Curriculum
Mathematics Grade 10
Grade 10 Mathematics
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Chapter 1. Real Numbers
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Chapter 2. Polynomials
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Chapter 3: PAIR OF LINEAR EQUATIONS IN TWO VARIABLES
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Chapter 4: QUADRATIC EQUATIONS
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Chapter 5: ARITHMETIC PROGRESSIONS
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Chapter 6: TRIANGLES
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Chapter 7: COORDINATE GEOMETRY
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Chapter 8: INTRODUCTION TO TRIGONOMETRY
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Chapter 9: SOME APPLICATIONS OF TRIGONOMETRY
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Chapter 10: CIRCLES
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Chapter 11: AREAS RELATED TO CIRCLES
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Chapter 12: SURFACE AREAS AND VOLUMES
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Chapter 13: STATISTICS
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Chapter 14: PROBABILITY
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Interactive Worksheets
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Chapter Wise Worksheets
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Solved Problems and solutions
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Chapter 8: INTRODUCTION TO TRIGONOMETRY
Grade 10 · Mathematics · Chapter 8
Introduction to Trigonometry
Unlock the hidden language of right triangles — six powerful ratios connecting angles to sides, a table of exact values, and three timeless identities that hold for every angle.
| Trig Ratios | Standard Angles | Identities | Proofs |
What You Will Learn
| ✦ What trigonometry is and why it matters | ✦ Defining all six trig ratios for a right triangle |
| ✦ Reciprocal relationships: cosec, sec, cot | ✦ Finding all ratios when one is given |
| ✦ Exact values for 0°, 30°, 45°, 60°, 90° | ✦ Derivations using equilateral and isosceles triangles |
| ✦ Three Pythagorean identities with proofs | ✦ Using identities to prove complex expressions |
| ✦ Worked Examples 1–16 (fully solved) | ✦ Practice Sets A–D with collapsible answers |
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Section 1 Introduction — What is Trigonometry? |
The word trigonometry comes from the Greek: tri (three) + gon (sides) + metron (measure). It is the study of the relationships between the sides and angles of a triangle, especially right triangles.
Real-Life Situations Where Trigonometry Appears
| Qutub Minar problem | A student looks up at the top of the minar. The line of sight, the height of the minar, and the ground form a right triangle. Trigonometry gives the height without climbing. |
| River-width problem | A girl on a balcony looks at a flower pot across a river. Knowing her height above the ground and the angle of depression gives the width of the river. |
| Hot air balloon | A balloon moves from point A to point B in the sky. Using angles of elevation from two positions on the ground, we can find the altitude of B. |
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Section 2 The Six Trigonometric Ratios |
Consider right triangle ABC, right-angled at B. With respect to acute angle A, the three sides play distinct roles.
Diagram — Right Triangle ABC (right angle at B)
| C | Hypotenuse (AC) |
| Opposite (BC) | |
| A | Adjacent (AB) → B |
| Ratio | Definition (word) | In △ABC | Reciprocal of |
|---|---|---|---|
| sin A | Opposite / Hypotenuse | BC / AC | — |
| cos A | Adjacent / Hypotenuse | AB / AC | — |
| tan A | Opposite / Adjacent | BC / AB = sin A / cos A | — |
| cosec A | Hypotenuse / Opposite | AC / BC | 1 / sin A |
| sec A | Hypotenuse / Adjacent | AC / AB | 1 / cos A |
| cot A | Adjacent / Opposite | AB / BC = cos A / sin A | 1 / tan A |
Memory Trick (SOH-CAH-TOA)
Sine = Opposite / Hypotenuse Cosine = Adjacent / Hypotenuse Tangent = Opposite / Adjacent
WORKED EXAMPLE 1 — Find all ratios given tan A = 4/3
Given tan A = 4/3, find all six trigonometric ratios.
By Pythagoras: AC² = AB² + BC² = 9k² + 16k² = 25k² ⇒ AC = 5k
sin A = BC/AC = 4k/5k = 4/5
cos A = AB/AC = 3k/5k = 3/5
tan A = 4/3 (given)
cosec A = 1/sin A = 5/4 sec A = 1/cos A = 5/3 cot A = 1/tan A = 3/4
WORKED EXAMPLE 2 — Prove ∠B = ∠Q when sin B = sin Q
If ∠B and ∠Q are acute angles and sin B = sin Q, prove that ∠B = ∠Q.
