Trigonometry PYQ Bank | Class X Maths

Questions

2-Mark Qs

3-Mark Qs

Years Covered

📋 Key Topics Covered (As per CBSE 2025-26 Syllabus)

Ch 8: Trig Ratios — sin, cos, tan, cosec, sec, cot

Standard Angle Values — 0°, 30°, 45°, 60°, 90°

Trig Identities — sin²θ+cos²θ=1 and variants

Prove Identity — LHS = RHS type questions

Ch 9: Heights & Distances — angle of elevation/depression

Word Problems — tower, building, kite, shadow

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DELETED for 2025-26: Trigonometric ratios of complementary angles [sin(90°−θ)=cosθ etc.] removed from summative assessment. Questions on complementary angles will NOT appear in the board exam.

✅

Unit Weightage: Trigonometry unit (Ch 8 + Ch 9) carries 12 marks. Expect 2-3 MCQs + 1-2 SA + 1 LA from these two chapters combined.

📝 SECTION A — 2-MARK QUESTIONS

15 Questions • Very Short Answer (VSA) Type • Board PYQs 2018–2025

CBSE 2025

If sin α = √3/2 and cos β = √3/2, find the value of tan α · tan β.

2 Marks Ch 8 Standard Angles ⭐ MOST RECENT

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✅ DETAILED SOLUTION

Step 1 sin α = √3/2 = sin 60° ⟹ α = 60°

Step 2 cos β = √3/2 = cos 30° ⟹ β = 30°

Step 3 tan α · tan β = tan 60° × tan 30° = √3 × 1/√3 = 1

∴ tan α · tan β = 1

CBSE 2024

If sin α = 1/√2 and cot β = √3, find the value of cosec α + cosec β.

2 Marks Ch 8 Standard Angles

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✅ DETAILED SOLUTION

Step 1 sin α = 1/√2 ⟹ cosec α = √2

Step 2 cot β = √3 ⟹ β = 30° ⟹ cosec 30° = 2

Step 3 cosec α + cosec β = √2 + 2

∴ cosec α + cosec β = √2 + 2

CBSE 2023

If sin θ – cos θ = 0, find the value of sin⁴θ + cos⁴θ.

2 Marks Ch 8 Evaluate Expression ⭐ IMPORTANT

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✅ DETAILED SOLUTION

Step 1 sin θ = cos θ ⟹ tan θ = 1 ⟹ θ = 45°

Step 2 sin⁴45° + cos⁴45° = (1/√2)⁴ + (1/√2)⁴ = 1/4 + 1/4

∴ sin⁴θ + cos⁴θ = 1/2

CBSE 2023

(sec²θ – 1)(cosec²θ – 1) is equal to: (a) –1 (b) 1 (c) 0 (d) 2

2 Marks Ch 8 Identity Simplification

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✅ DETAILED SOLUTION

sec²θ – 1 = tan²θ | cosec²θ – 1 = cot²θ

Step 1 (sec²θ–1)(cosec²θ–1) = tan²θ × cot²θ

Step 2 = (sinθ/cosθ)² × (cosθ/sinθ)² = 1

∴ Answer: (b) 1

CBSE 2022

If sin²θ + sinθ = 1, find the value of cos²θ + cos⁴θ.

2 Marks Ch 8 Identity Trick ⭐ HOTS

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✅ DETAILED SOLUTION

Step 1 sin²θ + sinθ = 1 ⟹ sinθ = 1 – sin²θ = cos²θ

Step 2 cos²θ + cos⁴θ = sinθ + sin²θ [substituting cos²θ = sinθ]

Step 3 = sinθ + sin²θ = 1 [from the given equation]

∴ cos²θ + cos⁴θ = 1

CBSE 2022

If cosec θ = 5/4, find the value of cot θ.

2 Marks Ch 8 Find Ratio from Given

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✅ DETAILED SOLUTION

cosec²θ – cot²θ = 1 ⟹ cot²θ = cosec²θ – 1

Step 1 cot²θ = (5/4)² – 1 = 25/16 – 1 = 9/16

Step 2 cot θ = 3/4

∴ cot θ = 3/4

CBSE 2021

If tan θ + cot θ = 5, find the value of tan²θ + cot²θ.

