Current Affairs updated daily
KalviGavaniKalviGavani
Class 9 · Tamil Nadu State Board

Samacheer Kalvi 9th Maths Trigonometry Exercise Solutions | KalviGavani

TN Class 9 Maths Trigonometry exercise solutions — trigonometric ratios, specific angle values, identities, complementary angles, full step-by-step answers, MCQ, word problems.

Updated 12 June 202628 min read

How to Use This Page

This page contains complete step-by-step solutions for all exercises in Samacheer Kalvi 9th Maths Chapter 6 (Trigonometry). Work through each exercise on your own first, then compare your method and answer. In trigonometry, showing every step is essential — small sign or ratio errors lead to wrong answers.

Key Concepts (Quick Recap)

  • Trigonometric Ratios: In a right triangle with angle theta: sin theta = opposite/hypotenuse, cos theta = adjacent/hypotenuse, tan theta = opposite/adjacent.
  • Reciprocal Ratios: cosec theta = 1/sin theta, sec theta = 1/cos theta, cot theta = 1/tan theta.
  • Ratio Relations: tan theta = sin theta / cos theta, cot theta = cos theta / sin theta.
  • Specific Angle Values:
  • sin 0° = 0, sin 30° = 1/2, sin 45° = 1/sqrt(2), sin 60° = sqrt(3)/2, sin 90° = 1
  • cos 0° = 1, cos 30° = sqrt(3)/2, cos 45° = 1/sqrt(2), cos 60° = 1/2, cos 90° = 0
  • tan 0° = 0, tan 30° = 1/sqrt(3), tan 45° = 1, tan 60° = sqrt(3), tan 90° = undefined
  • Trigonometric Identities:
  • sin²θ + cos²θ = 1
  • 1 + tan²θ = sec²θ
  • 1 + cot²θ = cosec²θ
  • Complementary Angles: sin(90° − θ) = cos θ, cos(90° − θ) = sin θ, tan(90° − θ) = cot θ.

Exercise 6.1 — Trigonometric Ratios

Q1. In a right triangle ABC, right-angled at B, if AB = 3, BC = 4, find all trigonometric ratios for angle A.

Solution
Given: Triangle ABC, right-angled at B. AB = 3 (adjacent to angle A) BC = 4 (opposite to angle A) Find AC (hypotenuse) using Pythagoras: AC = sqrt(AB² + BC²) = sqrt(9 + 16) = sqrt(25) = 5 For angle A: opposite = BC = 4 adjacent = AB = 3 hypotenuse = AC = 5 sin A = opposite/hypotenuse = 4/5 cos A = adjacent/hypotenuse = 3/5 tan A = opposite/adjacent = 4/3 cosec A = hypotenuse/opposite = 5/4 sec A = hypotenuse/adjacent = 5/3 cot A = adjacent/opposite = 3/4
Answersin A = 4/5, cos A = 3/5, tan A = 4/3, cosec A = 5/4, sec A = 5/3, cot A = 3/4.

Q2. In triangle PQR, right-angled at Q, PR = 13, PQ = 5. Find all trigonometric ratios for angle R.

Solution
Given: Triangle PQR, right-angled at Q. PR = 13 (hypotenuse) PQ = 5 (opposite to angle R) Find QR (adjacent to angle R) using Pythagoras: QR = sqrt(PR² − PQ²) = sqrt(169 − 25) = sqrt(144) = 12 For angle R: opposite = PQ = 5 adjacent = QR = 12 hypotenuse = PR = 13 sin R = opposite/hypotenuse = 5/13 cos R = adjacent/hypotenuse = 12/13 tan R = opposite/adjacent = 5/12 cosec R = hypotenuse/opposite = 13/5 sec R = hypotenuse/adjacent = 13/12 cot R = adjacent/opposite = 12/5
Answersin R = 5/13, cos R = 12/13, tan R = 5/12, cosec R = 13/5, sec R = 13/12, cot R = 12/5.

Q3. If sin theta = 3/5, find the other trigonometric ratios.

Solution
Given: sin theta = 3/5 = opposite/hypotenuse Let opposite = 3, hypotenuse = 5. Find adjacent using Pythagoras: adjacent = sqrt(hypotenuse² − opposite²) = sqrt(25 − 9) = sqrt(16) = 4 cos theta = adjacent/hypotenuse = 4/5 tan theta = opposite/adjacent = 3/4 cosec theta = hypotenuse/opposite = 5/3 sec theta = hypotenuse/adjacent = 5/4 cot theta = adjacent/opposite = 4/3
Answercos theta = 4/5, tan theta = 3/4, cosec theta = 5/3, sec theta = 5/4, cot theta = 4/3.

