CG Board Class 9 Mathematics Chapter 8 Solutions: Trigonometrical Ratio and Identities (त्रिकोणमितीय अनुपात एवं सर्वसमिकाएँ) – 2026-27

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CG SCERT (Chhattisgarh Board) Class 9 Mathematics Solutions | 2026-27 Official Edition

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Complete 100% textbook-grounded CG Board Class 9 Mathematics Chapter 8 Solutions: Trigonometrical Ratio and Identities (त्रिकोणमितीय अनुपात एवं सर्वसमिकाएँ) in Bilingual Format (English & Hindi Medium).

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Official Textbook Exercises

End-of-Chapter Exercises (Solutions)

Question 1

1 In a right angled triangle ABC B…


📝 Question:

/ 1 In a right angled triangle ABC, ∠B is a right angle; then find the value of sin A, cos C and tan A for the following given data

💡 Solution:

(i) Given: AC = 5, AB = 3, BC = 4**

  • (ii) Given: AB = 12, BC = 5, AC = 13
  • (iii) Given: AB = 5, AC = 13, BC = 12
  • (iv) Given: BC = 12, AB = 9, AC = 15

Question 2

1 Find the remaining trigonometric ratios


📝 Question:

/ 1 Find the remaining trigonometric ratios

💡 Solution:

(i) tan θ = 3 / 4:** sin θ = 3/5, cos θ = 4/5, cot θ = 4/3, sec θ = 5/4, cosec θ = 5/3.

  • (ii) sin θ = 5 / 13: cos θ = 12/13, tan θ = 5/12, cot θ = 12/5, sec θ = 13/12, cosec θ = 13/5.
  • (iii) cos α = 1 / 3: sin α = 2sqrt(2)/3, tan α = 2sqrt(2), cot α = 1/(2sqrt(2)), sec α = 3, cosec α = 3/(2sqrt(2)).
  • (iv) cot θ = 1: sin θ = 1/sqrt(2), cos θ = 1/sqrt(2), tan θ = 1, sec θ = sqrt(2), cosec θ = sqrt(2).
  • (v) cosec A = 5 / 4: sin A = 4/5, cos A = 3/5, tan A = 4/3, cot A = 3/4, sec A = 5/3.
  • (vi) sec β = 2: sin β = sqrt(3)/2, cos β = 1/2, tan β = sqrt(3), cot β = 1/sqrt(3), cosec β = 2/sqrt(3).
  • (vii) cosec A = sqrt(10): sin A = 1/sqrt(10), cos A = 3/sqrt(10), tan A = 1/3, cot A = 3, sec A = sqrt(10)/3.

Question 3

3 If cos A 4 5 find cot…


📝 Question:

/ 3 If cos A = 4 / 5, find (cot A - sin A) / (2 tan A). / (2 tan A)` )

💡 Solution:

sin A = 3/5, tan A = 3/4, cot A = 4/3.

Numerator = 4/3 - 3/5 = 11/15.

Denominator = 2 * (3/4) = 3/2.

Value = (11/15) / (3/2) = 22 / 45.

Question 4

4 If sec θ 5 3 find tan…


📝 Question:

/ 4 If sec θ = 5 / 3, find (tan θ - sin θ) / (1 + tan θ sin θ). / (1 + tan θ sin θ)` )

💡 Solution:

cos θ = 3/5, sin θ = 4/5, tan θ = 4/3.

Numerator = 4/3 - 4/5 = 8/15.

Denominator = 1 + (4/3)*(4/5) = 31/15.

Value = (8/15) / (31/15) = 8 / 31.

Question 5

5 If sin A 1 3 find cos…


📝 Question:

/ 5 If sin A = 1 / 3, find cos A, cosec A + tan A and sec A

💡 Solution:

Base = sqrt(32 - 12) = 2*sqrt(2).

cos A = 2*sqrt(2) / 3.

cosec A + tan A = 3 + 1/(2*sqrt(2)) = (12 + sqrt(2)) / 4.

sec A = 3 / (2sqrt(2)) = 3sqrt(2) / 4.

Question 6

6 In a right angled triangle ABC right…


📝 Question:

/ 6 In a right angled triangle ABC, right angle is at C and tan A = 1 / sqrt(3), find sin A cos B + cos A sin B. , sin A cos B + cos A sin B` )

💡 Solution:

Since ∠C = 90°, ∠A = 30° and ∠B = 60°.

sin A cos B + cos A sin B = (1/2)(1/2) + (sqrt(3)/2)(sqrt(3)/2) = 1/4 + 3/4 = 1.

