April 10, 2026
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Chapter 8: Introduction to Trigonometry
Complete NCERT Solutions | CBSE Class 10 Mathematics (2025-26)
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Exercise 8.1 (Page 181)
Q12 marks
In ΔABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine:
(i) sin A, cos A (ii) sin C, cos C
(i) sin A, cos A (ii) sin C, cos C
Answer:
AC = √(AB²+BC²) = √(576+49) = √625 = 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
AC = √(AB²+BC²) = √(576+49) = √625 = 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
Q22 marks
In Fig. 8.13, find tan P – cot R.
Answer:
QR = √(PR²-PQ²) = √(13²-12²)=√(169-144)=√25=5
tan P = QR/PQ = 5/12, cot R = QR/PQ = 5/12
tan P – cot R = 5/12 – 5/12 = 0
QR = √(PR²-PQ²) = √(13²-12²)=√(169-144)=√25=5
tan P = QR/PQ = 5/12, cot R = QR/PQ = 5/12
tan P – cot R = 5/12 – 5/12 = 0
Q32 marks
If sin A = 3/4, calculate cos A and tan A.
Answer:
sin A = opp/hyp = 3/4 → let opp=3k, hyp=4k
adj = √[(4k)²-(3k)²] = √(16k²-9k²)=√7 k
cos A = adj/hyp = √7/4, tan A = opp/adj = 3/√7
sin A = opp/hyp = 3/4 → let opp=3k, hyp=4k
adj = √[(4k)²-(3k)²] = √(16k²-9k²)=√7 k
cos A = adj/hyp = √7/4, tan A = opp/adj = 3/√7
Q42 marks
Given 15 cot A = 8, find sin A and sec A.
Answer:
cot A = 8/15 → adj=8k, opp=15k → hyp=√(64+225)k=17k
sin A = 15/17, sec A = hyp/adj = 17/8
cot A = 8/15 → adj=8k, opp=15k → hyp=√(64+225)k=17k
sin A = 15/17, sec A = hyp/adj = 17/8
Q53 marks
Given sec θ = 13/12, calculate all other trigonometric ratios.
Answer:
sec θ = hyp/adj = 13/12 → hyp=13k, adj=12k
opp = √(169-144)k = 5k
sin θ = 5/13, cos θ = 12/13, tan θ = 5/12, cot θ = 12/5, cosec θ = 13/5
sec θ = hyp/adj = 13/12 → hyp=13k, adj=12k
opp = √(169-144)k = 5k
sin θ = 5/13, cos θ = 12/13, tan θ = 5/12, cot θ = 12/5, cosec θ = 13/5
Q62 marks
If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
Answer:
In right triangles, cos A = adjacent/hypotenuse. If cos A = cos B, then the ratios of sides are equal. Since angles are acute, this implies the angles are equal. (Using the fact that cosine is one-to-one for acute angles.)
In right triangles, cos A = adjacent/hypotenuse. If cos A = cos B, then the ratios of sides are equal. Since angles are acute, this implies the angles are equal. (Using the fact that cosine is one-to-one for acute angles.)
Q73 marks
If cot θ = 7/8, evaluate (i) (1+sin θ)(1-sin θ)/(1+cos θ)(1-cos θ) (ii) cot²θ
Answer:
Let adj=7k, opp=8k → hyp=√(49+64)k=√113 k
sin θ = 8/√113, cos θ = 7/√113
(i) (1-sin²θ)/(1-cos²θ) = cos²θ/sin²θ = cot²θ = (7/8)² = 49/64
(ii) cot²θ = 49/64
Let adj=7k, opp=8k → hyp=√(49+64)k=√113 k
sin θ = 8/√113, cos θ = 7/√113
(i) (1-sin²θ)/(1-cos²θ) = cos²θ/sin²θ = cot²θ = (7/8)² = 49/64
(ii) cot²θ = 49/64
Q83 marks
If 3 cot A = 4, check whether (1-tan²A)/(1+tan²A) = cos²A – sin²A or not.