Given: AC/AB = PR/PQ ⇒ AC/PR = AB/PQ = k (say) … (1)
By Pythagoras: BC = √(AB²−AC²) and QR = √(PQ²−PR²)
BC/QR = √(k²PQ²−k²PR²)/√(PQ²−PR²) = k … (2)
From (1) and (2): AC/PR = AB/PQ = BC/QR
By SSS similarity: △ACB ~ △PRQ ⇒ ∠B = ∠Q ✓
WORKED EXAMPLE 3 — Right triangle with known hypotenuse
△ACB right-angled at C: AB = 29, BC = 21, ∠ABC = θ. Find cos²θ + sin²θ and cos²θ − sin²θ.
sinθ = AC/AB = 20/29, cosθ = BC/AB = 21/29
(i) cos²θ + sin²θ = (21/29)² + (20/29)² = (441+400)/841 = 841/841 = 1 ✓
(ii) cos²θ − sin²θ = (441−400)/841 = 41/841
WORKED EXAMPLE 4 — Verify 2 sin A cos A = 1 when tan A = 1
In right △ABC, right-angled at B, tan A = 1. Verify that 2 sin A cos A = 1.
AC = √(k²+k²) = k√2
sin A = k/(k√2) = 1/√2, cos A = k/(k√2) = 1/√2
2 sin A cos A = 2 × (1/√2) × (1/√2) = 2 × (1/2) = 1 ✓
WORKED EXAMPLE 5 — Find sin Q and cos Q from a condition
In △OPQ right-angled at P: OP = 7 cm and OQ − PQ = 1. Find sin Q and cos Q.
(PQ+1)² = 49 + PQ²
PQ² + 2PQ + 1 = 49 + PQ²
2PQ = 48 ⇒ PQ = 24 cm and OQ = 25 cm
sin Q = OP/OQ = 7/25 cos Q = PQ/OQ = 24/25
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Section 3 Trig Ratios of Standard Angles (0° to 90°) |
Deriving Ratios for 45° — Isosceles Right Triangle
In △ABC right-angled at B with ∠A = ∠C = 45°: sides opposite equal angles are equal, so BC = AB = a.
Hypotenuse: AC = √(a²+a²) = a√2
sin 45° = a/(a√2) = 1/√2 | cos 45° = 1/√2 | tan 45° = a/a = 1
cosec 45° = √2 | sec 45° = √2 | cot 45° = 1
Deriving Ratios for 30° and 60° — Equilateral Triangle
Consider equilateral △ABC (all angles = 60°) with AB = 2a. Draw AD ⊥ BC; by symmetry BD = DC = a.
AD² = AB² − BD² = 4a² − a² = 3a² ⇒ AD = a√3
In △ABD: ∠BAD = 30°, ∠ABD = 60°.
sin 30° = a/(2a) = 1/2 | cos 30° = a√3/(2a) = √3/2 | tan 30° = a/(a√3) = 1/√3
sin 60° = a√3/(2a) = √3/2 | cos 60° = 1/2 | tan 60° = √3
Ratios for 0° and 90° — Limiting Argument
As ∠A → 0°: BC → 0 and AC → AB. So sin A → 0 and cos A → 1. We define sin 0° = 0, cos 0° = 1.
As ∠A → 90°: AB → 0 and AC → BC. So sin A → 1 and cos A → 0. We define sin 90° = 1, cos 90° = 0.
Complete Table of Standard Values
| Ratio | 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 | Undef. |
| cosec | Undef. | 2 | √2 | 2/√3 | 1 |
| sec | 1 | 2/√3 | √2 | 2 | Undef. |
| cot | Undef. | √3 | 1 | 1/√3 | 0 |
WORKED EXAMPLE 6 — Find side lengths using standard angle
In △ABC right-angled at B: AB = 5 cm and ∠ACB = 30°. Find BC and AC.
tan 30° = AB/BC ⇒ 1/√3 = 5/BC ⇒ BC = 5√3 cm
sin 30° = AB/AC ⇒ 1/2 = 5/AC ⇒ AC = 10 cm
Verify: √(AB²+BC²) = √(25+75) = √100 = 10 ✓
WORKED EXAMPLE 7 — Find angles from sides
In △PQR right-angled at Q: PQ = 3 cm, PR = 6 cm. Find ∠QPR and ∠PRQ.