Q4. If cos theta = 12/13, find sin theta and tan theta.

Solution
Given: cos theta = 12/13 = adjacent/hypotenuse Let adjacent = 12, hypotenuse = 13. Find opposite using Pythagoras: opposite = sqrt(hypotenuse² − adjacent²) = sqrt(169 − 144) = sqrt(25) = 5 sin theta = opposite/hypotenuse = 5/13 tan theta = opposite/adjacent = 5/12
Answersin theta = 5/13, tan theta = 5/12.

Q5. If tan theta = 5/12, find all other trigonometric ratios.

Solution
Given: tan theta = 5/12 = opposite/adjacent Let opposite = 5, adjacent = 12. Find hypotenuse using Pythagoras: hypotenuse = sqrt(5² + 12²) = sqrt(25 + 144) = sqrt(169) = 13 sin theta = opposite/hypotenuse = 5/13 cos theta = adjacent/hypotenuse = 12/13 cosec theta = hypotenuse/opposite = 13/5 sec theta = hypotenuse/adjacent = 13/12 cot theta = adjacent/opposite = 12/5
Answersin theta = 5/13, cos theta = 12/13, cosec theta = 13/5, sec theta = 13/12, cot theta = 12/5.

Q6. If sin theta = 1/3, find cos theta, tan theta, and sec theta.

Solution
Given: sin theta = 1/3 = opposite/hypotenuse Let opposite = 1, hypotenuse = 3. Find adjacent: adjacent = sqrt(hyp² − opp²) = sqrt(9 − 1) = sqrt(8) = 2sqrt(2) cos theta = adjacent/hypotenuse = 2sqrt(2)/3 tan theta = opposite/adjacent = 1/(2sqrt(2)) = sqrt(2)/4 (Rationalised: 1/(2sqrt(2)) × sqrt(2)/sqrt(2) = sqrt(2)/4) sec theta = hypotenuse/adjacent = 3/(2sqrt(2)) = 3sqrt(2)/4 (Rationalised: 3/(2sqrt(2)) × sqrt(2)/sqrt(2) = 3sqrt(2)/4)
Answercos theta = 2sqrt(2)/3, tan theta = sqrt(2)/4, sec theta = 3sqrt(2)/4.

Q7. In triangle ABC, right-angled at C, AC = 8, BC = 15. Find sin A, cos A, tan A.

Solution
Given: Triangle ABC, right-angled at C. AC = 8 (adjacent to angle A) BC = 15 (opposite to angle A) Find AB (hypotenuse): AB = sqrt(AC² + BC²) = sqrt(64 + 225) = sqrt(289) = 17 For angle A: opposite = BC = 15 adjacent = AC = 8 hypotenuse = AB = 17 sin A = 15/17 cos A = 8/17 tan A = 15/8
Answersin A = 15/17, cos A = 8/17, tan A = 15/8.

Q8. If cot theta = 7/24, find sin theta and cos theta.

Solution
Given: cot theta = 7/24 = adjacent/opposite Let adjacent = 7, opposite = 24. Find hypotenuse: hypotenuse = sqrt(7² + 24²) = sqrt(49 + 576) = sqrt(625) = 25 sin theta = opposite/hypotenuse = 24/25 cos theta = adjacent/hypotenuse = 7/25
Answersin theta = 24/25, cos theta = 7/25.

Q9. If cosec theta = 2, find all other trigonometric ratios.

Solution
Given: cosec theta = 2 = hypotenuse/opposite Let hypotenuse = 2, opposite = 1. Find adjacent: adjacent = sqrt(hyp² − opp²) = sqrt(4 − 1) = sqrt(3) sin theta = 1/cosec theta = 1/2 cos theta = adjacent/hypotenuse = sqrt(3)/2 tan theta = opposite/adjacent = 1/sqrt(3) = sqrt(3)/3 sec theta = hypotenuse/adjacent = 2/sqrt(3) = 2sqrt(3)/3 cot theta = adjacent/opposite = sqrt(3)/1 = sqrt(3)
Answersin theta = 1/2, cos theta = sqrt(3)/2, tan theta = 1/sqrt(3), sec theta = 2/sqrt(3), cot theta = sqrt(3).

Q10. In a right triangle, one side is 8 and the hypotenuse is 17. Find all trigonometric ratios for the angle opposite to the side of length 8.

Solution
Given: opposite = 8, hypotenuse = 17 Find adjacent: adjacent = sqrt(17² − 8²) = sqrt(289 − 64) = sqrt(225) = 15 sin theta = opposite/hypotenuse = 8/17 cos theta = adjacent/hypotenuse = 15/17 tan theta = opposite/adjacent = 8/15 cosec theta = 17/8 sec theta = 17/15 cot theta = 15/8
Answersin theta = 8/17, cos theta = 15/17, tan theta = 8/15, cosec theta = 17/8, sec theta = 17/15, cot theta = 15/8.