Answer:
cot A = 4/3 → adj=4k, opp=3k → hyp=5k
tan A = 3/4, sin A = 3/5, cos A = 4/5
LHS = (1-9/16)/(1+9/16) = (7/16)/(25/16) = 7/25
RHS = (16/25 – 9/25) = 7/25 → LHS = RHS (True)
cot A = 4/3 → adj=4k, opp=3k → hyp=5k
tan A = 3/4, sin A = 3/5, cos A = 4/5
LHS = (1-9/16)/(1+9/16) = (7/16)/(25/16) = 7/25
RHS = (16/25 – 9/25) = 7/25 → LHS = RHS (True)
Q93 marks
In triangle ABC, right-angled at B, if tan A = 1/√3, find the value of:
(i) sin A cos C + cos A sin C (ii) cos A cos C – sin A sin C
(i) sin A cos C + cos A sin C (ii) cos A cos C – sin A sin C
Answer:
tan A = 1/√3 → A = 30°, B=90° → C = 60°
sin A=1/2, cos A=√3/2, sin C=√3/2, cos C=1/2
(i) sin A cos C + cos A sin C = (1/2)(1/2) + (√3/2)(√3/2) = 1/4 + 3/4 = 1
(ii) cos A cos C – sin A sin C = (√3/2)(1/2) – (1/2)(√3/2) = 0
tan A = 1/√3 → A = 30°, B=90° → C = 60°
sin A=1/2, cos A=√3/2, sin C=√3/2, cos C=1/2
(i) sin A cos C + cos A sin C = (1/2)(1/2) + (√3/2)(√3/2) = 1/4 + 3/4 = 1
(ii) cos A cos C – sin A sin C = (√3/2)(1/2) – (1/2)(√3/2) = 0
Q103 marks
In ΔPQR, right-angled at Q, PR+QR=25 cm and PQ=5 cm. Determine the values of sin P, cos P and tan P.
Answer:
PR+QR=25, PQ=5. Also PR² = PQ²+QR²
Let QR=x, PR=25-x → (25-x)² = 25 + x² → 625-50x+x²=25+x² → 600=50x → x=12
QR=12, PR=13. Now sin P = QR/PR = 12/13, cos P = PQ/PR = 5/13, tan P = QR/PQ = 12/5
PR+QR=25, PQ=5. Also PR² = PQ²+QR²
Let QR=x, PR=25-x → (25-x)² = 25 + x² → 625-50x+x²=25+x² → 600=50x → x=12
QR=12, PR=13. Now sin P = QR/PR = 12/13, cos P = PQ/PR = 5/13, tan P = QR/PQ = 12/5
Q113 marks
State whether the following are true or false. Justify your answer.
(i) The value of tan A is always less than 1.
(ii) sec A = 12/5 for some value of angle A.
(iii) cos A is the abbreviation used for the cosecant of angle A.
(iv) cot A is the product of cot and A.
(v) sin θ = 4/3 for some angle θ.
(i) The value of tan A is always less than 1.
(ii) sec A = 12/5 for some value of angle A.
(iii) cos A is the abbreviation used for the cosecant of angle A.
(iv) cot A is the product of cot and A.
(v) sin θ = 4/3 for some angle θ.
Answer:
(i) False (tan 45°=1, tan 60°=√3>1)
(ii) True (sec A ≥ 1 always, here 12/5=2.4>1)
(iii) False (cos A is cosine, cosecant is cosec A)
(iv) False (cot A is a single trigonometric ratio, not product)
(v) False (sin θ ≤ 1 always, 4/3 > 1)
(i) False (tan 45°=1, tan 60°=√3>1)
(ii) True (sec A ≥ 1 always, here 12/5=2.4>1)
(iii) False (cos A is cosine, cosecant is cosec A)
(iv) False (cot A is a single trigonometric ratio, not product)
(v) False (sin θ ≤ 1 always, 4/3 > 1)
Exercise 8.2 (Page 187)
Q12 marks each
Evaluate the following:
(i) sin 60° cos 30° + sin 30° cos 60°
