NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry

Exercise 8.1 11 Questions – Finding Trigonometric Ratios

Q 1Basic Ratios

In �"ABC right-angled at B, AB = 24 cm, BC = 7 cm. Determine: (i) sin A, cos A (ii) sin C, cos C

By Pythagoras: AC² = AB² + BC² = 24² + 7² = 576 + 49 = 625 �' AC = 25 cm

For ∠A: Opp=BC=7, Adj=AB=24, Hyp=25. sin A=7/25, cos A=24/25

For ∠C: Opp=AB=24, Adj=BC=7, Hyp=25. sin C=24/25, cos C=7/25

Q 2–5Finding Ratios

Q2: If sin A = 3/4, find cos A and tan A. �' cos A = √(1�'9/16) = √7/4. tan A = 3/√7

Q3: If cot θ = 7/8, find (1+sinθ)(1�'sinθ)/(1+cosθ)(1�'cosθ). �' sinθ=8/√113, cosθ=7/√113. Answer = 49/64

Q5: Given sec θ = 13/12, find all other trigonometric ratios. �' cos θ=12/13, sin θ=5/13, tan θ=5/12, cot θ=12/5, cosec θ=13/5

Q 6–11Verification

Q6: If ∠A and ∠B are acute such that cos A = cos B, prove ∠A = ∠B. �' Since cosine is decreasing in (0,90°), equal cosine �' equal angle.

Q8: If 3 cot A = 4, check whether (1�'tan²A)/(1+tan²A) = cos²A �' sin²A. �' tan A = 3/4, sin A = 3/5, cos A = 4/5. LHS = (1�'9/16)/(1+9/16) = 7/25. RHS = 16/25�'9/25 = 7/25 �"

Q10: In �"PQR right at Q, PR+QR=25cm, PQ=5cm. Find sin P, cos P, tan P. �' PR²=PQ²+QR². PR=QR+25�'PR. Solve: PR=13, QR=12. sin P=12/13, cos P=5/13, tan P=12/5

Exercise 8.2 4 Questions – Standard Angle Values

�"� Standard Angle Table: sin 0°=0, sin 30°=1/2, sin 45°=1/√2, sin 60°=√3/2, sin 90°=1. cos 0°=1, cos 30°=√3/2, cos 45°=1/√2, cos 60°=1/2, cos 90°=0. tan 0°=0, tan 30°=1/√3, tan 45°=1, tan 60°=√3, tan 90°=undefined.

Q 1Evaluate

Evaluate: (i) sin 60° cos 30° + sin 30° cos 60° = (√3/2)(√3/2) + (1/2)(1/2) = 3/4+1/4 = 1

(ii) 2 tan² 45° + cos² 30° �' sin² 60° = 2(1)+(3/4)�'(3/4) = 2

Q 2–4MCQ & Summary

Q2: Choose correct option: (i) 2 tan 30°/(1+tan²30°) = 2(1/√3)/(1+1/3) = (2/√3)�—(3/4) = √3/2 = sin 60°

Q4: State True/False: (i) sin(A+B)=sinA+sinB �' False. (ii) sinθ increases as θ increases from 0° to 90° �' True. (iii) cosθ decreases as θ increases �' True. (iv) sinθ=cosθ for all θ �' False (only at 45°). (v) cot A is not defined for A=0° �' True.

Exercise 8.3 7 Questions – Trigonometric Identities

�"� Fundamental Identities: sin²θ+cos²θ=1 · sec²θ=1+tan²θ · cosec²θ=1+cot²θ

Q 1Proofs

(i) (cosec θ �' cot θ)² = (1 �' cos θ)/(1 + cos θ)

LHS = (1/sinθ �' cosθ/sinθ)² = ((1�'cosθ)/sinθ)² = (1�'cosθ)²/sin²θ = (1�'cosθ)²/(1�'cos²θ) = (1�'cosθ)/(1+cosθ) = RHS �"

(ii) cos A/(1+sin A) + (1+sin A)/cos A = 2 sec A

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 �"

Q 2–5More Proofs

Q4: Prove (1+secA)/secA = sin²A/(1�'cosA). Q5: (sinA+cosecA)²+(cosA+secA)² = 7+tan²A+cot²A

Q7: Express sin 67°+cos 75° in terms of ratios of angles between 0° and 45°. �' sin 67° = cos 23°, cos 75° = sin 15°. So cos 23° + sin 15°

Exercise 8.4 5 Questions – Advanced Identities

Q 1Express in Terms

Express the trigonometric ratios sin A, sec A, and tan A in terms of cot A. �' sin A = 1/√(1+cot²A), sec A = √(1+cot²A)/cot A, tan A = 1/cot A

Q 2–5Identity Proofs

Q2: Write all other ratios of ∠A in terms of sec A.

Q4: Prove (1+tan²A)/(1+cot²A) = (1�'tanA/1�'cotA)² = tan²A. Choose correct option �' (D) tan²A

Q5: Prove: √(1+sinA)/(1�'sinA) = secA + tanA. LHS = √(1+sinA)²/(1�'sin²A) = (1+sinA)/cosA = secA + tanA = RHS �"

�"� Proving Strategy: Start from the more complex side, convert everything to sin and cos, use sin²θ+cos²θ=1 to simplify, and factor to reach the other side.

Practice This Chapter

Generate unlimited practice questions for Introduction to Trigonometry with Examzo � AI-powered, faculty-validated, zero hallucinations.

Try PaperGenie → Take a Quick Test →

Class 10 Maths – Chapter 8: Introduction to Trigonometry

Exercise 8.1 11 Questions – Finding Trigonometric Ratios

Exercise 8.2 4 Questions – Standard Angle Values

Exercise 8.3 7 Questions – Trigonometric Identities

Exercise 8.4 5 Questions – Advanced Identities

Practice This Chapter

Trigonometry Practice?