Q11. If sin A = 3/4 and cos B = 3/4, what can you say about angles A and B?

Solution
Given: sin A = 3/4 and cos B = 3/4. If sin A = cos B, then: sin A = cos B = sin(90° − B) Therefore: A = 90° − B, or A + B = 90°. Angles A and B are complementary (they add up to 90°).
AnswerAngles A and B are complementary: A + B = 90°.

Q12. If sec theta = 5/3, find the other trigonometric ratios.

Solution
Given: sec theta = 5/3 = hypotenuse/adjacent Let hypotenuse = 5, adjacent = 3. Find opposite: opposite = sqrt(hyp² − adj²) = sqrt(25 − 9) = sqrt(16) = 4 cos theta = 1/sec theta = 3/5 sin theta = opposite/hypotenuse = 4/5 tan theta = opposite/adjacent = 4/3 cosec theta = hypotenuse/opposite = 5/4 cot theta = adjacent/opposite = 3/4
Answercos theta = 3/5, sin theta = 4/5, tan theta = 4/3, cosec theta = 5/4, cot theta = 3/4.

Exercise 6.2 — Trigonometric Ratios of Specific Angles

Q1. Find the value of: sin 30° + cos 60°

Solution
sin 30° = 1/2 cos 60° = 1/2 sin 30° + cos 60° = 1/2 + 1/2 = 1
Answer1

Q2. Evaluate: 2 sin 60° × cos 30°

Solution
sin 60° = sqrt(3)/2 cos 30° = sqrt(3)/2 2 × (sqrt(3)/2) × (sqrt(3)/2) = 2 × 3/4 = 6/4 = 3/2
Answer3/2

Q3. Evaluate: sin² 45° + cos² 45°

Solution
sin 45° = 1/sqrt(2), cos 45° = 1/sqrt(2) sin² 45° + cos² 45° = (1/sqrt(2))² + (1/sqrt(2))² = 1/2 + 1/2 = 1 This verifies the identity sin²θ + cos²θ = 1 for theta = 45°.
Answer1

Q4. Find the value of: tan 45° × sec 60° × cosec 30°

Solution
tan 45° = 1 sec 60° = 2 cosec 30° = 2 tan 45° × sec 60° × cosec 30° = 1 × 2 × 2 = 4
Answer4

Q5. Evaluate: (sin 30° + cos 30°)²

Solution
sin 30° = 1/2 cos 30° = sqrt(3)/2 (sin 30° + cos 30°)² = (1/2 + sqrt(3)/2)² = ((1 + sqrt(3))/2)² = (1 + sqrt(3))²/4 = (1 + 2sqrt(3) + 3)/4 = (4 + 2sqrt(3))/4 = (2 + sqrt(3))/2
Answer(2 + sqrt(3))/2

Q6. Find the value of: sin 60° cos 30° + cos 60° sin 30°

Solution
sin 60° = sqrt(3)/2, cos 30° = sqrt(3)/2 cos 60° = 1/2, sin 30° = 1/2 sin 60° cos 30° + cos 60° sin 30° = (sqrt(3)/2)(sqrt(3)/2) + (1/2)(1/2) = 3/4 + 1/4 = 4/4 = 1 Note: This equals sin(60° + 30°) = sin 90° = 1 ✓
Answer1

Q7. Find the value of: 4 sin² 30° + cos² 60° − 2 tan² 45°

Solution
sin 30° = 1/2, cos 60° = 1/2, tan 45° = 1 4 sin² 30° + cos² 60° − 2 tan² 45° = 4 × (1/2)² + (1/2)² − 2 × (1)² = 4 × 1/4 + 1/4 − 2 = 1 + 1/4 − 2 = 5/4 − 2 = 5/4 − 8/4 = −3/4
Answer−3/4

Q8. Find the value of: sin² 30° + cos² 30° + tan² 45°

Solution
sin 30° = 1/2, cos 30° = sqrt(3)/2, tan 45° = 1 sin² 30° + cos² 30° + tan² 45° = (1/2)² + (sqrt(3)/2)² + (1)² = 1/4 + 3/4 + 1 = 1 + 1 = 2
Answer2

Q9. Evaluate: 2 tan² 60° + 3 sec² 30° − 2 cos² 30°

Solution
tan 60° = sqrt(3), sec 30° = 2/sqrt(3), cos 30° = sqrt(3)/2 2 tan² 60° + 3 sec² 30° − 2 cos² 30° = 2(sqrt(3))² + 3(2/sqrt(3))² − 2(sqrt(3)/2)² = 2(3) + 3(4/3) − 2(3/4) = 6 + 4 − 3/2 = 10 − 3/2 = 17/2
Answer17/2

Q10. If sin A = cos A, find the value of A.

