Trigonometry is one of the most scoring chapters in Class 10 Maths — but only if your basics of trigonometric ratios, specific angle values, and identities are crystal clear. In this blog, our faculty at Convex Classes, Jaipur have solved every single question from NCERT Chapter 8 – Introduction to Trigonometry (Exercise 8.1, 8.2, and 8.3) with full, step-by-step solutions.
Use this as your go-to revision guide before exams, class tests, and board preparation.
Quick Formula Recap (Keep This Handy While Solving)
For a right triangle, right-angled at B, with angle A:
- sin A = Opposite / Hypotenuse
- cos A = Adjacent / Hypotenuse
- tan A = Opposite / Adjacent = sin A / cos A
- cosec A = 1 / sin A
- sec A = 1 / cos A
- cot A = 1 / tan A = cos A / sin A
Trigonometric Identities:
- sin²A + cos²A = 1
- 1 + tan²A = sec²A
- 1 + cot²A = cosec²A
Standard Angle Table:
| ∠A | 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 |
Exercise 8.1 — Full Solutions
Q1. In ∆ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine: (i) sin A, cos A (ii) sin C, cos C
Solution: By Pythagoras theorem: AC² = AB² + BC² = 24² + 7² = 576 + 49 = 625, so AC = 25 cm
(i) sin A = BC/AC = 7/25, cos A = AB/AC = 24/25
(ii) sin C = AB/AC = 24/25, cos C = BC/AC = 7/25
Q2. In the given figure, PR = 13 cm, PQ = 12 cm (right angle at Q). Find tan P − cot R.
Solution: QR² = PR² − PQ² = 13² − 12² = 169 − 144 = 25, so QR = 5 cm
tan P = QR/PQ = 5/12 cot R = QR/PQ = 5/12 (cot R = adjacent/opposite = QR/PQ)
tan P − cot R = 5/12 − 5/12 = 0
Q3. If sin A = 3/4, calculate cos A and tan A.
Solution: Let opposite = 3k, hypotenuse = 4k. Adjacent = √(4k² − 3k²)… i.e. √(16k² − 9k²) = √7 k
cos A = √7 k / 4k = √7/4 tan A = 3k / √7k = 3/√7
Q4. Given 15 cot A = 8, find sin A and sec A.
Solution: cot A = 8/15 → tan A = 15/8 So opposite = 15k, adjacent = 8k Hypotenuse = √(15² + 8²)k = √(225+64)k = √289 k = 17k
sin A = 15k/17k = 15/17 sec A = 17k/8k = 17/8
Q5. Given sec θ = 13/12, calculate all other trigonometric ratios.
Solution: sec θ = hyp/adjacent = 13/12, so hypotenuse = 13k, adjacent = 12k Opposite = √(13² − 12²)k = √(169−144)k = √25 k = 5k
sin θ = 5/13, cos θ = 12/13, tan θ = 5/12, cosec θ = 13/5, cot θ = 12/5
Q6. If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
Solution: Consider two right triangles ABC and PQR, right-angled at B and Q respectively, such that cos A = cos P.
cos A = AB/AC and cos P = PQ/PR
Since cos A = cos P: AB/AC = PQ/PR = k (say)
By Pythagoras theorem: BC = √(AC² − AB²) and QR = √(PR² − PQ²)
BC/QR = √(AC²−AB²) / √(PR²−PQ²) = k (same ratio, using AB = k·AC and PQ = k·PR)
So AB/PQ = AC/PR = BC/QR
By SSS similarity, ∆ABC ~ ∆PQC, and therefore ∠A = ∠P (i.e. ∠A = ∠B). Hence proved.
Q7. If cot θ = 7/8, evaluate: (i) (1+sinθ)(1−sinθ) / (1+cosθ)(1−cosθ) (ii) cot²θ
Solution: cot θ = 7/8 → adjacent = 7k, opposite = 8k, hypotenuse = √(49+64)k = √113 k
sin θ = 8/√113, cos θ = 7/√113
(i) (1+sinθ)(1−sinθ) = 1 − sin²θ (1+cosθ)(1−cosθ) = 1 − cos²θ
So expression = (1−sin²θ)/(1−cos²θ) = cos²θ/sin²θ = cot²θ = (7/8)² = 49/64
(ii) cot²θ = 49/64
Q8. If 3 cot A = 4, check whether (1−tan²A)/(1+tan²A) = cos²A − sin²A or not.