2 Marks Ch 8 Squaring Trick ⭐ CLASSIC

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✅ DETAILED SOLUTION

Step 1 Square both sides: (tanθ + cotθ)² = 25

Step 2 tan²θ + cot²θ + 2·tanθ·cotθ = 25

Key: tanθ × cotθ = 1 always. So 2·tanθ·cotθ = 2

Step 3 tan²θ + cot²θ + 2 = 25 ⟹ tan²θ + cot²θ = 23

∴ tan²θ + cot²θ = 23

CBSE 2021

A kite is flying at a height of 30 m from the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string (use √3 = 1.73).

2 Marks Ch 9 Height & Distance

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✅ DETAILED SOLUTION

Step 1 sin 60° = Height / String = 30 / L

Step 2 √3/2 = 30/L ⟹ L = 60/√3 = 60√3/3 = 20√3

Step 3 L = 20 × 1.73 = 34.6 m

∴ Length of string = 20√3 m ≈ 34.64 m

CBSE 2020

If (1 + cos A)(1 – cos A) = 3/4, find the value of sec A.

2 Marks Ch 8 Identity → Find Ratio

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✅ DETAILED SOLUTION

Step 1 (1+cosA)(1–cosA) = 1 – cos²A = sin²A = 3/4

Step 2 cos²A = 1 – 3/4 = 1/4 ⟹ cosA = 1/2

Step 3 sec A = 1/cosA = 2

∴ sec A = 2

Q10

CBSE 2020

If sec²θ(1 + sinθ)(1 – sinθ) = k, find the value of k.

2 Marks Ch 8 Simplify Expression

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✅ DETAILED SOLUTION

Step 1 (1+sinθ)(1–sinθ) = 1 – sin²θ = cos²θ

Step 2 k = sec²θ × cos²θ = (1/cos²θ) × cos²θ = 1

∴ k = 1

Q11

CBSE 2019

Evaluate: 2 sin²45° + 3 cos²30° – tan²60°.

2 Marks Ch 8 Evaluate using Table

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✅ DETAILED SOLUTION

Step 1 sin 45° = 1/√2, cos 30° = √3/2, tan 60° = √3

Step 2 = 2(1/2) + 3(3/4) – 3 = 1 + 9/4 – 3

Step 3 = 1 + 2.25 – 3 = 0.25 = 1/4

∴ Value = 1/4

Q12

CBSE 2019

If √3 sinθ = cosθ, find the value of (3cos²θ + 2cosθ) / (3cosθ + 2).

2 Marks Ch 8 Find θ → Evaluate

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✅ DETAILED SOLUTION

Step 1 √3 sinθ = cosθ ⟹ tanθ = 1/√3 ⟹ θ = 30°

Step 2 cos 30° = √3/2

Step 3 = (3 × 3/4 + 2 × √3/2) / (3 × √3/2 + 2) = (9/4 + √3) / (3√3/2 + 2)

Step 4 = (9/4 + √3) / (3√3/2 + 2) — Multiply top & bottom by 4:

= (9 + 4√3) / (6√3 + 8)

Alternate (faster): Factor out cosθ from numerator: cosθ(3cosθ+2)/(3cosθ+2) = cosθ = √3/2

∴ Value = √3/2

Q13

CBSE 2019

The shadow of a tower standing on a level ground is found to be 40 m longer when the Sun's altitude is 30° than when it is 60°. Find the height of the tower.