Solution
Given: sin A = cos A Dividing both sides by cos A: sin A/cos A = cos A/cos A tan A = 1 We know tan 45° = 1 Therefore A = 45°.
AnswerA = 45°

Q11. If sin(A + B) = 1 and cos(A − B) = 1, find A and B where 0° ≤ A, B ≤ 90°.

Solution
Given: sin(A + B) = 1 → A + B = 90° (since sin 90° = 1) Given: cos(A − B) = 1 → A − B = 0° (since cos 0° = 1) From the second equation: A = B Substituting in the first: A + A = 90° → 2A = 90° → A = 45° Therefore B = 45°. Verification: sin(45° + 45°) = sin 90° = 1 ✓ cos(45° − 45°) = cos 0° = 1 ✓
AnswerA = 45°, B = 45°.

Q12. Find the value of: (sin 30° × cos 45° + cos 30° × sin 45°)

Solution
sin 30° = 1/2, cos 45° = 1/sqrt(2) cos 30° = sqrt(3)/2, sin 45° = 1/sqrt(2) = (1/2)(1/sqrt(2)) + (sqrt(3)/2)(1/sqrt(2)) = 1/(2sqrt(2)) + sqrt(3)/(2sqrt(2)) = (1 + sqrt(3))/(2sqrt(2)) Rationalising: = (1 + sqrt(3)) × sqrt(2) / (2 × 2) = sqrt(2)(1 + sqrt(3))/4 = (sqrt(2) + sqrt(6))/4
Answer(sqrt(2) + sqrt(6))/4

Q13. Evaluate: 3 cot² 60° + 4 cosec² 30° − 2 sec² 45°

Solution
cot 60° = 1/sqrt(3), cosec 30° = 2, sec 45° = sqrt(2) 3 cot² 60° + 4 cosec² 30° − 2 sec² 45° = 3(1/sqrt(3))² + 4(2)² − 2(sqrt(2))² = 3(1/3) + 4(4) − 2(2) = 1 + 16 − 4 = 13
Answer13

Q14. Evaluate: sin 0° + cos 90° + tan 0°

Solution
sin 0° = 0 cos 90° = 0 tan 0° = 0 sin 0° + cos 90° + tan 0° = 0 + 0 + 0 = 0
Answer0

Q15. Show that: sin 60° = 2 sin 30° cos 30°

Solution
RHS: 2 sin 30° cos 30° = 2 × (1/2) × (sqrt(3)/2) = sqrt(3)/2 LHS: sin 60° = sqrt(3)/2 LHS = RHS ✓
AnswerProved. Both sides equal sqrt(3)/2.

Exercise 6.3 — Trigonometric Identities

Q1. Prove the identity: sin²θ + cos²θ = 1

Solution
Let ABC be a right triangle with angle theta at A. By Pythagoras theorem: AB² + BC² = AC² Dividing both sides by AC²: AB²/AC² + BC²/AC² = AC²/AC² = 1 (AB/AC)² + (BC/AC)² = 1 Since cos theta = AB/AC (adjacent/hypotenuse) and sin theta = BC/AC (opposite/hypotenuse): cos²θ + sin²θ = 1 sin²θ + cos²θ = 1 Proved.
AnswerIdentity proved.

Q2. If sin theta = 4/5, find cos theta using identity.

Solution
Using identity: sin²θ + cos²θ = 1 (4/5)² + cos²θ = 1 16/25 + cos²θ = 1 cos²θ = 1 − 16/25 cos²θ = 9/25 cos theta = 3/5 (cos theta > 0 for acute angles)
Answercos theta = 3/5

Q3. If cos theta = 5/13, find sin theta using identity.

Solution
Using identity: sin²θ + cos²θ = 1 sin²θ + (5/13)² = 1 sin²θ + 25/169 = 1 sin²θ = 1 − 25/169 sin²θ = 144/169 sin theta = 12/13 (sin theta > 0 for acute angles)
Answersin theta = 12/13

Q4. If tan theta = 3/4, find sec theta using identity.

Solution
Using identity: 1 + tan²θ = sec²θ tan theta = 3/4 tan²θ = 9/16 sec²θ = 1 + 9/16 = 25/16 sec theta = 5/4 (sec theta > 0 for acute angles)
Answersec theta = 5/4

Q5. If cot theta = 12/5, find cosec theta using identity.