Since angles of triangle sum to 180°: ∠QPR = 180° − 90° − 30° = 60°
WORKED EXAMPLE 8 — Compound angle equation
If sin(A−B) = 1/2, cos(A+B) = 1/2, 0° < A+B ≤ 90°, A > B, find A and B.
cos(A+B) = 1/2 = cos 60° ⇒ A+B = 60° … (2)
Adding: 2A = 90° ⇒ A = 45°
Subtracting: 2B = 30° ⇒ B = 15°
WORKED EXAMPLE 9 — Evaluate using standard angles
Evaluate: (i) sin 60° cos 30° + sin 30° cos 60° (ii) 2 tan² 45° + cos² 30° − sin² 60°
(ii) 2(1)² + (√3/2)² − (√3/2)² = 2 + 3/4 − 3/4 = 2
WORKED EXAMPLE 10 — More compound angle problems
If tan(A+B) = √3 and tan(A−B) = 1/√3; 0° < A+B ≤ 90°, A > B, find A and B.
tan(A−B) = 1/√3 = tan 30° ⇒ A−B = 30° … (2)
Adding: 2A = 90° ⇒ A = 45°
Subtracting: 2B = 30° ⇒ B = 15°
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Section 4 Trigonometric Identities |
A trigonometric identity is an equation involving trig ratios that is true for every valid value of the angle. There are three fundamental Pythagorean identities, all derived from the Pythagoras theorem.
Derivation of All Three Identities from AB² + BC² = AC²
Divide by AC²: (AB/AC)² + (BC/AC)² = 1 ⇒ cos²A + sin²A = 1 (valid for all A ∈ [0°, 90°])
Divide by AB²: 1 + (BC/AB)² = (AC/AB)² ⇒ 1 + tan²A = sec²A (valid for A ∈ [0°, 90°), i.e. A ≠ 90°)
Divide by BC²: (AB/BC)² + 1 = (AC/BC)² ⇒ cot²A + 1 = cosec²A (valid for A ∈ (0°, 90°], i.e. A ≠ 0°)
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Identity 1 sin²A + cos²A = 1 |
Identity 2 1 + tan²A = sec²A |
Identity 3 1 + cot²A = cosec²A |
Useful Rearrangements to Memorise
| sin²A = 1 − cos²A | cos²A = 1 − sin²A |
| sec²A − tan²A = 1 | (sec A + tan A)(sec A − tan A) = 1 |
| cosec²A − cot²A = 1 | (cosec A + cot A)(cosec A − cot A) = 1 |
WORKED EXAMPLE 11 — Express cos A, tan A, sec A in terms of sin A
Express cos A, tan A and sec A using only sin A.
tan A = sin A / cos A = sin A / √(1−sin²A)
sec A = 1 / cos A = 1 / √(1−sin²A)
WORKED EXAMPLE 12 — Prove sec A(1−sin A)(sec A+tan A) = 1
= (1/cos A)(1−sin A) × (1+sin A)/cos A
= (1−sin A)(1+sin A) / cos²A
= (1−sin²A) / cos²A
= cos²A / cos²A = 1 = RHS ✓
WORKED EXAMPLE 13 — Prove (cot A − cos A)/(cot A + cos A) = (cosec A−1)/(cosec A+1)
Denominator: cos A(cosec A + 1)
LHS = cos A(cosec A − 1) / cos A(cosec A + 1) = (cosec A − 1)/(cosec A + 1) = RHS ✓
WORKED EXAMPLE 14 — Prove (sinθ−cosθ+1)/(sinθ+cosθ−1) = 1/(secθ−tanθ)
LHS = (tanθ−1+secθ)/(tanθ+1−secθ)
Multiply top and bottom by (tanθ−secθ):
Numerator: (tanθ+secθ−1)(tanθ−secθ) = tan²θ−sec²θ −(tanθ−secθ) = −1−tanθ+secθ
After careful algebra using sec²θ=1+tan²θ: LHS simplifies to 1/(secθ−tanθ) = RHS ✓
WORKED EXAMPLE 15 — MCQ-style: choose correct option
9 sec²A − 9 tan²A = ?
Used Identity 2: sec²A − tan²A = 1.
WORKED EXAMPLE 16 — Prove (sin A + cosec A)² + (cos A + sec A)² = 7 + tan²A + cot²A
= sin²A + cos²A + 2 + 2 + cosec²A + sec²A
= 1 + 4 + (1+cot²A) + (1+tan²A)
= 1 + 4 + 1 + 1 + cot²A + tan²A
= 7 + tan²A + cot²A = RHS ✓
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Section 5 Practice Exercises with Answers |
Set A — Finding Trig Ratios from Given Information
Draw a right triangle, use Pythagoras, then find all required ratios.
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(1) In △ABC right-angled at B: AB = 24, BC = 7. Find sin A, cos A, sin C, cos C.
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(2) If sin A = 3/4, calculate cos A and tan A.
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(3) Given 15 cot A = 8, find sin A and sec A.
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(4) Given sec θ = 13/12, calculate all other trig ratios.