Solution: cot A = 4/3 → tan A = 3/4
LHS: tan²A = 9/16 (1 − 9/16)/(1 + 9/16) = (7/16)/(25/16) = 7/25
RHS: With tan A = 3/4, take opposite = 3k, adjacent = 4k, hypotenuse = 5k cos A = 4/5, sin A = 3/5 cos²A − sin²A = 16/25 − 9/25 = 7/25
Since LHS = RHS = 7/25, the identity holds true.
Q9. In triangle ABC, right-angled at B, if tan A = 1/√3, find: (i) sin A cos C + cos A sin C (ii) cos A cos C − sin A sin C
Solution: Since tan A = 1/√3, ∠A = 30°. Since ∠B = 90°, ∠A + ∠C = 90°, so ∠C = 60°.
(i) sin A cos C + cos A sin C = sin(A+C) = sin 90° = 1
(ii) cos A cos C − sin A sin C = cos(A+C) = cos 90° = 0
Q10. In ∆PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine sin P, cos P and tan P.
Solution: Let QR = x, so PR = 25 − x
By Pythagoras theorem: PR² = PQ² + QR² (25−x)² = 5² + x² 625 − 50x + x² = 25 + x² 625 − 50x = 25 50x = 600 x = 12, so QR = 12 cm and PR = 25 − 12 = 13 cm
sin P = QR/PR = 12/13 cos P = PQ/PR = 5/13 tan P = QR/PQ = 12/5
Q11. State whether the following are true or false. Justify your answer.
(i) The value of tan A is always less than 1. False. For example, tan 60° = √3 ≈ 1.732, which is greater than 1.
(ii) sec A = 12/5 for some value of angle A. True. Since sec A ≥ 1 always (for acute A), and 12/5 = 2.4 ≥ 1, this is a valid value.
(iii) cos A is the abbreviation used for the cosecant of angle A. False. “cos A” stands for cosine of angle A. Cosecant is abbreviated as “cosec A”.
(iv) cot A is the product of cot and A. False. “cot A” is one single symbol for cotangent of angle A; it is not “cot” multiplied by “A” (just like sin A is not sin × A).
(v) sin θ = 4/3 for some angle θ. False. Since the hypotenuse is always the longest side of a right triangle, sin θ can never exceed 1. As 4/3 > 1, this value is not possible.
Exercise 8.2 — Full Solutions
Q1. Evaluate the following:
(i) sin 60° cos 30° + sin 30° cos 60°
= (√3/2)(√3/2) + (1/2)(1/2) = 3/4 + 1/4 = 1
(ii) 2 tan²45° + cos²30° − sin²60°
= 2(1)² + (√3/2)² − (√3/2)² = 2 + 3/4 − 3/4 = 2
(iii) cos45° / (sec30° + cosec30°)
= (1/√2) / (2/√3 + 2) = (1/√2) / [(2+2√3)/√3] = √3 / [√2(2+2√3)] = (3√2 − √6) / 8 (after rationalising)
Answer ≈ 0.224, exact form = (3√2 − √6)/8
(iv) (sin30° + tan45° − cosec60°) / (sec30° + cos60° + cot45°)
Numerator = 1/2 + 1 − 2/√3 = 3/2 − 2/√3 Denominator = 2/√3 + 1/2 + 1 = 2/√3 + 3/2
On simplifying (multiplying throughout by 2√3): Numerator → 3√3 − 4 Denominator → 3√3 + 4
Answer = (3√3 − 4) / (3√3 + 4) ≈ 0.13
(v) [5cos²60° + 4sec²30° − tan²45°] / [sin²30° + cos²30°]
Numerator: 5(1/2)² + 4(2/√3)² − (1)² = 5/4 + 16/3 − 1 LCM = 12: 15/12 + 64/12 − 12/12 = 67/12
Denominator: sin²30° + cos²30° = 1 (identity)
Answer = 67/12
Q2. Choose the correct option and justify your choice:
(i) 2tan30° / (1+tan²30°) = ? Using the identity 2tanθ/(1+tan²θ) = sin2θ, with θ = 30°: = sin 60° Answer: (A) sin 60°
(ii) (1−tan²45°) / (1+tan²45°) = ? Using (1−tan²θ)/(1+tan²θ) = cos2θ, with θ = 45°: = cos 90° = 0 (Direct check: tan45°=1, so (1−1)/(1+1) = 0/2 = 0) Answer: (D) 0
(iii) sin 2A = 2 sin A is true when A = ? Check A = 0°: sin 0° = 0 and 2 sin 0° = 0. Both sides equal. For other options (30°, 45°, 60°), the two sides don’t match. Answer: (A) 0°
(iv) 2tan30° / (1−tan²30°) = ? Using 2tanθ/(1−tan²θ) = tan2θ, with θ = 30°: = tan 60° Answer: (C) tan 60°
Q3. If tan(A+B) = √3 and tan(A−B) = 1/√3; 0° < A+B ≤ 90°; A > B, find A and B.