Solution
Using identity: 1 + cot²θ = cosec²θ cot theta = 12/5 cot²θ = 144/25 cosec²θ = 1 + 144/25 = 169/25 cosec theta = 13/5 (cosec theta > 0 for acute angles)
Answercosec theta = 13/5

Q6. Prove that: (sin theta + cos theta)² + (sin theta − cos theta)² = 2

Solution
LHS = (sin theta + cos theta)² + (sin theta − cos theta)² Expanding each: (sin theta + cos theta)² = sin²θ + 2 sin theta cos theta + cos²θ (sin theta − cos theta)² = sin²θ − 2 sin theta cos theta + cos²θ Adding: = sin²θ + 2 sin theta cos theta + cos²θ + sin²θ − 2 sin theta cos theta + cos²θ = 2 sin²θ + 2 cos²θ = 2(sin²θ + cos²θ) = 2 × 1 [using identity sin²θ + cos²θ = 1] = 2 = RHS Proved.
AnswerIdentity proved.

Q7. Prove that: (1 − sin²θ) sec²θ = 1

Solution
LHS = (1 − sin²θ) sec²θ Since sin²θ + cos²θ = 1, we have 1 − sin²θ = cos²θ = cos²θ × sec²θ = cos²θ × (1/cos²θ) = 1 = RHS Proved.
AnswerIdentity proved.

Q8. Prove that: (1 − cos²θ) cosec²θ = 1

Solution
LHS = (1 − cos²θ) cosec²θ Since sin²θ + cos²θ = 1, we have 1 − cos²θ = sin²θ = sin²θ × cosec²θ = sin²θ × (1/sin²θ) = 1 = RHS Proved.
AnswerIdentity proved.

Q9. Prove that: sec²θ − tan²θ = 1

Solution
LHS = sec²θ − tan²θ Using identity: 1 + tan²θ = sec²θ Therefore: sec²θ − tan²θ = 1 = 1 = RHS Proved.
AnswerIdentity proved.

Q10. Prove that: cosec²θ − cot²θ = 1

Solution
LHS = cosec²θ − cot²θ Using identity: 1 + cot²θ = cosec²θ Therefore: cosec²θ − cot²θ = 1 = 1 = RHS Proved.
AnswerIdentity proved.

Q11. Prove that: sin theta / (1 + cos theta) + sin theta / (1 − cos theta) = 2 cosec theta

Solution
LHS = sin theta/(1 + cos theta) + sin theta/(1 − cos theta) Taking common denominator: = sin theta × [(1 − cos theta) + (1 + cos theta)] / [(1 + cos theta)(1 − cos theta)] = sin theta × [1 − cos theta + 1 + cos theta] / [1 − cos²θ] = sin theta × 2 / sin²θ = 2 sin theta / sin²θ = 2 / sin theta = 2 cosec theta = RHS Proved.
AnswerIdentity proved.

Q12. Prove that: (tan theta + cot theta) = sec theta cosec theta

Solution
LHS = tan theta + cot theta = sin theta/cos theta + cos theta/sin theta = (sin²θ + cos²θ) / (sin theta cos theta) = 1 / (sin theta cos theta) = (1/sin theta) × (1/cos theta) = cosec theta × sec theta = sec theta cosec theta = RHS Proved.
AnswerIdentity proved.

Q13. If sin A + cos A = sqrt(2), find sin A cos A.

Solution
Given: sin A + cos A = sqrt(2) Squaring both sides: (sin A + cos A)² = (sqrt(2))² sin²A + 2 sin A cos A + cos²A = 2 (sin²A + cos²A) + 2 sin A cos A = 2 1 + 2 sin A cos A = 2 2 sin A cos A = 1 sin A cos A = 1/2
Answersin A cos A = 1/2

Q14. If tan A = 1 and tan B = sqrt(3), find A + B.

Solution
tan A = 1 → A = 45° (since tan 45° = 1) tan B = sqrt(3) → B = 60° (since tan 60° = sqrt(3)) A + B = 45° + 60° = 105°
AnswerA + B = 105°

Q15. Prove that: cos theta / (1 + sin theta) = (1 − sin theta) / cos theta

Solution
LHS = cos theta / (1 + sin theta) Multiply numerator and denominator by (1 − sin theta): = cos theta (1 − sin theta) / [(1 + sin theta)(1 − sin theta)] = cos theta (1 − sin theta) / [1 − sin²θ] = cos theta (1 − sin theta) / cos²θ = (1 − sin theta) / cos theta = RHS Proved.
AnswerIdentity proved.

Q16. If sec theta + tan theta = 5, find sec theta − tan theta.