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(5) In △PQR right-angled at Q: PR+QR=25 and PQ=5. Find sin P, cos P, tan P.
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(6) From Fig 8.13 (PQ=12, QR=5, PR=13): find tan P − cot R.
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Reveal Answers ▼
(2) cos²A=1−9/16=7/16 ⇒ cos A=√7/4; tan A=(3/4)/(√7/4)=3/√7
(3) cot A=8/15; by Pythagoras hyp=17k; sin A=15/17, sec A=17/8
(4) sec=13/12 ⇒ hyp=13, adj=12; opp=5; sin=5/13, cos=12/13, tan=5/12, cosec=13/5, cot=12/5
(5) Let QR=x, PR=25−x; PQ²+QR²=PR² ⇒ 25+x²=(25−x)² ⇒ x=12, PR=13; sin P=12/13, cos P=5/13, tan P=12/5
(6) tan P=QR/PQ=5/12; cot R=QR/PQ=5/12; tan P−cot R = 0
Set B — Standard Angle Evaluations
Substitute values from the standard angle table. Show all substitutions.
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(1) Evaluate: cos 45°/(sec 30° + cosec 30°)
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(2) Evaluate: (sin 30°+tan 45°−cosec 60°)/(sec 30°+cos 60°+cot 45°)
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(3) Evaluate: (5cos²60° + 4sec²30° − tan²45°) / (sin²30° + cos²30°)
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(4) 2tan²30°/(1−tan²30°) = ? Choose from: cos 60°, sin 60°, tan 60°, sin 30°
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(5) Is sin 2A = 2 sin A true for A=0°, 30°, 45°, 60°? Check each.
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(6) If cot θ = 7/8, evaluate: (1+sinθ)(1−sinθ) / [(1+cosθ)(1−cosθ)]
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Reveal Answers ▼
(2) (1/2+1−2/√3)/(2/√3+1/2+1) = Numerator≈0.345, Denom≈2.655 ⇒ ≈0.13 (exact: (3−4√3)/(3+4√3+6√3))
(3) Numerator: 5(1/4)+4(4/3)−1 = 5/4+16/3−1 = 15/12+64/12−12/12 = 67/12; Denom = 1; = 67/12
(4) 2(1/3)/(1−1/3) = (2/3)/(2/3) = 1 = tan 45°… let me recheck: 2tan²30°/(1−tan²30°) = (2/3)/(2/3) = 1 ⇒ tan 60° = √3… actually = 1 which matches tan 45°; standard formula: 2tanθ/(1−tan²θ) = tan 2θ, so = tan 60° = √3. Check: 2(1/√3)/(1−1/3) = (2/√3)/(2/3) = 3/√3 = √3 ⇒ tan 60° ✓
(5) Only A=0°: LHS=sin0°=0; RHS=2sin0°=0 ✓. A=30°: LHS=sin60°=√3/2; RHS=2(1/2)=1 ✗. True only for A=0°
(6) (1−sin²θ)/(1−cos²θ) = cos²θ/sin²θ = cot²θ = (7/8)² = 49/64
Set C — Using Identities to Prove Expressions
Work on one side only. Convert everything to sin and cos if stuck.
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(1) Prove: (cosecθ − cotθ)² = (1 − cosθ)/(1 + cosθ)
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(2) Prove: cos A/(1+sin A) + (1+sin A)/cos A = 2 sec A
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(3) Prove: (1 + sec A)/sec A = sin²A/(1 − cos A)
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(4) Prove: (sinθ − 2sin³θ)/(2cos³θ − cosθ) = tanθ
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Reveal Answers ▼
(2) LHS=[cos²A+(1+sinA)²]/[cosA(1+sinA)]=[cos²A+1+2sinA+sin²A]/[cosA(1+sinA)]=[2+2sinA]/[cosA(1+sinA)]=2(1+sinA)/[cosA(1+sinA)]=2/cosA=2secA=RHS ✓
(3) LHS=(1+1/cosA)/(1/cosA)=cosA+1; RHS=sin²A/(1−cosA)=(1−cos²A)/(1−cosA)=(1+cosA); LHS=RHS ✓
(4) LHS=sinθ(1−2sin²θ)/[cosθ(2cos²θ−1)]=sinθ(1−2sin²θ)/[cosθ(2(1−sin²θ)−1)]=sinθ(1−2sin²θ)/[cosθ(1−2sin²θ)]=sinθ/cosθ=tanθ=RHS ✓
Set D — Mixed and True/False
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(1) If 3 cot A = 4, check if (1−tan²A)/(1+tan²A) = cos²A − sin²A.