Solution: tan(A+B) = √3 = tan 60° → A + B = 60° … (1) tan(A−B) = 1/√3 = tan 30° → A − B = 30° … (2)
Adding (1) and (2): 2A = 90° → A = 45° Substituting in (1): 45° + B = 60° → B = 15°
Q4. State whether the following are true or false. Justify your answer.
(i) sin(A+B) = sin A + sin B. False. Take A = B = 30°: sin(60°) = √3/2 ≈ 0.866, but sin30° + sin30° = 0.5+0.5 = 1. Not equal.
(ii) The value of sin θ increases as θ increases. True. In the range 0° to 90°, sin θ increases steadily from 0 to 1 (as shown in the standard angle table).
(iii) The value of cos θ increases as θ increases. False. In the range 0° to 90°, cos θ actually decreases from 1 to 0 as θ increases.
(iv) sin θ = cos θ for all values of θ. False. This is only true at θ = 45° (where both equal 1/√2), not for all values of θ.
(v) cot A is not defined for A = 0°. True. cot A = cos A/sin A, and sin 0° = 0, making the denominator zero — hence cot 0° is undefined.
Exercise 8.3 — Full Solutions
Q1. Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
Solution: Using cosec²A = 1 + cot²A → cosec A = √(1+cot²A)
sin A = 1/cosec A = 1/√(1+cot²A)
cos A = cot A × sin A = cot A/√(1+cot²A)
sec A = 1/cos A = √(1+cot²A)/cot A
tan A = 1/cot A
Q2. Write all the other trigonometric ratios of ∠A in terms of sec A.
Solution: cos A = 1/sec A
Using sin²A = 1−cos²A = 1 − 1/sec²A = (sec²A−1)/sec²A sin A = √(sec²A−1)/sec A
tan A = √(sec²A−1) (from sec²A = 1+tan²A)
cosec A = sec A/√(sec²A−1)
cot A = 1/√(sec²A−1)
Q3. Choose the correct option. Justify your choice.
(i) 9sec²A − 9tan²A = ? = 9(sec²A − tan²A) = 9(1) = 9 [using sec²A − tan²A = 1] Answer: (B) 9
(ii) (1+tanθ+secθ)(1+cotθ−cosecθ) = ? This is a standard identity that simplifies to 2. Answer: (C) 2
(iii) (secA+tanA)(1−sinA) = ? secA + tanA = (1+sinA)/cosA Multiplying by (1−sinA): = (1−sin²A)/cosA = cos²A/cosA = cos A Answer: (D) cos A
(iv) (1+tan²A)/(1+cot²A) = ? = sec²A/cosec²A = (1/cos²A) × (sin²A/1) = sin²A/cos²A = tan²A Answer: (D) tan²A
Q4. Prove the following identities:
(i) (cosecθ − cotθ)² = (1−cosθ)/(1+cosθ)
LHS = (1/sinθ − cosθ/sinθ)² = [(1−cosθ)/sinθ]² = (1−cosθ)²/sin²θ = (1−cosθ)²/(1−cos²θ) [since sin²θ=1−cos²θ] = (1−cosθ)²/[(1−cosθ)(1+cosθ)] = (1−cosθ)/(1+cosθ) = RHS ✔
(ii) cosA/(1+sinA) + (1+sinA)/cosA = 2secA
LHS = [cos²A + (1+sinA)²] / [cosA(1+sinA)] Numerator = cos²A + 1 + 2sinA + sin²A = (sin²A+cos²A) + 1 + 2sinA = 1+1+2sinA = 2(1+sinA)
LHS = 2(1+sinA)/[cosA(1+sinA)] = 2/cosA = 2secA = RHS ✔
(iii) tanθ/(1−cotθ) + cotθ/(1−tanθ) = 1 + secθ·cosecθ
Writing everything in sinθ, cosθ and simplifying:
tanθ/(1−cotθ) = sin²θ / [cosθ(sinθ−cosθ)] cotθ/(1−tanθ) = −cos²θ / [sinθ(sinθ−cosθ)]
Adding: [sin³θ − cos³θ] / [sinθcosθ(sinθ−cosθ)]
Using a³−b³ = (a−b)(a²+ab+b²): sin³θ−cos³θ = (sinθ−cosθ)(1+sinθcosθ) [since sin²θ+cos²θ=1]
So the sum = (1+sinθcosθ)/(sinθcosθ) = 1/(sinθcosθ) + 1 = secθ·cosecθ + 1 = RHS ✔
(iv) (1+secA)/secA = sin²A/(1−cosA)
LHS = (1+secA)/secA = 1/secA + 1 = cosA + 1
RHS = sin²A/(1−cosA) = (1−cos²A)/(1−cosA) = (1−cosA)(1+cosA)/(1−cosA) = 1+cosA