Solution
We know: sec²θ − tan²θ = 1 (sec theta + tan theta)(sec theta − tan theta) = 1 Given: sec theta + tan theta = 5 5 × (sec theta − tan theta) = 1 sec theta − tan theta = 1/5
Answersec theta − tan theta = 1/5

Q17. If cosec theta + cot theta = 7, find cosec theta − cot theta.

Solution
We know: cosec²θ − cot²θ = 1 (cosec theta + cot theta)(cosec theta − cot theta) = 1 Given: cosec theta + cot theta = 7 7 × (cosec theta − cot theta) = 1 cosec theta − cot theta = 1/7
Answercosec theta − cot theta = 1/7

Q18. Prove that: sqrt((1 + sin theta)/(1 − sin theta)) = sec theta + tan theta

Solution
LHS = sqrt((1 + sin theta)/(1 − sin theta)) Multiply numerator and denominator inside by (1 + sin theta): = sqrt((1 + sin theta)² / [(1 − sin theta)(1 + sin theta)]) = sqrt((1 + sin theta)² / (1 − sin²θ)) = sqrt((1 + sin theta)² / cos²θ) = (1 + sin theta) / cos theta = 1/cos theta + sin theta/cos theta = sec theta + tan theta = RHS Proved.
AnswerIdentity proved.

Exercise 6.4 — Complementary Angles and Applications

Q1. If sin A = cos B, then A + B = ?

Solution
Given: sin A = cos B We know: cos B = sin(90° − B) Therefore: sin A = sin(90° − B) This gives: A = 90° − B Therefore: A + B = 90°
AnswerA + B = 90° (the angles are complementary).

Q2. If tan A = cot B, find A + B.

Solution
Given: tan A = cot B We know: cot B = tan(90° − B) Therefore: tan A = tan(90° − B) This gives: A = 90° − B Therefore: A + B = 90°
AnswerA + B = 90°

Q3. Evaluate: sin 72° / cos 18°

Solution
We know: cos theta = sin(90° − theta) cos 18° = sin(90° − 18°) = sin 72° sin 72° / cos 18° = sin 72° / sin 72° = 1
Answer1

Q4. Evaluate: tan 55° / cot 35°

Solution
We know: cot theta = tan(90° − theta) cot 35° = tan(90° − 35°) = tan 55° tan 55° / cot 35° = tan 55° / tan 55° = 1
Answer1

Q5. Evaluate: sec 52° / cosec 38°

Solution
We know: cosec theta = sec(90° − theta) cosec 38° = sec(90° − 38°) = sec 52° sec 52° / cosec 38° = sec 52° / sec 52° = 1
Answer1

Q6. Evaluate: sin 35° cos 55° + cos 35° sin 55°

Solution
We know: cos 55° = sin(90° − 55°) = sin 35° sin 55° = cos(90° − 55°) = cos 35° Substituting: sin 35° × sin 35° + cos 35° × cos 35° = sin² 35° + cos² 35° = 1 [using identity sin²θ + cos²θ = 1]
Answer1

Q7. Evaluate: cos 48° cos 42° − sin 48° sin 42°

Solution
We know: sin 42° = cos(90° − 42°) = cos 48° cos 42° = sin(90° − 42°) = sin 48° Substituting: cos 48° × sin 48° − sin 48° × cos 48° = cos 48° sin 48° − sin 48° cos 48° = 0
Answer0

Q8. Evaluate: tan 1° tan 2° tan 3° ... tan 89°

Solution
Pair terms using complementary angles: tan 1° × tan 89° = tan 1° × cot 1° = 1 tan 2° × tan 88° = tan 2° × cot 2° = 1 tan 3° × tan 87° = 1 ... tan 44° × tan 46° = 1 And tan 45° = 1 (the middle term) There are 44 pairs that each equal 1, plus tan 45° = 1. Product = 1 × 1 × 1 × ... × 1 = 1 Note: tan theta × tan(90° − theta) = tan theta × cot theta = 1 for each complementary pair, and tan 45° = 1.
Answer1

Q9. If tan 2A = cot(3A − 18°), find A where 2A is an acute angle.

Solution
Given: tan 2A = cot(3A − 18°) Since tan theta = cot(90° − theta): tan 2A = cot(90° − 2A) Therefore: cot(90° − 2A) = cot(3A − 18°) 90° − 2A = 3A − 18° 90° + 18° = 3A + 2A 108° = 5A A = 108/5 = 21.6° = 21°36'
AnswerA = 21.6° (or 21°36')

Q10. If sin(A + 20°) = cos(A − 10°), find A.