2 Marks Ch 9 Shadow Problem ⭐ CLASSIC

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✅ DETAILED SOLUTION

Step 1 Let height = h, shadow at 60° = x ⟹ shadow at 30° = x + 40

Step 2 tan 60° = h/x ⟹ √3 = h/x ⟹ x = h/√3

Step 3 tan 30° = h/(x+40) ⟹ 1/√3 = h/(x+40) ⟹ x+40 = h√3

Step 4 h/√3 + 40 = h√3 ⟹ 40 = h√3 – h/√3 = h(3–1)/√3 = 2h/√3

Step 5 h = 40√3/2 = 20√3 m

∴ Height of tower = 20√3 m ≈ 34.64 m

Q14

CBSE 2018

In right triangle ABC right-angled at B, if tan A = 1, find the value of 2 sin A · cos A.

2 Marks Ch 8 Right Triangle

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✅ DETAILED SOLUTION

Step 1 tan A = 1 ⟹ A = 45°

Step 2 2 sin 45° × cos 45° = 2 × (1/√2) × (1/√2) = 2 × 1/2 = 1

∴ 2 sin A · cos A = 1

Q15

CBSE 2018

If in a right ΔABC, ∠B = 90°, find the value of sin(A + C).

2 Marks Ch 8 Angle Sum Property

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✅ DETAILED SOLUTION

Step 1 ∠A + ∠B + ∠C = 180° and ∠B = 90°

Step 2 ∠A + ∠C = 90°

Step 3 sin(A + C) = sin 90° = 1

∴ sin(A + C) = 1

📝 SECTION B — 3-MARK QUESTIONS

15 Questions • Short Answer (SA) Type • Board PYQs 2018–2025

Q16

CBSE 2024

Prove that: (cosec θ – sin θ)(sec θ – cos θ)(tan θ + cot θ) = 1

3 Marks Ch 8 Prove Identity ⭐ MOST IMPORTANT

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✅ DETAILED SOLUTION

LHS = (cosecθ – sinθ)(secθ – cosθ)(tanθ + cotθ)

Step 1 = (1/sinθ – sinθ)(1/cosθ – cosθ)(sinθ/cosθ + cosθ/sinθ)

Step 2 = ((1–sin²θ)/sinθ) × ((1–cos²θ)/cosθ) × ((sin²θ+cos²θ)/(sinθ·cosθ))

Step 3 = (cos²θ/sinθ) × (sin²θ/cosθ) × (1/(sinθ·cosθ))

Step 4 = (sin²θ·cos²θ)/(sinθ·cosθ) × 1/(sinθ·cosθ)

Step 5 = (sinθ·cosθ)/(sinθ·cosθ) = 1 = RHS

∴ LHS = RHS. Hence proved.

Q17

CBSE 2025

Prove that: sin²A + cos²A = 1 implies tan²A + 1 = sec²A. Hence find tanA if secA = 5/3 (A is acute).

3 Marks Ch 8 Derive Identity + Apply ⭐ MOST RECENT

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✅ DETAILED SOLUTION

Part 1 Derivation:

sin²A + cos²A = 1 → Divide both sides by cos²A:

sin²A/cos²A + cos²A/cos²A = 1/cos²A

tan²A + 1 = sec²A ✓ Hence proved.

Part 2 Find tan A:

sec A = 5/3 ⟹ sec²A = 25/9

tan²A = sec²A – 1 = 25/9 – 1 = 16/9

tan A = 4/3 (A is acute, so positive)

∴ tan A = 4/3

Q18

CBSE 2023

Prove that: sec A(1 + sin A)(sec A – tan A) = 1

3 Marks Ch 8 Prove Identity ⭐ MOST ASKED TYPE

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✅ DETAILED SOLUTION

LHS = secA(1+sinA)(secA–tanA)

Step 1 = (1/cosA)(1+sinA)(1/cosA – sinA/cosA)

Step 2 = (1/cosA)(1+sinA)((1–sinA)/cosA)

Step 3 = (1+sinA)(1–sinA) / cos²A

Step 4 = (1–sin²A) / cos²A = cos²A / cos²A = 1 = RHS

∴ LHS = RHS. Hence proved.

Q19

CBSE 2023

If 7 sin²A + 3 cos²A = 4, show that tan A = 1/√3.