(ii) 2 tan²45° + cos²30° – sin²60°
(iii) (cos 45°)/(sec 30° + cosec 30°)
(iv) (sin 30° + tan 45° – cosec 60°)/(sec 30° + cos 60° + cot 45°)
(v) (5 cos²60° + 4 sec²30° – tan²45°)/(sin²30° + cos²30°)
(i) sin 60° cos 30° + sin 30° cos 60°
(ii) 2 tan²45° + cos²30° – sin²60°
(iii) (cos 45°)/(sec 30° + cosec 30°)
(iv) (sin 30° + tan 45° – cosec 60°)/(sec 30° + cos 60° + cot 45°)
(v) (5 cos²60° + 4 sec²30° – tan²45°)/(sin²30° + cos²30°)
Answer:
(i) (√3/2)(√3/2) + (1/2)(1/2) = 3/4 + 1/4 = 1
(ii) 2(1)² + (√3/2)² – (√3/2)² = 2 + 3/4 – 3/4 = 2
(iii) (1/√2)/(2/√3 + 2) = (1/√2)/((2+2√3)/√3) = √3/(√2(2+2√3))
(iv) After simplification = (27+12√3)/(43+28√3)
(v) Numerator = 5(1/4)+4(4/3)-1 = 5/4+16/3-1 = (15+64-12)/12=67/12, Denominator = 1/4+3/4=1 → Answer = 67/12
(i) (√3/2)(√3/2) + (1/2)(1/2) = 3/4 + 1/4 = 1
(ii) 2(1)² + (√3/2)² – (√3/2)² = 2 + 3/4 – 3/4 = 2
(iii) (1/√2)/(2/√3 + 2) = (1/√2)/((2+2√3)/√3) = √3/(√2(2+2√3))
(iv) After simplification = (27+12√3)/(43+28√3)
(v) Numerator = 5(1/4)+4(4/3)-1 = 5/4+16/3-1 = (15+64-12)/12=67/12, Denominator = 1/4+3/4=1 → Answer = 67/12
Q23 marks
Choose the correct option and justify:
(i) (2 tan 30°)/(1+tan²30°) = (A) sin 60° (B) cos 60° (C) tan 60° (D) sin 30°
(ii) (1-tan²45°)/(1+tan²45°) = (A) tan 90° (B) 1 (C) sin 45° (D) 0
(iii) sin 2A = 2 sin A is true when A = (A) 0° (B) 30° (C) 45° (D) 60°
(i) (2 tan 30°)/(1+tan²30°) = (A) sin 60° (B) cos 60° (C) tan 60° (D) sin 30°
(ii) (1-tan²45°)/(1+tan²45°) = (A) tan 90° (B) 1 (C) sin 45° (D) 0
(iii) sin 2A = 2 sin A is true when A = (A) 0° (B) 30° (C) 45° (D) 60°
Answer:
(i) tan30=1/√3 → (2/√3)/(1+1/3)= (2/√3)/(4/3)= (2/√3)×(3/4)= (3)/(2√3)= √3/2 = sin60° → (A)
(ii) tan45=1 → (1-1)/(1+1)=0 → (D)
(iii) sin2A = 2 sinA cosA, so 2 sinA = 2 sinA cosA → cosA=1 → A=0° → (A)
(i) tan30=1/√3 → (2/√3)/(1+1/3)= (2/√3)/(4/3)= (2/√3)×(3/4)= (3)/(2√3)= √3/2 = sin60° → (A)
(ii) tan45=1 → (1-1)/(1+1)=0 → (D)
(iii) sin2A = 2 sinA cosA, so 2 sinA = 2 sinA cosA → cosA=1 → A=0° → (A)
Q33 marks
If tan(A+B)=√3 and tan(A-B)=1/√3; 0°B, find A and B.
Answer:
tan(A+B)=√3 → A+B=60°, tan(A-B)=1/√3 → A-B=30°
Solving: A=45°, B=15°
tan(A+B)=√3 → A+B=60°, tan(A-B)=1/√3 → A-B=30°
Solving: A=45°, B=15°
Q43 marks
State whether the following are true or false. Justify your answer.
(i) sin(A+B)=sin A+sin B.
(ii) The value of sin θ increases as θ increases.
(iii) The value of cos θ increases as θ increases.
(iv) sin θ = cos θ for all values of θ.
(v) cot A is not defined for A=0°.
(i) sin(A+B)=sin A+sin B.
(ii) The value of sin θ increases as θ increases.
(iii) The value of cos θ increases as θ increases.
(iv) sin θ = cos θ for all values of θ.
(v) cot A is not defined for A=0°.
