Class 10 · Chapter 8 · Trigonometry unit (12 of 80 marks)

Introduction to Trigonometry Class 10: notes and important questions

Q: What is in CBSE Class 10 Maths Chapter 8, Introduction to Trigonometry?

The topics are: Trigonometric ratios, Ratios of specific angles, Trigonometric identities. This page has revision notes for each, and 38 board-style practice questions with step-marked solutions.

Q: How many marks is Introduction to Trigonometry worth?

It is in the Trigonometry unit, which carries 12 of the 80 theory marks. CBSE gives marks by unit, not by chapter.

Q: Where do students lose marks in Introduction to Trigonometry?

Working on both sides of an identity at once, as if solving an equation. Fix: start from one side (usually the more complicated one) and reach the other; write LHS = … = RHS. Given one ratio, jumping to the others without the triangle. Fix: draw the right triangle, use 3k, 4k, 5k and show the Pythagoras step; it earns a mark.

Q: Are the Introduction to Trigonometry important questions free?

Yes. The 10 important questions on this page are free to try, from 1-mark MCQs to a case study. 5 worked answers are open on the page; the others open with a free account (no card).

Q: Are these NCERT questions?

No. Every practice question here is original, written in the board's question styles with step marks. For the textbook itself, use the NCERT link on this page.

Revision notes, 10 free important questions with step-marked answers, and a four-step route from the basics to 95+ through 38 board-style questions.

Practise this chapter Join freeFree account, no card: +5 practice questions on every chapter, with instant feedback.

38 questions
10 multiple choice, 4 assertion–reason, 8 very short answer, 8 short answer, 5 long answer, 3 case study
About 12 hours to master

Trigonometry unit: 12 of 80 theory marks (Introduction to Trigonometry and Some Applications of Trigonometry).

NCERT textbook: Chapter 8 on ncert.nic.in (PDF)

Revision notes

1. Trigonometric ratios

In a triangle $A B C$ right-angled at $B$ , for the acute angle $A$ : side opposite $= B C$ , side adjacent $= A B$ , hypotenuse $= A C$ .

$\sin A = \frac{B C}{A C}, \cos A = \frac{A B}{A C}, \tan A = \frac{B C}{A B}$
$cosec A = \frac{1}{\sin A}, \sec A = \frac{1}{\cos A}, \cot A = \frac{1}{\tan A}$ , and $\tan A = \frac{\sin A}{\cos A}, \cot A = \frac{\cos A}{\sin A}$
The ratios depend only on the angle, not on the size of the triangle (similar triangles).
For an acute angle, $0 < \sin A < 1$ and $0 < \cos A < 1$ , because the hypotenuse is the longest side. So $\sec A > 1$ and $cosec A > 1$ .

2. Ratios of specific angles

$A$	$0^{\circ}$	$30^{\circ}$	$45^{\circ}$	$60^{\circ}$	$90^{\circ}$
$\sin A$	0	$\frac{1}{2}$	$\frac{1}{\sqrt{2}}$	$\frac{\sqrt{3}}{2}$	1
$\cos A$	1	$\frac{\sqrt{3}}{2}$	$\frac{1}{\sqrt{2}}$	$\frac{1}{2}$	0
$\tan A$	0	$\frac{1}{\sqrt{3}}$	1	$\sqrt{3}$	not defined

Memory aid: $\sin A$ runs $\sqrt{0} / 2, \sqrt{1} / 2, \sqrt{2} / 2, \sqrt{3} / 2, \sqrt{4} / 2$ ; $\cos A$ runs the same list backwards. $cosec 0^{\circ}$ and $\cot 0^{\circ}$ are also not defined. As $A$ increases from $0^{\circ}$ to $90^{\circ}$ , $\sin A$ increases from 0 to 1 and $\cos A$ decreases from 1 to 0.