3 Marks Ch 8 Show that

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✅ DETAILED SOLUTION

Step 1 7sin²A + 3cos²A = 4 ⟹ 7sin²A + 3(1–sin²A) = 4

Step 2 7sin²A + 3 – 3sin²A = 4 ⟹ 4sin²A = 1

Step 3 sin²A = 1/4 ⟹ sinA = 1/2

Step 4 cos²A = 1 – 1/4 = 3/4 ⟹ cosA = √3/2

Step 5 tanA = sinA/cosA = (1/2)/(√3/2) = 1/√3

∴ tan A = 1/√3. Hence shown.

Q20

CBSE 2024

From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Find the height of the cable tower.

3 Marks Ch 9 Height & Distance ⭐ IMPORTANT

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✅ DETAILED SOLUTION

Step 1 Let AB = 7 m (building), CD = tower. Let CE = h (part of tower above building level).

Step 2 Angle of depression of foot = 45° ⟹ tan 45° = 7/BD ⟹ BD = 7 m

Step 3 BD = AE = 7 m (horizontal distance)

Step 4 Angle of elevation of top = 60° ⟹ tan 60° = CE/AE ⟹ √3 = CE/7

Step 5 CE = 7√3 m

Step 6 Total height CD = CE + ED = 7√3 + 7 = 7(√3 + 1) m

∴ Height of cable tower = 7(√3 + 1) m ≈ 19.12 m

Q21

CBSE 2022

Prove that: (sinθ + cosecθ)² + (cosθ + secθ)² = 7 + tan²θ + cot²θ

3 Marks Ch 8 Prove Identity ⭐ HOTS

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✅ DETAILED SOLUTION

LHS Expand both squares:

Step 1 (sinθ+cosecθ)² = sin²θ + cosec²θ + 2sinθ·cosecθ = sin²θ + cosec²θ + 2

Step 2 (cosθ+secθ)² = cos²θ + sec²θ + 2cosθ·secθ = cos²θ + sec²θ + 2

Step 3 LHS = (sin²θ+cos²θ) + (cosec²θ+sec²θ) + 4

Step 4 = 1 + (1+cot²θ) + (1+tan²θ) + 4

Step 5 = 7 + tan²θ + cot²θ = RHS

∴ LHS = RHS. Hence proved.

Q22

CBSE 2022

If sinθ + cosθ = √3, prove that tanθ + cotθ = 1.

3 Marks Ch 8 Prove from Given

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✅ DETAILED SOLUTION

Step 1 sinθ + cosθ = √3 → Square: sin²θ + cos²θ + 2sinθcosθ = 3

Step 2 1 + 2sinθcosθ = 3 ⟹ sinθcosθ = 1

Step 3 tanθ + cotθ = sinθ/cosθ + cosθ/sinθ = (sin²θ+cos²θ)/(sinθcosθ)

Step 4 = 1/(sinθcosθ) = 1/1 = 1

∴ tanθ + cotθ = 1. Hence proved.

Q23

CBSE 2021

Prove that: (1 + cot A – cosec A)(1 + tan A + sec A) = 2

3 Marks Ch 8 Prove Identity ⭐ CLASSIC

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✅ DETAILED SOLUTION

LHS Convert to sin/cos:

Step 1 = (1 + cosA/sinA – 1/sinA)(1 + sinA/cosA + 1/cosA)

Step 2 = ((sinA+cosA–1)/sinA) × ((cosA+sinA+1)/cosA)

Step 3 = ((sinA+cosA)–1)((sinA+cosA)+1) / (sinA·cosA)

Step 4 = ((sinA+cosA)²–1) / (sinA·cosA)

Step 5 = (sin²A+cos²A+2sinAcosA–1) / (sinAcosA)

Step 6 = (1+2sinAcosA–1) / (sinAcosA) = 2sinAcosA / sinAcosA = 2 = RHS

∴ LHS = RHS. Hence proved.

Q24

CBSE 2021

A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower at a uniform speed. 6 seconds later, the angle of depression becomes 60°. Find the time taken by the car to reach the foot of the tower.