Question 7

7 If cot A 3 4 find sin…


📝 Question:

/ 7 If cot A = 3 / 4, find (sin A + cos A) / (sin A - cos A). / (sin A – cos A)` )

💡 Solution:

Divide numerator and denominator by cos A

(tan A + 1) / (tan A - 1) = (4/3 + 1) / (4/3 - 1) = (7/3) / (1/3) = 7.

Question 8

8 If sin θ 4 5 find 4…


📝 Question:

/ 8 If sin θ = 4 / 5, find (4 tan θ - 5 cos θ) / (sec θ + 4 cot θ). / (sec θ + 4 cot θ)` )

💡 Solution:

cos θ = 3/5, tan θ = 4/3, sec θ = 5/3, cot θ = 3/4.

Numerator = 4(4/3) - 5(3/5) = 16/3 - 3 = 7/3.

Denominator = 5/3 + 4*(3/4) = 5/3 + 3 = 14/3.

Value = (7/3) / (14/3) = 1 / 2.

Question 9

1 Choose the correct option


📝 Question:

/ 1: Choose the correct option /

💡 Solution:

(i) (1 - tan2 45°) / (1 + tan2 45°) =**

(1 - 1) / (1 + 1) = 0 / 2 = 0. Correct option is (c) 0.

  • (ii) (2 tan 30°) / (1 - tan2 30°) =

Formula tan(2θ) = tan(60°) = sqrt(3). Correct option is (c) tan 60°.

Question 10

2 Evaluate the following


📝 Question:

/ 2: Evaluate the following /

  1. sin θ = cos θ => tan θ = 1 => θ = 45°.
  2. 2 cos θ = 1 => cos θ = 1/2 => θ = 60°.
  3. 2 sin2 θ = 1 / 2 => sin2 θ = 1/4 => sin θ = 1/2 => θ = 30°.
  4. 3 tan2 θ - 1 = 0 => tan2 θ = 1/3 => tan θ = 1/sqrt(3) => θ = 30°.
  5. 2 sin θ = sqrt(3) => sin θ = sqrt(3)/2 => θ = 60°.
  6. tan θ = 0 => θ = 0°.
  7. 3 cosec2 θ = 4 => cosec2 θ = 4/3 => cosec θ = 2/sqrt(3) => θ = 60°.
  8. 2 cos2 θ = 1 / 2 => cos2 θ = 1/4 => cos θ = 1/2 => θ = 60°.
  9. 4 sin2 θ - 3 = 0 => sin2 θ = 3/4 => sin θ = sqrt(3)/2 => θ = 60°.
  10. 4 sec2 θ - 1 = 3 => 4 sec2 θ = 4 => sec θ = 1 => θ = 0°.
  11. cot2 θ = 3 => cot θ = sqrt(3) => θ = 30°.
💡 Solution:

(i) cos 30° cos 45° - sin 30° sin 45°* = (sqrt(3)/2)(1/sqrt(2)) - (1/2)(1/sqrt(2)) = (sqrt(3) - 1) / (2sqrt(2)).

  • (ii) tan 30° sec 45° + tan 60° sec 30° = (1/sqrt(3))sqrt(2) + sqrt(3)(2/sqrt(3)) = (2*sqrt(3) + sqrt(6)) / sqrt(3).
  • (iii) cosec 30° + cot 45° = 2 + 1 = 3.
  • (iv) (sec 30° - tan 45°) / cot 60° = (2/sqrt(3) - 1) / (1/sqrt(3)) = 2 - sqrt(3).
  • (v) tan2 60° + tan2 45° = (sqrt(3))2 + 12 = 3 + 1 = 4.
  • (vi) (cos 30° + sin 60°) / (1 + cos 60° + sin 30°) = (sqrt(3)/2 + sqrt(3)/2) / (1 + 1/2 + 1/2) = sqrt(3) / 2.
  • (vii) (sin2 45° + cos2 45°) / tan2 60° = 1 / (sqrt(3))2 = 1 / 3.
  • (viii) (sin 30° - sin 90° + 2 cos 0°) / (tan 30° tan 60°) = (1/2 - 1 + 2) / ((1/sqrt(3))sqrt(3)) = 3 / 2.
  • (i) sin 30° cos 60° + cos 30° sin 60° = tan 90°
  • (ii) 1 - 2 sin2 30° = cos2 60°
  • (iii) 2 cos2 45° - 1 = cos 90°
  • (iv) sin2 45° = 1 - cos2 45°
  • (v) sin2 60° + cos2 60° = 1

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