3. Trigonometric identities

$\sin^{2} A + \cos^{2} A = 1$ (for $0^{\circ} \leq A \leq 90^{\circ}$ )
$1 + \tan^{2} A = \sec^{2} A$ (for $0^{\circ} \leq A < 90^{\circ}$ )
$1 + \cot^{2} A = {cosec}^{2} A$ (for $0^{\circ} < A \leq 90^{\circ}$ )
Useful factorised forms: $\sec^{2} A - \tan^{2} A = (\sec A - \tan A) (\sec A + \tan A) = 1$ , so $\sec A + \tan A = \frac{1}{\sec A - \tan A}$ . Similarly $cosec A + \cot A = \frac{1}{cosec A - \cot A}$ .

Worked example 1

If $4 \tan θ = 3$ , find $\frac{\sin θ + \cos θ}{\sin θ - \cos θ}$ .

$\tan θ = \frac{3}{4}$ . Take opposite $= 3 k$ , adjacent $= 4 k$ , hypotenuse $= 5 k$ . Then $\sin θ = \frac{3}{5}, \cos θ = \frac{4}{5}$ , and the expression $= \frac{7 / 5}{- 1 / 5} = - 7$ . (Quicker: divide numerator and denominator by $\cos θ$ to get $\frac{\tan θ + 1}{\tan θ - 1} = \frac{7 / 4}{- 1 / 4} = - 7$ .)

Worked example 2

Evaluate $\frac{3 \tan^{2} 30^{\circ} + \sec^{2} 45^{\circ}}{{cosec}^{2} 60^{\circ}}$ .

$= \frac{3 \cdot \frac{1}{3} + 2}{\frac{4}{3}} = \frac{3}{4 / 3} = \frac{9}{4}$ .

Worked example 3

Prove that $\frac{\cos A}{1 + \sin A} = \sec A - \tan A$ .

RHS $= \frac{1 - \sin A}{\cos A} = \frac{(1 - \sin A) (1 + \sin A)}{\cos A (1 + \sin A)} = \frac{\cos^{2} A}{\cos A (1 + \sin A)} = \frac{\cos A}{1 + \sin A} =$ LHS.

Common errors

Writing $\sin A$ as $\sin \times A$ , or $\sin^{2} A$ as $\sin A^{2}$ .
Assuming $\sin (A + B) = \sin A + \sin B$ — check with $A = B = 30^{\circ}$ : $\sin 60^{\circ} \neq 1$ .
Mixing up opposite and adjacent sides when the angle is at a different vertex.
Working on both sides of an identity at once as if it were an equation. Start from one side (usually the more complicated one) and reach the other.
Forgetting that $\tan 90^{\circ}$ and $\sec 90^{\circ}$ are not defined.

Board-exam tips

When one ratio is given, draw a right triangle, use $k$ for the sides, and find the third side by Pythagoras. Show this step: it earns a mark.
In identity proofs, convert everything to $\sin$ and $\cos$ if you are stuck, and look for $1 - \sin^{2} A = \cos^{2} A$ style substitutions.
Write the values used (for example $\tan 60^{\circ} = \sqrt{3}$ ) before simplifying. Examiners give a mark for correct substitution.
Trigonometric ratios of complementary angles (for example $\sin (90^{\circ} - A) = \cos A$ ) are not in the current syllabus. You will not be asked to use them.

Topics in this chapter: Trigonometric ratios · Ratios of specific angles · Trigonometric identities.

Formulas for this chapter: Chapter 8 on the Class 10 formula sheet (free PDF and A3 poster).

Introduction to Trigonometry Class 10: important questions

10 original questions in the board paper's question types, from 1-mark MCQs to a case study, each with a step-marked worked answer. Try each one first. 5 worked answers open here; the other 5 open with a free account (no card).