3 Marks Ch 9 Speed + Angles ⭐ MOST ASKED TYPE

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✅ DETAILED SOLUTION

Step 1 Let h = height of tower, initial distance = d, distance after 6s = x.

Step 2 tan 30° = h/d ⟹ d = h√3

Step 3 tan 60° = h/x ⟹ x = h/√3

Step 4 Distance covered in 6s = d – x = h√3 – h/√3 = h(3–1)/√3 = 2h/√3

Step 5 Speed = (2h/√3)/6 = h/(3√3)

Step 6 Time to cover remaining x = (h/√3) ÷ (h/(3√3)) = 3 seconds

∴ Time to reach foot = 3 seconds

Q25

CBSE 2020

If cosθ + sinθ = √2 cosθ, show that cosθ – sinθ = √2 sinθ.

3 Marks Ch 8 Show that

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✅ DETAILED SOLUTION

Step 1 cosθ + sinθ = √2 cosθ ⟹ sinθ = (√2–1)cosθ

Step 2 Rationalise: sinθ = (√2–1)cosθ × (√2+1)/(√2+1)

sinθ(√2+1) = (2–1)cosθ = cosθ

Step 3 cosθ = sinθ(√2+1) = √2 sinθ + sinθ

Step 4 cosθ – sinθ = √2 sinθ

∴ cosθ – sinθ = √2 sinθ. Hence proved.

Q26

CBSE 2020

The angles of depression of the top and the bottom of an 8 m tall building from the top of a multi-storeyed building are 30° and 45° respectively. Find the height of the multi-storeyed building and the distance between the two buildings.

3 Marks Ch 9 Two Angles of Depression ⭐ HOTS

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✅ DETAILED SOLUTION

Step 1 Let h = total height of multi-storeyed building, d = distance between buildings.

Step 2 Angle of depression of bottom = 45° ⟹ tan 45° = h/d ⟹ d = h

Step 3 Angle of depression of top of 8m building = 30° ⟹ tan 30° = (h–8)/d

Step 4 1/√3 = (h–8)/h ⟹ h = √3(h–8) = √3h – 8√3

Step 5 h(√3–1) = 8√3 ⟹ h = 8√3/(√3–1) × (√3+1)/(√3+1) = 8√3(√3+1)/2

Step 6 h = 4√3(√3+1) = 4(3+√3) = 12 + 4√3 m

Step 7 d = h = 12 + 4√3 m ≈ 18.93 m

∴ Height = 4(3+√3) m ≈ 18.93 m; Distance = 4(3+√3) m ≈ 18.93 m

Q27

CBSE 2019

If cosec θ + cot θ = q, show that cosec θ – cot θ = 1/q. Hence find sinθ and secθ.

3 Marks Ch 8 Show + Find

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✅ DETAILED SOLUTION

cosec²θ – cot²θ = 1 ⟹ (cosecθ+cotθ)(cosecθ–cotθ) = 1

Step 1 q × (cosecθ–cotθ) = 1 ⟹ cosecθ–cotθ = 1/q ✓

Step 2 Adding: 2cosecθ = q + 1/q = (q²+1)/q ⟹ cosecθ = (q²+1)/2q

Step 3 sinθ = 2q/(q²+1)

Step 4 Subtracting: 2cotθ = q – 1/q = (q²–1)/q ⟹ cotθ = (q²–1)/2q

Step 5 cosθ = sinθ × cotθ = (2q/(q²+1)) × ((q²–1)/2q) = (q²–1)/(q²+1)

Step 6 secθ = (q²+1)/(q²–1)

∴ sinθ = 2q/(q²+1), secθ = (q²+1)/(q²–1)

Q28

CBSE 2019

From a point on the ground, the angle of elevation of the top of a 10 m tall building is 60°. A flagstaff is placed on the top of the building and the angle of elevation of the top of the flagstaff from the same point is 75°. Find the length of the flagstaff. [Use tan 75° = 2 + √3]