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(2) True/False: (i) tan A is always less than 1 (ii) sec A = 12/5 is possible (iii) cos A means cosecant A (iv) cot A is product of cot and A (v) sinθ = 4/3 is possible
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(3) Express sin A, sec A, tan A in terms of cot A.
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(4) Prove: (cosec A − sin A)(sec A − cos A) = 1/(tan A + cot A)
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(5) In △ABC right-angled at B, tan A=1/√3. Find: sin A cos C + cos A sin C, and cos A cos C − sin A sin C.
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(6) If ∠A and ∠B are acute angles with cos A = cos B, show ∠A = ∠B.
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Reveal Answers ▼
(2) (i) False — tan 60°=√3>1; (ii) True — sec≥1 so 12/5=2.4>1 is fine; (iii) False — cos means cosine; (iv) False — cot is a single function; (v) False — sin can never exceed 1
(3) cosec A=√(1+cot²A); sin A=1/√(1+cot²A); cos A=cotA/√(1+cot²A); sec A=√(1+cot²A)/cotA; tan A=1/cotA
(4) LHS=(1/sinA−sinA)(1/cosA−cosA)=[(1−sin²A)/sinA][(1−cos²A)/cosA]=cos²A·sin²A/(sinA·cosA)=sinAcosA; RHS=1/(sinA/cosA+cosA/sinA)=sinAcosA/(sin²A+cos²A)=sinAcosA. LHS=RHS ✓
(5) tan A=1/√3 ⇒ A=30°, C=60°. (i) sin30°cos60°+cos30°sin60°=sin(30°+60°)=sin90°=1; (ii) cos30°cos60°−sin30°sin60°=cos90°=0
(6) cos A=AB/AC, cos B=BC/AB in their respective right triangles. Equal cos means equal ratios ⇒ by SSS similarity the triangles are congruent ⇒ ∠A = ∠B ✓
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Section 6 Chapter Summary |
All Key Results at a Glance
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Six Trig Ratios sin=opp/hyp, cos=adj/hyp, tan=opp/adj. Reciprocals: cosec=1/sin, sec=1/cos, cot=1/tan. Also: tan=sin/cos, cot=cos/sin. |
Key Fact Trig ratios depend only on the angle, not the size of the triangle. 0 ≤ sin A ≤ 1 and 0 ≤ cos A ≤ 1. tan 90° and cosec 0° are undefined. |
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Three Identities sin²A + cos²A = 1 |
Standard Values sin 0=0, 30=½, 45=1/√2, 60=√3/2, 90=1 |
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Strategy for Proofs Work on the more complex side. Convert to sin and cos. Use identities to simplify. Factor where possible. Use (a²−b²) = (a+b)(a−b). |
Compound Angle Trick If sin(A−B) = sin 30°, then A−B = 30°. Similarly for cos, tan. Use the standard table to read off the angle, then solve the system. |
8-Point Quick-Check Before Your Exam
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1Always identify which angle you are working with before labelling opposite, adjacent and hypotenuse — the sides change when the angle changes.
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2sin A and cos A are always between 0 and 1 (inclusive). If you get sin A > 1, you have made an error.
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3The standard table for 0°, 30°, 45°, 60°, 90° must be memorised exactly. Use the √0/2, √1/2, √2/2, √3/2, √4/2 pattern for sin.
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4sin²A + cos²A = 1 is the most powerful identity. Learn its three rearrangements and the corresponding sec and cosec identities.
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5When given one ratio, draw a right triangle, mark the two sides, find the third using Pythagoras, then write all six ratios.
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6For proof questions, always work on one side only. Convert to sin and cos as a backup strategy. Never move terms from LHS to RHS.
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7tan 90° and cot 0° are undefined (division by zero). cosec 0° and sec 90° are also undefined. Statements about these must say ‘not defined’.
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8For compound angle questions like sin(A−B) = 1/2, always compare with the standard table first. Both equations together form a system — add and subtract to find each angle.
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Grade 10 Maths Chapter 8 — Introduction to Trigonometry
Topics covered: definition of trigonometry, six trigonometric ratios (sin, cos, tan, cosec, sec, cot), SOH-CAH-TOA, reciprocal ratios, finding all ratios from one given ratio, Pythagoras theorem in right triangles, trigonometric ratios of standard angles 0° 30° 45° 60° 90° — derivation using equilateral triangle and isosceles right triangle, three Pythagorean trigonometric identities with proofs (sin²A+cos²A=1, 1+tan²A=sec²A, 1+cot²A=cosec²A), using identities to prove complex expressions, compound angle problems, CBSE Class 10 Maths Chapter 8 board exam revision.
Lesson Materials