Solution
Given: sin(A + 20°) = cos(A − 10°) Since sin theta = cos(90° − theta): sin(A + 20°) = cos(90° − (A + 20°)) = cos(70° − A) Therefore: cos(70° − A) = cos(A − 10°) 70° − A = A − 10° 70° + 10° = 2A 80° = 2A A = 40° Verification: sin(40° + 20°) = sin 60° = sqrt(3)/2 cos(40° − 10°) = cos 30° = sqrt(3)/2 ✓
AnswerA = 40°

Q11. A ladder leans against a wall making an angle of 60° with the ground. If the foot of the ladder is 2.5 m from the wall, find the length of the ladder.

Solution
Let the length of ladder = L (hypotenuse) Distance from wall = 2.5 m (adjacent to 60°) cos 60° = adjacent/hypotenuse = 2.5/L 1/2 = 2.5/L L = 2.5 × 2 = 5 m Also, height reached on wall: sin 60° = opposite/hypotenuse = height/L sqrt(3)/2 = height/5 height = 5 × sqrt(3)/2 = 5sqrt(3)/2 ≈ 4.33 m
AnswerLength of ladder = 5 m. Height reached on wall = 5sqrt(3)/2 m.

Q12. A tree breaks due to a storm and the broken part bends so that the top of the tree touches the ground making an angle of 30° with it. The distance between the foot of the tree and the point where the top touches the ground is 10 m. Find the original height of the tree.

Solution
Let the broken part of the tree be L (hypotenuse). The distance on the ground = 10 m (adjacent to 30°). cos 30° = adjacent/hypotenuse = 10/L sqrt(3)/2 = 10/L L = 20/sqrt(3) = 20sqrt(3)/3 m Height of the standing part (opposite to 30°): tan 30° = opposite/adjacent = height/10 1/sqrt(3) = height/10 height = 10/sqrt(3) = 10sqrt(3)/3 m Original height of tree = standing part + broken part = 10sqrt(3)/3 + 20sqrt(3)/3 = 30sqrt(3)/3 = 10sqrt(3) m ≈ 17.32 m
AnswerOriginal height of the tree = 10sqrt(3) m.

Q13. From the top of a 15 m high tower, the angle of depression of a car is 30°. Find the distance of the car from the tower.

Solution
Height of tower = 15 m Angle of depression = 30° Angle of elevation from car to tower top = 30° (alternate interior angles) tan 30° = opposite/adjacent = height/distance 1/sqrt(3) = 15/distance distance = 15 × sqrt(3) = 15sqrt(3) m ≈ 25.98 m
AnswerDistance of the car from the tower = 15sqrt(3) m.

Q14. A kite is flying at a height of 60 m above the ground. The string attached to the kite is tied to a point on the ground and makes an angle of 60° with the ground. Find the length of the string.

Solution
Height = 60 m (opposite to 60°) Length of string = L (hypotenuse) sin 60° = opposite/hypotenuse = 60/L sqrt(3)/2 = 60/L L = 60 × 2/sqrt(3) L = 120/sqrt(3) L = 120sqrt(3)/3 L = 40sqrt(3) m ≈ 69.28 m
AnswerLength of the string = 40sqrt(3) m.

Q15. The shadow of a tower standing on level ground is 40 m long when the sun's altitude is 30°. Find the height of the tower.

Solution
Shadow length = 40 m (adjacent to 30°) Height of tower = h (opposite to 30°) tan 30° = opposite/adjacent = h/40 1/sqrt(3) = h/40 h = 40/sqrt(3) = 40sqrt(3)/3 m ≈ 23.09 m
AnswerHeight of tower = 40sqrt(3)/3 m.

Q16. Evaluate: sin² 25° + sin² 65° + cos² 35° + cos² 55°

Solution
Using complementary angle relations: sin 65° = cos(90° − 65°) = cos 25° cos 55° = sin(90° − 55°) = sin 35° sin² 25° + sin² 65° + cos² 35° + cos² 55° = sin² 25° + cos² 25° + cos² 35° + sin² 35° = (sin² 25° + cos² 25°) + (cos² 35° + sin² 35°) = 1 + 1 = 2
Answer2

Q17. Evaluate: (sin 20°/cos 70°) + (sin 50°/cos 40°)

Solution
sin 20° = cos(90° − 20°) = cos 70° sin 50° = cos(90° − 50°) = cos 40° (sin 20°/cos 70°) + (sin 50°/cos 40°) = (cos 70°/cos 70°) + (cos 40°/cos 40°) = 1 + 1 = 2
Answer2

Q18. If sec 4A = cosec(A + 20°), where 4A is an acute angle, find A.