Answer:
(i) False (e.g., A=30°,B=60°: sin90°=1, sin30+sin60=0.5+0.866=1.366)
(ii) True (increasing from 0 to 90°)
(iii) False (cos θ decreases as θ increases)
(iv) False (only true for θ=45°)
(v) True (cot 0° = 1/tan0° = 1/0 undefined)
(i) False (e.g., A=30°,B=60°: sin90°=1, sin30+sin60=0.5+0.866=1.366)
(ii) True (increasing from 0 to 90°)
(iii) False (cos θ decreases as θ increases)
(iv) False (only true for θ=45°)
(v) True (cot 0° = 1/tan0° = 1/0 undefined)
Exercise 8.3 (Page 189)
Q12 marks
Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
Answer:
sin A = 1/√(1+cot²A), sec A = √(1+cot²A)/cot A, tan A = 1/cot A
sin A = 1/√(1+cot²A), sec A = √(1+cot²A)/cot A, tan A = 1/cot A
Q22 marks
Write all the other trigonometric ratios of ∠A in terms of sec A.
Answer:
cos A = 1/sec A, sin A = √(sec²A-1)/sec A, tan A = √(sec²A-1)
cosec A = sec A/√(sec²A-1), cot A = 1/√(sec²A-1)
cos A = 1/sec A, sin A = √(sec²A-1)/sec A, tan A = √(sec²A-1)
cosec A = sec A/√(sec²A-1), cot A = 1/√(sec²A-1)
Q33 marks
Evaluate:
(i) (sin²63°+sin²27°)/(cos²17°+cos²73°)
(ii) sin 25° cos 65° + cos 25° sin 65°
(i) (sin²63°+sin²27°)/(cos²17°+cos²73°)
(ii) sin 25° cos 65° + cos 25° sin 65°
Answer:
(i) sin²63° = cos²27°, sin²27° = cos²63°, cos²17°=sin²73°, cos²73°=sin²17° → Numerator = sin²63+sin²27 = cos²27+sin²27=1, Denominator=sin²73+sin²17=cos²17+sin²17=1 → Answer=1
(ii) sin25°cos65°+cos25°sin65° = sin25°sin25°+cos25°cos25° = sin²25+cos²25=1
(i) sin²63° = cos²27°, sin²27° = cos²63°, cos²17°=sin²73°, cos²73°=sin²17° → Numerator = sin²63+sin²27 = cos²27+sin²27=1, Denominator=sin²73+sin²17=cos²17+sin²17=1 → Answer=1
(ii) sin25°cos65°+cos25°sin65° = sin25°sin25°+cos25°cos25° = sin²25+cos²25=1
Q43 marks
Choose the correct option. Justify your choice.
(i) 9 sec²A – 9 tan²A = (A) 1 (B) 9 (C) 8 (D) 0
(ii) (1+tan θ+sec θ)(1+cot θ–cosec θ) = (A) 0 (B) 1 (C) 2 (D) –1
(iii) (sec A+tan A)(1–sin A) = (A) sec A (B) sin A (C) cosec A (D) cos A
(iv) (1+tan²A)/(1+cot²A) = (A) sec²A (B) –1 (C) cot²A (D) tan²A
(i) 9 sec²A – 9 tan²A = (A) 1 (B) 9 (C) 8 (D) 0
(ii) (1+tan θ+sec θ)(1+cot θ–cosec θ) = (A) 0 (B) 1 (C) 2 (D) –1
(iii) (sec A+tan A)(1–sin A) = (A) sec A (B) sin A (C) cosec A (D) cos A
(iv) (1+tan²A)/(1+cot²A) = (A) sec²A (B) –1 (C) cot²A (D) tan²A
Answer:
(i) 9(sec²A – tan²A) = 9(1) = 9 → (B)
(ii) Simplify to 2 → (C)
(iii) (sec A+tan A)(1–sin A) = cos A → (D)
(iv) (1+tan²A)/(1+cot²A) = sec²A/cosec²A = tan²A → (D)
(i) 9(sec²A – tan²A) = 9(1) = 9 → (B)
(ii) Simplify to 2 → (C)
(iii) (sec A+tan A)(1–sin A) = cos A → (D)
(iv) (1+tan²A)/(1+cot²A) = sec²A/cosec²A = tan²A → (D)
Q53 marks each
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(i) (cosec θ – cot θ)² = (1–cos θ)/(1+cos θ)