Download these questions (PDF) 10 questions, 5 worked answers · 5 pages · 246 KB

Section A style · Multiple choice and assertion–reason (1 mark each)

Q1·1 mark·Multiple choiceTrigonometric ratios

If $\sin A = \frac{3}{5}$ , where $A$ is acute, then $\cot A$ equals

Show worked answer

$\begin{aligned} \cos A & = \sqrt{1 - \frac{9}{25}} \\ = \frac{4}{5} \end{aligned}$ , so $\begin{aligned} \cot A & = \frac{\cos A}{\sin A} \\ = \frac{4}{3} \end{aligned}$ A1
Correct option: (a) $\frac{4}{3}$

Q2·1 mark·Multiple choiceRatios of specific angles

The value of $\sin^{2} 60^{\circ} - \cos^{2} 45^{\circ} + \tan^{2} 30^{\circ}$ is

Q3·1 mark·Assertion–reasonTrigonometric ratios

In the following question, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option.

Assertion (A): There is an acute angle $θ$ with $\sin θ = \frac{4}{3}$ .

Reason (R): For every acute angle $θ$ , $0 < \sin θ < 1$ .

Section B style · Short answer (2 marks each)

Q4·2 marks·Short answerTrigonometric ratios

Given $\tan θ = \frac{5}{12}$ , where $θ$ is acute, find $\sin θ$ and $\cos θ$ .

Show worked answer

Answer:

\sin θ = \frac{5}{13}

\cos θ = \frac{12}{13}

Take opposite $= 5 k$ , adjacent $= 12 k$ ; hypotenuse $= \sqrt{25 k^{2} + 144 k^{2}} = 13 k$ M1
$\sin θ = \frac{5}{13}$ , $\cos θ = \frac{12}{13}$ A1

Q5·2 marks·Short answerRatios of specific angles

Evaluate $2 \cos^{2} 45^{\circ} + 3 \tan^{2} 60^{\circ} - 4 \sin^{2} 30^{\circ}$ .

Section C style · Short answer (3 marks each)

Q6·3 marks·Short answerTrigonometric identities

Prove that $\sec^{4} θ - \sec^{2} θ = \tan^{4} θ + \tan^{2} θ$ .

Show worked answer

Answer: Proved

LHS $= \sec^{2} θ (\sec^{2} θ - 1)$ M1
$= (1 + \tan^{2} θ) \tan^{2} θ$ , using $\sec^{2} θ - 1 = \tan^{2} θ$ A1
$= \tan^{2} θ + \tan^{4} θ =$ RHS A1

Q7·3 marks·Short answerTrigonometric ratios

If $4 \tan θ = 3$ , find the value of $\frac{2 \sin θ + \cos θ}{2 \sin θ - \cos θ}$ .

Section D style · Long answer (5 marks each)

Q8·5 marks·Long answerTrigonometric ratios

In $△ A B C$ , right-angled at $B$ , $A B = 15$ cm and $B C = 8$ cm. (a) Find $A C$ . (b) Find $\sin A$ , $\cos A$ and $\tan A$ . (c) Show that $\sin A \cos C + \cos A \sin C = 1$ .

Show worked answer

Answer:

A C = 17

cm;

\frac{8}{17}, \frac{15}{17}, \frac{8}{15}

; shown

(a) $A C = \sqrt{15^{2} + 8^{2}} = \sqrt{289} = 17$ cm A1
(b) $\sin A = \frac{B C}{A C} = \frac{8}{17}$ , $\cos A = \frac{15}{17}$ , $\tan A = \frac{8}{15}$ A1
(c) For angle $C$ : $\sin C = \frac{A B}{A C} = \frac{15}{17}$ , $\cos C = \frac{B C}{A C} = \frac{8}{17}$ M1
$\sin A \cos C + \cos A \sin C = \frac{8}{17} \cdot \frac{8}{17} + \frac{15}{17} \cdot \frac{15}{17}$ M1
$= \frac{64 + 225}{289} = 1$ A1

Q9·5 marks·Long answerTrigonometric identities

(a) Prove that $\frac{\cos θ}{1 - \tan θ} + \frac{\sin θ}{1 - \cot θ} = \sin θ + \cos θ$ . (b) Hence find the value of the left-hand side when $θ = 30^{\circ}$ .

Section E style · Case study (4 marks)

Q10·4 marks·Case studyTrigonometric ratios

Set square. A carpenter's set square is a right triangle $P Q R$ , right-angled at $Q$ , with $P Q = 12$ cm and $Q R = 5$ cm.