LHS = RHS = 1+cosA ✔
(v) (cosA−sinA+1)/(cosA+sinA−1) = cosecA + cotA
Divide numerator and denominator by sinA: Numerator becomes: cotA + cosecA − 1 Denominator becomes: cotA − cosecA + 1
Let x = cotA, y = cosecA (note y²−x² = 1)
We need to show: (x+y−1)/(x−y+1) = x+y
Cross-multiplying the RHS: (x+y)(x−y+1) = x²−y² + x+y = −1 + x+y = x+y−1
This exactly matches the numerator, so LHS = (x+y)(x−y+1)/(x−y+1) = x+y = cosecA+cotA = RHS ✔
(vi) √[(1+sinA)/(1−sinA)] = secA + tanA
Multiply numerator and denominator inside the root by (1+sinA):
= √[(1+sinA)² / (1−sin²A)] = √[(1+sinA)²/cos²A] = (1+sinA)/cosA
= 1/cosA + sinA/cosA = secA + tanA = RHS ✔
(vii) (sinθ − 2sin³θ) / (2cos³θ − cosθ) = tanθ
LHS = sinθ(1−2sin²θ) / [cosθ(2cos²θ−1)]
Since 1−2sin²θ = 1−2(1−cos²θ) = 2cos²θ−1, both brackets are identical:
LHS = sinθ(2cos²θ−1) / [cosθ(2cos²θ−1)] = sinθ/cosθ = tanθ = RHS ✔
(viii) (sinA+cosecA)² + (cosA+secA)² = 7 + tan²A + cot²A
LHS = sin²A + 2·sinA·cosecA + cosec²A + cos²A + 2·cosA·secA + sec²A
Since sinA·cosecA = 1 and cosA·secA = 1:
= (sin²A+cos²A) + 2 + 2 + cosec²A + sec²A = 1 + 4 + (1+cot²A) + (1+tan²A) = 7 + tan²A + cot²A = RHS ✔
(ix) (cosecA−sinA)(secA−cosA) = 1/(tanA+cotA)
cosecA − sinA = (1−sin²A)/sinA = cos²A/sinA secA − cosA = (1−cos²A)/cosA = sin²A/cosA
Product = (cos²A/sinA) × (sin²A/cosA) = sinA·cosA
RHS: 1/(tanA+cotA) = 1/[(sin²A+cos²A)/(sinAcosA)] = sinA·cosA
LHS = RHS = sinA·cosA ✔
(x) (1+tan²A)/(1+cot²A) = [(1−tanA)/(1−cotA)]² = tan²A
Part 1: (1+tan²A)/(1+cot²A) = sec²A/cosec²A = (1/cos²A)/(1/sin²A) = sin²A/cos²A = tan²A
Part 2: 1−cotA = (sinA−cosA)/sinA 1−tanA = (cosA−sinA)/cosA = −(sinA−cosA)/cosA
(1−tanA)/(1−cotA) = [−(sinA−cosA)/cosA] × [sinA/(sinA−cosA)] = −sinA/cosA = −tanA
Squaring: [(1−tanA)/(1−cotA)]² = tan²A
Since both parts equal tan²A, the identity is proved: (1+tan²A)/(1+cot²A) = [(1−tanA)/(1−cotA)]² = tan²A ✔
Common Mistakes Students Make in This Chapter
- Mixing up opposite and adjacent sides — always identify the angle first, then decide which side is “opposite” and which is “adjacent” to it.
- Forgetting that sin A and cos A can never exceed 1 — a quick sanity check for MCQs and true/false questions.
- Treating sin A, cos A, tan A as sin × A — these are single inseparable symbols, not products.
- Using the wrong identity — remember: sin²A+cos²A=1 works for all A; 1+tan²A=sec²A fails at A=90°; 1+cot²A=cosec²A fails at A=0°.
- Not simplifying using standard angle values (0°, 30°, 45°, 60°, 90°) before attempting long identity proofs — always substitute known values first if the question allows it.
Need More Help With Trigonometry?
This chapter forms the base for Chapter 9 (Applications of Trigonometry — heights and distances), so getting your concepts and identities rock-solid here is essential for board exams.
At Convex Classes, Jaipur, our Class 10 Maths batches focus on exactly this — strong conceptual clarity, exam-pattern practice, and doubt-solving for every NCERT chapter. Reach out to us to join a batch or book a free demo class.
