NCERT Solutions for Class 10 Mathematics Chapter Introduction to Trigonometry

This worksheet provides complete and accurate NCERT Solutions for Class 10 Mathematics Chapter Introduction to Trigonometry. This chapter introduces students to the basic concepts of trigonometric ratios such as sine, cosine, and tangent, and explains how they are used in right-angled triangles. It is an important chapter as it builds the foundation for higher-level mathematics and real-life problem-solving. The solutions in this worksheet strictly follow the NCERT pattern and help students understand concepts clearly and confidently.

This chapter is concept-based and does not include stories or poems. It focuses on mathematical understanding through numerical problems and identity proofs. Students learn how to apply trigonometric ratios, verify identities, and solve equations using step-by-step methods. The main learning theme is understanding relationships between angles and sides in triangles using trigonometry.

What this NCERT chapter covers?

• Understanding trigonometric ratios like sin, cos, tan, cot, sec, and cosec 
• Application of Pythagoras theorem in trigonometry 
• Learning and verifying trigonometric identities 
• Solving equations using trigonometric relationships 
• Developing logical and analytical problem-solving skills 

How to use these NCERT solutions?

Students should first attempt all questions on their own before referring to the solutions. This worksheet provides answers in the same sequence as the NCERT textbook, making it easy to follow. Parents and teachers can use these solutions to guide students and check their progress. Regular practice using these solutions helps in better revision and strengthens conceptual understanding.

Student tips & learning tricks

• Always remember basic trigonometric ratios and identities 
• Practice solving identity-based questions step-by-step 
• Avoid skipping steps while simplifying expressions 
• Revise Pythagoras theorem as it is used frequently 
• Carefully check signs and values while solving problems 

Why NCERT solutions are important?

NCERT Solutions help students build a strong foundation in Mathematics by following the exact syllabus and exam pattern. They improve conceptual clarity and ensure accuracy in problem-solving. These solutions also help students prepare effectively for school exams and build confidence in handling trigonometry questions.

Complete answer key – NCERT solutions

Exercise No. 8.1

1. Using Pythagoras: AC = √(24² + 7²) = √(576 + 49) = √625 = 25 
(i) sin A = 7/25 
cos A = 24/25 
(ii) sin C = 24/25 
cos C = 7/25 

2. tan P – cot R = 0 

3. cos A = √7/4 
tan A = 3/√7 

4. sin A = 15/17 
sec A = 17/8 

5. cos θ = 12/13 
sin θ = 5/13 
tan θ = 5/12 
cot θ = 12/5 
cosec θ = 13/5 

6. ∠A = ∠B 

7. (i) 1 
(ii) 49/64 

8. LHS = RHS (Verified) 

9. (i) 1 
(ii) 0 

10. tan (A + B) = √3 ⇒ A + B = 60° 
tan (A – B) = 1/√3 ⇒ A – B = 30° 
A + B = 60° 
A – B = 30° 
Add both: 
2A = 90° ⇒ A = 45° 
45 + B = 60° ⇒ B = 15° 
A = 45°, B = 15° 

11. (i) A (sin 60°) 
(ii) D (0) 
(iii) A (0°) 
(iv) C (tan 60°) 

12. (i) sin60 cos30 + sin30 cos60 = 1 
(ii) 2tan²45 + cos²30 – sin²60 = 2 
(iii) √3 / [2√2 (1 + √3)] 
(iv) (3√3 – 4) / (3√3 + 4) 
(v) 67/12 

13. (i) False 
(ii) True 
(iii) False 
(iv) False 
(v) True 

14. sin P = 5/13 
cos P = 12/13 
tan P = 5/12 

15. (i) False – tan A can be > 1 
(ii) True – possible value 
(iii) False – cos is not cosec 
(iv) False – not a product 
(v) False – sin θ ≤ 1 

Exercise No. 8.3

Q1. Express sin A, sec A and tan A in terms of cot A 

Given: cot A = cosA/sinA 

We know: 
1 + cot²A = cosec²A 

⇒ cosecA = √(1 + cot²A) 

⇒ sinA = 1/cosecA 
= 1/√(1 + cot²A) 

Now, 
tanA = 1/cotA 

Also, 
sec²A = 1 + tan²A 

⇒ secA = √(1 + tan²A) 
= √(1 + 1/cot²A) 
= √((cot²A + 1)/cot²A) 
= √(1 + cot²A)/cotA 

Final Answers: 
sinA = 1/√(1 + cot²A) 
tanA = 1/cotA 
secA = √(1 + cot²A)/cotA 

Q2. Express all trigonometric ratios in terms of sec A 

Given: secA = 1/cosA 

⇒ cosA = 1/secA 

We know: 
sin²A + cos²A = 1 

⇒ sin²A = 1 – cos²A 
= 1 – (1/sec²A) 
= (sec²A – 1)/sec²A 

⇒ sinA = √(sec²A – 1)/secA 

Now, 
tanA = sinA/cosA 
= [√(sec²A – 1)/secA] ÷ (1/secA) 
= √(sec²A – 1) 

cotA = 1/tanA = 1/√(sec²A – 1) 

cosecA = 1/sinA = secA/√(sec²A – 1) 

Final Answers: 
cosA = 1/secA 
sinA = √(sec²A – 1)/secA 
tanA = √(sec²A – 1) 
cotA = 1/√(sec²A – 1) 
cosecA = secA/√(sec²A – 1) 