Find the length of $P R$ . [1 mark]
Find $\sin P$ and $\cos R$ . [1 mark]
Find $\tan P \times \tan R$ and $\sin^{2} P + \sin^{2} R$ . [2 marks]

Show worked answer

$P R = \sqrt{144 + 25} = 13$ cm A1
$\sin P = \frac{Q R}{P R} = \frac{5}{13}$ and $\cos R = \frac{Q R}{P R} = \frac{5}{13}$ A1
$\tan P = \frac{5}{12}$ , $\tan R = \frac{12}{5}$ , so $\tan P \times \tan R = 1$ M1
$\begin{aligned} \sin^{2} P + \sin^{2} R & = \frac{25}{169} + \frac{144}{169} \\ = 1 \end{aligned}$ A1

Next: practise by Route to 95 step · the revision notes · free MCQs and case studies · all Class 10 assertion–reason questions · the Class 10 syllabus checklist.

Route to 95: four steps

Work through the steps in order. Take each checkpoint closed book, about 1.5 minutes per mark; pass at 80% to move on. Two misses in a row means going back one step.

Step 1

Secure the basics

You can find all six trigonometric ratios from one given ratio and know the values for 0°, 30°, 45°, 60° and 90°.

Read first: 1. Trigonometric ratios; 2. Ratios of specific angles 6 practice questions · checkpoint: 3 questions, 5 marks, pass 80%

Practise step 1

Step 2

Board standard

You can evaluate expressions with standard angles and prove the common identities by working from one side.

Read first: 2. Ratios of specific angles; 3. Trigonometric identities; Worked examples 1-3 18 practice questions · checkpoint: 3 questions, 7 marks, pass 80%

Practise step 2

Step 3

Full marks on long answers

You can write long answers that combine ratios, standard values and identities, with every substitution shown.

Read first: 3. Trigonometric identities; Board-exam tips 7 practice questions · checkpoint: 3 questions, 15 marks, pass 80%

Practise step 3

Step 4

95+ stretch (HOTS)

You can prove the tougher identities and solve problems where the ratio is given in terms of letters such as a and b.

Read first: 3. Trigonometric identities; Common errors 7 practice questions · checkpoint: 3 questions, 9 marks, pass 80%

Practise step 4

Practice questions

Original questions in the board's styles. Multiple-choice answers are checked as you go; for written answers, compare your working with the step mark scheme and record your marks. 3 of the 38 are competency-based (case studies and questions set in a real-life situation): the "Competency-based" button shows just those.

Q1·1 mark·Multiple choiceTrigonometric ratios

In $△ P Q R$ , right-angled at $Q$ , $P Q = 20$ cm and $Q R = 21$ cm. The value of $\sin R$ is

(a) $\frac{20}{29}$
(b) $\frac{21}{29}$
(c) $\frac{20}{21}$
(d) $\frac{21}{20}$

Q2·1 mark·Multiple choiceTrigonometric ratios

If $\cos A = \frac{5}{13}$ , where $A$ is acute, then $\tan A$ equals

(a) $\frac{5}{12}$
(b) $\frac{12}{13}$
(c) $\frac{12}{5}$
(d) $\frac{13}{12}$

Q3·2 marks·Very short answerTrigonometric ratios

If $12 \sin A = 5 \cos A$ , where $A$ is acute, find the value of $\sec A + cosec A$ .

Worked solution·2 marks·Very short answerTrigonometric identities

Show that $\frac{\sin^{2} θ}{1 - \cos θ} - \cos θ = 1$ , for $0^{\circ} < θ < 90^{\circ}$ .

