 + 49) = √625 = 25 cm. So sin A = 7/25 and cos A = 24/25, while sin C = 24/25 and cos C = 7/25. The sine of one acute angle is the cosine of the other.
Exercise 8.1, Q11 (v): is sin θ = 4/3 possible? No. sin θ is opposite/hypotenuse, and the hypotenuse is the longest side, so sin θ is at most 1.
Exercise 8.2, Q3. tan (A + B) = √3 and tan (A − B) = 1/√3, with 0° < A + B ≤ 90° and A > B. So A + B = 60° and A − B = 30°, giving A = 45° and B = 15°. This is the same question as the sample paper's Q8: the textbook exercise is the board question.
Exercise 8.3, Q3 (i): 9 sec²A − 9 tan²A = ? Take 9 out: 9(sec²A − tan²A) = 9 × 1 = 9, option (B).
5. Five things that cost marks
- Treating sin A as a product. sin A is one symbol. "Cancelling" the sin in sin A / sin B to get A/B is wrong.
- Assuming sin (A + B) = sin A + sin B. Try A = B = 30°: sin 60° is about 0.866, but sin 30° + sin 30° = 1. Exercise 8.2 Q4 (i) asks exactly this.
- Squaring (1 − sin θ) wrongly. (1 − sin θ)² = 1 − 2 sin θ + sin²θ. Losing the middle term is the sample paper's Q31 mistake.
- Using a value that does not exist. tan 90°, sec 90°, cosec 0° and cot 0° are not defined. Write "not defined", not 0 or ∞.
- Working both sides of an identity as an equation. Start from one side and end at the other; each line should follow from the one before.
6. Self-check: 6 questions with answers
Try them on paper first. Answers are below.
- Q1. In triangle ABC right-angled at B, AB = 12 cm and BC = 5 cm. Find sin A, cos A and tan A.
- Q2. Evaluate sin 30° cos 60° + cos 30° sin 60°.
- Q3. Evaluate 2 tan²45° + cos²60° − sin²30°.
- Q4. If cos A = 3/5, find tan A and cosec A.
- Q5. Prove that (sec A − tan A)(sec A + tan A) = 1.
- Q6. A ramp 6 m long makes an angle of 30° with the ground. How high above the ground is its top end?
Answers:
- Q1: AC = √(144 + 25) = 13 cm, so sin A = 5/13, cos A = 12/13, tan A = 5/12.
- Q2: (1/2)(1/2) + (√3/2)(√3/2) = 1/4 + 3/4 = 1.
- Q3: 2 × 1 + 1/4 − 1/4 = 2.
- Q4: The sides are 3k (adjacent) and 5k (hypotenuse), so the opposite side is 4k. tan A = 4/3 and cosec A = 5/4.
- Q5: LHS = sec²A − tan²A = 1 (from 1 + tan²A = sec²A) = RHS.
- Q6: height = 6 × sin 30° = 6 × 1/2 = 3 m.
7. How to check an AI's answer for this chapter
- Put an angle in. To test a claimed identity, try A = 30° or 60° on both sides using the table. If the two sides give different numbers, the identity is wrong.
- Check the size. Any sin or cos bigger than 1, or any sec or cosec smaller than 1, is impossible.
- Check the method is in your book. The current chapter has four working sections: the ratios (8.2), the special angles (8.3), the identities (8.4) and the summary. Older guides, and AIs trained on them, also teach a section on the ratios of complementary angles, such as sin (90° − A) = cos A, which is not one of this chapter's sections now. If an answer depends on a rule you cannot find in the book, ask your teacher.
- Look for "not defined". An AI that gives a number for tan 90° has made a mistake.
More on using AI well: how to use ChatGPT for homework without cheating yourself.
Practise this chapter
- Printable worksheet: Introduction to Trigonometry, board practice for Class 10, in English, Hindi and Marathi, with answers. The next chapter: Applications of Trigonometry, board practice.
- Online, marked instantly: on the practice page, pick Class 10 and "Introduction to Trigonometry (board practice)".
- Timed: the free Class 10 revision papers include trigonometry questions in the Maths paper and the mixed paper.
- The whole Maths paper, section by section: CBSE Class 10 Maths sample paper 2026-27. Other chapters in the same style: Real Numbers and Quadratic Equations.
Where CMDLearn fits
CMDLearn is building an AI tutor for Classes 1 to 10 that explains a step and checks the child's working, but leaves the writing to the child. It is still in development, and no child uses it yet. If you would like to hear when there is something to try, you can join the waitlist.
Sources
All opened and read on 21 October 2026.
- NCERT, Mathematics, Textbook for Class X, Chapter 8 "Introduction to Trigonometry" (Reprint 2026-27), printed pages 113-132 (PDF pages 1-20): sections 8.1-8.2 and the history of sine (pp. 113-120), Exercise 8.1 (p. 121), section 8.3 (pp. 121-126), Table 8.1 (p. 125), Exercise 8.2 (p. 127), section 8.4 (pp. 128-131), Exercise 8.3 (pp. 131-132), summary 8.5 (p. 132). ncert.nic.in/textbook/pdf/jemh108.pdf
- CBSE, Mathematics curriculum, Class X, 2026-27: course structure (PDF page 3) and Unit V, Trigonometry (PDF page 6). cbseacademic.nic.in/web_material/CurriculumMain27/SecPart1/Maths_SecP1X_2026-27.pdf
- CBSE, Class X Mathematics (Standard, 041) sample paper for 2026-27: Q8 and Q17 (page 2), Q23 (page 4), Q31 (page 4), Q35 (page 5). cbseacademic.nic.in/web_material/SQP/ClassX_2026_27/MathsStandard-SQP.pdf
- CBSE, Class X Mathematics (Standard, 041) marking scheme for 2026-27: Q8 and Q17 (pages 1-2), Q23 (page 4), Q31 (pages 6-7). cbseacademic.nic.in/web_material/SQP/ClassX_2026_27/MathsStandard-MS.pdf
The definitions, the table of values, the identities, the section and exercise numbers, the page numbers and the board questions are from the documents above. All explanations, worked examples, the memory trick and the self-check questions are our own, and every number was checked by script.
Practise this class: Class 10 worksheets, chapter by chapter · free practice, marked instantly · timed revision papers