Solution
Given: sec 4A = cosec(A + 20°) Since sec theta = cosec(90° − theta): sec 4A = cosec(90° − 4A) Therefore: cosec(90° − 4A) = cosec(A + 20°) 90° − 4A = A + 20° 90° − 20° = 4A + A 70° = 5A A = 14° Verification: sec(4 × 14°) = sec 56° cosec(14° + 20°) = cosec 34° sec 56° = cosec 34° ✓ (since 56° + 34° = 90°)
AnswerA = 14°

Q19. Express sin 67° in terms of cosine.

Solution
sin 67° = cos(90° − 67°) = cos 23°
Answersin 67° = cos 23°

Q20. Express cos 38° in terms of sine.

Solution
cos 38° = sin(90° − 38°) = sin 52°
Answercos 38° = sin 52°

Common Mistakes

1. Confusing opposite and adjacent sides — When the angle changes, the opposite and adjacent sides swap. For angle A, the side opposite to A is the "opposite" side. For angle C (the other acute angle), the same side becomes "adjacent." Always identify which angle you are working with first.

2. Wrong reciprocal ratios — cosec = 1/sin, NOT 1/cos. sec = 1/cos, NOT 1/sin. Use the mnemonic: the reciprocal pairs are (sin, cosec), (cos, sec), (tan, cot).

3. Forgetting to rationalise denominators — When tan theta = 1/sqrt(3), write it as sqrt(3)/3 after rationalisation. Similarly, 1/sqrt(2) should be written as sqrt(2)/2. Examiners expect rationalised forms.

4. Sign errors in identities — When using sin²θ = 1 − cos²θ, students sometimes write sin²θ = cos²θ − 1. Always write the identity correctly: sin²θ + cos²θ = 1, then rearrange.

5. Mixing up complementary angle conversions — sin(90° − theta) = cos theta, NOT sin theta. A common error is writing sin(90° − theta) = sin theta. Draw the triangle and verify the relationship.

6. Incorrect Pythagorean triple identification — When given one side and the hypotenuse, always compute the third side using c² = a² + b² before finding trigonometric ratios. Do not assume the sides form a 3-4-5 or 5-12-13 triple without checking.

7. Forgetting that tan 90° is undefined — cos 90° = 0, so tan 90° = sin 90°/cos 90° = 1/0 = undefined. Do not write tan 90° = infinity or any numerical value in exams.

Exam Checklist

  • [ ] Know all six trigonometric ratio definitions (sin, cos, tan, cosec, sec, cot)
  • [ ] Can find all six ratios given one ratio (draw a right triangle, use Pythagoras)
  • [ ] Can find the third side of a right triangle given two sides
  • [ ] Memorise trigonometric values for 0°, 30°, 45°, 60°, 90°
  • [ ] Can evaluate expressions involving specific angle values
  • [ ] Know the three fundamental identities: sin²θ+cos²θ=1, 1+tan²θ=sec²θ, 1+cot²θ=cosec²θ
  • [ ] Can prove simple identities using the three fundamental identities
  • [ ] Can find sin theta, cos theta, or tan theta given one of the others using identities
  • [ ] Know complementary angle relations: sin(90°−θ)=cosθ, cos(90°−θ)=sinθ, tan(90°−θ)=cotθ
  • [ ] Can evaluate expressions using complementary angle substitutions
  • [ ] Can solve for unknown angles when given relationships like sin A = cos B
  • [ ] Can apply trigonometry to real-world problems (heights, distances, ladders, shadows)

Internal Links

  • [Class 9 Maths — Trigonometry Chapter Guide](/school/tamil-nadu-state-board/class/9/maths/trigonometry)
  • [All Class 9 Maths Solutions](/school/tamil-nadu-state-board/class/9/maths/solutions)
  • [Class 9 Maths Subject Guide](/school/tamil-nadu-state-board/class/9/maths)

Frequently Asked Questions

In a right triangle with angle theta: sin theta = opposite/hypotenuse, cos theta = adjacent/hypotenuse, tan theta = opposite/adjacent, cosec theta = hypotenuse/opposite, sec theta = hypotenuse/adjacent, cot theta = adjacent/opposite. Remember: tan = sin/cos, cosec = 1/sin, sec = 1/cos, cot = 1/tan.
← Browse all Class 9 Mathematics Exercise Solutions

Other question types — same chapter

Chapter Notes

Tamil Nadu Class 9 Maths Trigonometry — Samacheer Kalvi

Exercise Solutions

Samacheer Kalvi 9th Maths Algebra Exercise Solutions | KalviGavani

Exercise Solutions

Samacheer Kalvi 9th Maths Coordinate Geometry Exercise Solutions | KalviGavani