Full worked solution Free, no sign-in

$\frac{\sin^{2} θ}{1 - \cos θ} = \frac{1 - \cos^{2} θ}{1 - \cos θ} = \frac{(1 - \cos θ) (1 + \cos θ)}{1 - \cos θ} = 1 + \cos θ$ M1

LHS $= 1 + \cos θ - \cos θ = 1 =$ RHS A1

Every question here has a full worked solution. See them all →

Stretch yourself: original geometry problems at Intermediate level (ages 13 to 16), with hints and full solutions: Problem I98 · Problem I16 · Problem I84 · Problem I91. All 28 →

Knowledge organiser: Introduction to Trigonometry on one page (key facts, formulas and mistakes to avoid; printable PDF)

Formulas in this topic: Trig identities. All formulas A–Z

Where marks are lost in Introduction to Trigonometry

Working on both sides of an identity at once, as if solving an equation. Fix: start from one side (usually the more complicated one) and reach the other; write LHS = … = RHS.
Given one ratio, jumping to the others without the triangle. Fix: draw the right triangle, use 3k, 4k, 5k and show the Pythagoras step; it earns a mark.
Not writing the standard values before simplifying. Fix: write tan 60° = √3 and so on first; the substitution is marked separately.
Treating sin(A + B) as sin A + sin B. Fix: test with A = B = 30°; if the question needs it, use the given values only.
Using tan 90° or sec 90° as if defined. Fix: remember they are not defined, and say so in AR and MCQ items.
Opposite and adjacent swapped when the angle is at a different vertex (e.g. tan C instead of tan A). Fix: mark the angle on your figure before writing any ratio.

Exam strategy: how answers are step-marked · the proofs, written out · competency-based questions

Introduction to Trigonometry in our sample papers

Once step 3 is passed, test the chapter inside a full timed paper on the 2026-27 pattern (original papers by us, not official CBSE papers).

Sample paper 1 (Board standard): questions 10, 24, 30
Sample paper 2 (Board standard): questions 11, 12, 24, 30
Sample paper 3 (Above board): questions 13, 14, 23, 30
Sample paper 4 (Above board): questions 12, 24, 31
Sample paper 5 (95+ challenge): questions 13, 14, 20, 23, 30

Want it against the clock? Take a timed 30-mark chapter test on Introduction to Trigonometry, new questions each time, or climb the Difficulty ladder, levels 1 to 10, three questions a level (CBSE Pro or the free trial).

Introduction to Trigonometry: common questions

What is in CBSE Class 10 Maths Chapter 8, Introduction to Trigonometry?

The topics are: Trigonometric ratios, Ratios of specific angles, Trigonometric identities. This page has revision notes for each, and 38 board-style practice questions with step-marked solutions.

How many marks is Introduction to Trigonometry worth?

It is in the Trigonometry unit, which carries 12 of the 80 theory marks. CBSE gives marks by unit, not by chapter.

Where do students lose marks in Introduction to Trigonometry?

Working on both sides of an identity at once, as if solving an equation. Fix: start from one side (usually the more complicated one) and reach the other; write LHS = … = RHS. Given one ratio, jumping to the others without the triangle. Fix: draw the right triangle, use 3k, 4k, 5k and show the Pythagoras step; it earns a mark.

Are the Introduction to Trigonometry important questions free?

Yes. The 10 important questions on this page are free to try, from 1-mark MCQs to a case study. 5 worked answers are open on the page; the others open with a free account (no card).

Are these NCERT questions?

No. Every practice question here is original, written in the board's question styles with step marks. For the textbook itself, use the NCERT link on this page.

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Revision notes

1. Trigonometric ratios

2. Ratios of specific angles

3. Trigonometric identities

Worked example 1

Worked example 2

Worked example 3

Common errors

Board-exam tips

Introduction to Trigonometry Class 10: important questions

Section A style · Multiple choice and assertion–reason (1 mark each)

Section B style · Short answer (2 marks each)

Section C style · Short answer (3 marks each)

Section D style · Long answer (5 marks each)

Section E style · Case study (4 marks)

Route to 95: four steps

Secure the basics

Board standard

Full marks on long answers

95+ stretch (HOTS)

Practice questions

Full worked solution Free, no sign-in

Where marks are lost in Introduction to Trigonometry

Introduction to Trigonometry in our sample papers

Introduction to Trigonometry: common questions

Work through the Route to 95 with your progress saved