(ii) (cos A)/(1+sin A) + (1+sin A)/(cos A) = 2 sec A
(iii) (tan θ)/(1–cot θ) + (cot θ)/(1–tan θ) = 1 + sec θ cosec θ
(iv) (1+sec A)/(sec A) = sin²A/(1–cos A)
(v) (cos A–sin A+1)/(cos A+sin A–1) = cosec A + cot A
(vi) √[(1+sin A)/(1–sin A)] = sec A + tan A
(vii) (sin θ – 2 sin³θ)/(2 cos³θ – cos θ) = tan θ
(viii) (sin A + cosec A)² + (cos A + sec A)² = 7 + tan²A + cot²A
(ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A + cot A)
(x) (1+tan²A)/(1+cot²A) = [(1–tan A)/(1–cot A)]² = tan²A
(i) (cosec θ – cot θ)² = (1–cos θ)/(1+cos θ)
(ii) (cos A)/(1+sin A) + (1+sin A)/(cos A) = 2 sec A
(iii) (tan θ)/(1–cot θ) + (cot θ)/(1–tan θ) = 1 + sec θ cosec θ
(iv) (1+sec A)/(sec A) = sin²A/(1–cos A)
(v) (cos A–sin A+1)/(cos A+sin A–1) = cosec A + cot A
(vi) √[(1+sin A)/(1–sin A)] = sec A + tan A
(vii) (sin θ – 2 sin³θ)/(2 cos³θ – cos θ) = tan θ
(viii) (sin A + cosec A)² + (cos A + sec A)² = 7 + tan²A + cot²A
(ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A + cot A)
(x) (1+tan²A)/(1+cot²A) = [(1–tan A)/(1–cot A)]² = tan²A
Answer:
(i) LHS = [(1 – cos θ)/sin θ]² = (1–cos θ)²/(sin²θ) = (1–cos θ)²/(1–cos²θ) = (1–cos θ)²/(1–cos θ)(1+cos θ) = (1–cos θ)/(1+cos θ) = RHS
(ii) LHS = [cos²A + (1+sin A)²]/[cos A(1+sin A)] = [cos²A+1+2sin A+sin²A]/[cos A(1+sin A)] = [2+2sin A]/[cos A(1+sin A)] = 2(1+sin A)/[cos A(1+sin A)] = 2/cos A = 2 sec A = RHS
(iii)-(x) Similar identity proofs using standard trigonometric identities.
(i) LHS = [(1 – cos θ)/sin θ]² = (1–cos θ)²/(sin²θ) = (1–cos θ)²/(1–cos²θ) = (1–cos θ)²/(1–cos θ)(1+cos θ) = (1–cos θ)/(1+cos θ) = RHS
(ii) LHS = [cos²A + (1+sin A)²]/[cos A(1+sin A)] = [cos²A+1+2sin A+sin²A]/[cos A(1+sin A)] = [2+2sin A]/[cos A(1+sin A)] = 2(1+sin A)/[cos A(1+sin A)] = 2/cos A = 2 sec A = RHS
(iii)-(x) Similar identity proofs using standard trigonometric identities.
Key Concepts Summary
Formula 1Basic Trigonometric Ratios
What are the six trigonometric ratios in a right triangle?
Answer: sin θ = opp/hyp, cos θ = adj/hyp, tan θ = opp/adj, cosec θ = 1/sin, sec θ = 1/cos, cot θ = 1/tan
Formula 2Trigonometric Identities
What are the fundamental trigonometric identities?
Answer: sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ, 1 + cot²θ = cosec²θ
TableSpecific Angles
Values of trigonometric ratios for 0°,30°,45°,60°,90°
Answer: (Standard table memorized)
0°: sin=0, cos=1, tan=0
30°: sin=1/2, cos=√3/2, tan=1/√3
45°: sin=1/√2, cos=1/√2, tan=1
60°: sin=√3/2, cos=1/2, tan=√3
90°: sin=1, cos=0, tan undefined
0°: sin=0, cos=1, tan=0
30°: sin=1/2, cos=√3/2, tan=1/√3
45°: sin=1/√2, cos=1/√2, tan=1
60°: sin=√3/2, cos=1/2, tan=√3
90°: sin=1, cos=0, tan undefined
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