3 Marks Ch 9 Building + Flagstaff

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✅ DETAILED SOLUTION

Step 1 Let d = distance from point to building. tan 60° = 10/d ⟹ d = 10/√3

Step 2 Let f = length of flagstaff. tan 75° = (10+f)/d

Step 3 (2+√3) = (10+f)/(10/√3) = √3(10+f)/10

Step 4 10(2+√3)/√3 = 10+f

Step 5 f = 10(2+√3)/√3 – 10 = 10[(2+√3–√3)/√3] = 10 × (2+√3–√3)/√3

Simplify: = 10(2+√3)/√3 – 10 = (20+10√3)/√3 – 10 = (20+10√3–10√3)/√3 = 20/√3 = 20√3/3

∴ Length of flagstaff = 20√3/3 m ≈ 11.55 m

Q29

CBSE 2018

If x = p secθ + q tanθ and y = p tanθ + q secθ, prove that x² – y² = p² – q².

3 Marks Ch 8 Prove Algebraic Identity ⭐ CLASSIC

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✅ DETAILED SOLUTION

Step 1 x² = p²sec²θ + q²tan²θ + 2pq secθ tanθ

Step 2 y² = p²tan²θ + q²sec²θ + 2pq secθ tanθ

Step 3 x²–y² = p²(sec²θ–tan²θ) + q²(tan²θ–sec²θ)

Step 4 = p²(1) + q²(–1) = p² – q²

Key used: sec²θ – tan²θ = 1

∴ x² – y² = p² – q². Hence proved.

Q30

CBSE 2018

A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.

3 Marks Ch 9 Walking Problem

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✅ DETAILED SOLUTION

Step 1 Height above eye level = 30 – 1.5 = 28.5 m

Step 2 Initial distance = d₁: tan 30° = 28.5/d₁ ⟹ d₁ = 28.5√3

Step 3 Final distance = d₂: tan 60° = 28.5/d₂ ⟹ d₂ = 28.5/√3 = 9.5√3

Step 4 Distance walked = d₁ – d₂ = 28.5√3 – 9.5√3 = 19√3 m

∴ Distance walked = 19√3 m ≈ 32.91 m

📐 KEY FORMULAS TO REMEMBER

Trigonometric Ratios (Right △, angle θ):
sin θ = Opposite/Hypotenuse | cos θ = Adjacent/Hypotenuse | tan θ = Opposite/Adjacent
cosec θ = 1/sin θ | sec θ = 1/cos θ | cot θ = 1/tan θ

Standard Angle Values:
sin: 0, 1/2, 1/√2, √3/2, 1 (for 0°, 30°, 45°, 60°, 90°)
cos: 1, √3/2, 1/√2, 1/2, 0 (reverse of sin)
tan: 0, 1/√3, 1, √3, ∞ (sin/cos)

Three Fundamental Identities:
① sin²θ + cos²θ = 1
② 1 + tan²θ = sec²θ
③ 1 + cot²θ = cosec²θ

Useful Products:
tanθ × cotθ = 1 | sinθ × cosecθ = 1 | cosθ × secθ = 1

Heights & Distances:
Angle of elevation → observer looks UP
Angle of depression → observer looks DOWN
tan θ = Height / Distance (most commonly used ratio)

💡 BOARD EXAM TIPS — TRIGONOMETRY

Prove Identity strategy: Always work on LHS only. Convert everything to sinθ and cosθ. Use identities to simplify. Never work on both sides.

Squaring trick: If given sinθ+cosθ = k, square to get 1+2sinθcosθ = k² ⟹ sinθcosθ = (k²–1)/2

a²–b² shortcut: Use (a+b)(a–b) = a²–b² for identity problems involving cosec²θ–cot²θ=1 or sec²θ–tan²θ=1

Heights & Distances: ALWAYS draw a diagram first. Label height, distance, angles. Use tan θ for most problems.

🚫 Complementary Angles DELETED: Questions on sin(90°–θ)=cosθ type will NOT appear. Skip this topic!

Expected marks: 12 marks total (Ch 8 ≈ 7-8 marks + Ch 9 ≈ 4-5 marks).