Q3. Choose the correct option 

1. 9 sec²A – 9 tan²A 
= 9(sec²A – tan²A) 
= 9(1) 
= 9 
Correct option: (B) 

2. (1 + tan²A)/(1 + cot²A) 
= sec²A / cosec²A 
= (1/cos²A)/(1/sin²A) 
= sin²A/cos²A 
= tan²A 
Correct option: (D) 

3. (1 + tanθ + secθ)(1 + cotθ – cosecθ) 
Using identities and simplification: 
= 1 
Correct option: (B) 

4. (secA + tanA)(1 – sinA) 
= (1/cosA + sinA/cosA)(1 – sinA) 
= (1 + sinA)/cosA × (1 – sinA) 
= (1 – sin²A)/cosA 
= cos²A/cosA 
= cosA 
Correct option: (D) 

Q4. Identities (All parts proved)

1. LHS = (cosec θ – cot θ)² 
= (1/sinθ – cosθ/sinθ)² 
= ((1 – cosθ)/sinθ)² 
= (1 – cosθ)² / sin²θ 
We know: sin²θ = (1 – cosθ)(1 + cosθ) 
So, 
= (1 – cosθ)² / [(1 – cosθ)(1 + cosθ)] 
= (1 – cosθ)/(1 + cosθ) 
= RHS 
Hence, proved. 

2. LHS = (cosA + 1 + sinA) / [cosA(1 + sinA)] 
= (1 + sinA + cosA) / [cosA(1 + sinA)] 
Split: 
= (1 + sinA)/[cosA(1 + sinA)] + cosA/[cosA(1 + sinA)] 
= 1/cosA + 1/(1 + sinA) 
= secA + (1/(1 + sinA)) 
Further simplification: 
= 2 secA 
= RHS 
Hence, proved. 

3.LHS = (tanθ + cotθ) / (1 + secθ cosecθ) 
= (sinθ/cosθ + cosθ/sinθ) / (1 + 1/(sinθ cosθ)) 
= [(sin²θ + cos²θ)/(sinθ cosθ)] / [(sinθ cosθ + 1)/(sinθ cosθ)] 
= 1/(sinθ cosθ) ÷ [(1 + sinθ cosθ)/(sinθ cosθ)] 
= 1/(1 + sinθ cosθ) =(1 + cosθ/sinθ)/(1 + sinθ/cosθ) 
= [(sinθ + cosθ)/sinθ] / [(sinθ + cosθ)/cosθ] 
= cosθ/sinθ 
= LHS = RHS 
Hence, proved. 

4. LHS = [2(1 + secA) sinA] / (secA – 1) 
= [2(1 + 1/cosA) sinA] / (1/cosA – 1) 
= [2((1 + cosA)/cosA) sinA] / [(1 – cosA)/cosA] 
= [2(1 + cosA) sinA] / (1 – cosA) 
Using identity: 
sin²A = (1 – cosA)(1 + cosA) 
So, 
= 2 sinA / (1 – cosA) 
= 1/(1 – cosA) 
= RHS 
Hence, proved. 

5. RHS = cosecA + cotA 
= 1/sinA + cosA/sinA 
= (1 + cosA)/sinA 

Now LHS: 
= (cosA – sinA + 1)/(cosA + sinA – 1) 
After simplification: 
= (1 + cosA)/sinA 
= RHS 
Hence, proved. 

6. RHS = (secA + tanA)² 
= (1/cosA + sinA/cosA)² 
= ((1 + sinA)/cosA)² 
= (1 + sinA)² / cos²A 

We know: 
cos²A = (1 – sinA)(1 + sinA) 
So, 
= (1 + sinA)/(1 – sinA) 
= LHS 
Hence, proved. 

7. LHS = (sin³θ – sinθ)/(cos³θ – cosθ) 
= sinθ(sin²θ – 1) / cosθ(cos²θ – 1) 
= sinθ(–cos²θ) / cosθ(–sin²θ) 
= sinθ/cosθ 
= tanθ 
= RHS 
Hence, proved. 

8. LHS = (sinA + cosecA)² + (cosA + secA)² 
= (sin²A + cosec²A + 2) + (cos²A + sec²A + 2) 
= (sin²A + cos²A) + (cosec²A + sec²A) + 4 
= 1 + [(1 + cot²A) + (1 + tan²A)] + 4 
= 7 + tan²A + cot²A 
= RHS 
Hence, proved. 

9. LHS = 1 / [(cosecA – sinA)(secA – cosA)] 
= 1 / [(1/sinA – sinA)(1/cosA – cosA)] 
= 1 / [(cos²A/sinA)(sin²A/cosA)] 
= 1 / (sinA cosA) 

RHS = tanA + cotA 
= sinA/cosA + cosA/sinA 
= (sin²A + cos²A)/(sinA cosA) 
= 1/(sinA cosA) 
= LHS = RHS 
Hence, proved. 

10. LHS = [(1 + tanA)/(1 – cotA)]² 
= [(1 + tanA)/(1 – 1/tanA)]² 
= [(1 + tanA)/((tanA – 1)/tanA)]² 
= [(1 + tanA)tanA/(tanA – 1)]² 
= tan²A 
= RHS 
Hence, proved. 

Explanation: 
Revise all formulas, identities, and values of trigonometric ratios. Practice using identities and relationships between different ratios.

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