CG Board Class 10 Maths Chapter 9 Solutions: Trigonometric Equations and Identities (त्रिकोणमितीय समीकरण एवं सर्वसमिकाएँ) (2026-27)

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CG Board (CGBSE / SCERT Chhattisgarh) Class 10 Mathematics Solutions | 2026-27 Edition

📚 Chapter Overview:

  • Board: Chhattisgarh Board of Secondary Education (CGBSE) / CG SCERT Raipur
  • Subject & Unit: Class 10 Mathematics – Unit 4: Trigonometry (2026-27 Edition)
  • Chapter 9: Trigonometric Equations and Identities (त्रिकोणमितीय समीकरण एवं सर्वसमिकाएँ)
  • Medium: Bilingual Step-by-Step Solutions (English Medium + हिंदी माध्यम)
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Q1

Fundamental Trigonometric Identities A trigonometric identity is an…


📝 Question:

Fundamental Trigonometric Identities A trigonometric identity is an equation involving trigonometric ratios of an angle which is true for all values of the angle. First Identity: sin ^2θ + cos ^2θ = 1 Derived forms: sin ^2θ = 1 – cos ^2θ, cos ^2θ = 1 – sin ^2θ. Second Identity: 1 + tan ^2θ = sec ^2θ (where 0° ≤ θ < 90°) Derived form: sec ^2θ – tan ^2θ = 1 Third Identity: 1 + cot ^2θ = cosec2θ (where 0°< θ ≤ 90°) Derived form: cosec2θ – cot ^2θ = 1 2. Reciprocal & Quotient Ratios sin θ = (1)/(cosecθ) ⇔ cosecθ = (1)/(sin θ) cos θ = (1)/(sec θ) ⇔ sec θ = (1)/(cos θ) tan θ = (sin θ)/(cos θ) = (1)/(cot θ) cot θ = (cos θ)/(sin θ) = (1)/(tan θ) 3. Trigonometric Ratios of Complementary Angles Two angles are complementary if their sum is 90°. For acute angle θ: sin (90°- θ) = cos θ cos (90°- θ) = sin θ tan (90°- θ) = cot θ cot (90°- θ) = tan θ sec (90°- θ) = cosecθ cosec(90°- θ) = sec θ Part 2: Exercise – 1* Prove the following identities: Question 1: (1)/(sec θ – 1) – (1)/(sec θ + 1) = 2cot ^2θ

💡 Solution:

Step 1:** Taking Left Hand Side (L.H.S.)

L.H.S. = (1)/(sec θ – 1) – (1)/(sec θ + 1)

  • Step 2: Taking LCM in denominator

= ((sec θ + 1) – (sec θ – 1))/((sec θ – 1)(sec θ + 1))

  • Step 3: Simplify numerator and denominator using (a – b)(a + b) = a2 – b2

= (sec θ + 1 – sec θ + 1)/(sec ^2θ – 1) = (2)/(sec ^2θ – 1)

  • Step 4: Using identity sec ^2θ – 1 = tan ^2θ

= (2)/(tan ^2θ) = 2cot ^2θ = R.H.S.

Question 2

sec 2θ cosecsup2supθ sec 2θ cosecsup2supθ


📝 Question:

sec ^2θ + cosec2θ = sec ^2θ · cosec2θ

💡 Solution:

Step 1:** Express in terms of sin θ and cos θ

L.H.S. = (1)/(cos ^2θ) + (1)/(sin ^2θ)

  • Step 2: Taking LCM

= (sin ^2θ + cos ^2θ)/(cos ^2θ · sin ^2θ)

  • Step 3: Using identity sin ^2θ + cos ^2θ = 1

= (1)/(cos ^2θ · sin ^2θ) = (1)/(cos ^2θ) · (1)/(sin ^2θ) = sec ^2θ · cosec2θ = R.H.S.

Question 3

sin 4 A cos 4 A 1 2sin…


📝 Question:

sin ^4 A + cos ^4 A = 1 – 2sin ^2 A cos ^2 A

💡 Solution:

Step 1:** Rewrite L.H.S. using a2 + b2 = (a + b)2 – 2ab

L.H.S. = (sin ^2 A)2 + (cos ^2 A)2 = (sin ^2 A + cos ^2 A)2 – 2sin ^2 A cos ^2 A

  • Step 2: Using identity sin ^2 A + cos ^2 A = 1

= (1)2 – 2sin ^2 A cos ^2 A = 1 – 2sin ^2 A cos ^2 A = R.H.S.

Question 4

1 cos θ1 cos θ cosecθ cot θ


📝 Question:

√((1 – cos θ)/(1 + cos θ)) = cosecθ – cot θ

💡 Solution:

Step 1:** Rationalize the fraction inside the square root by multiplying numerator and denominator by (1 – cos θ)

L.H.S. = √(((1 – cos θ)(1 – cos θ))/((1 + cos θ)(1 – cos θ))) = √(((1 – cos θ)2)/(1 – cos ^2θ))

  • Step 2: Using 1 – cos ^2θ = sin ^2θ

= √(((1 – cos θ)2)/(sin ^2θ)) = (1 – cos θ)/(sin θ)

  • Step 3: Separate into two fractions

= (1)/(sin θ) – (cos θ)/(sin θ) = cosecθ – cot θ = R.H.S.

Question 5

1 cot θ cosecθ1 tan θ sec θ…


📝 Question:

(1 + cot θ – cosecθ)(1 + tan θ + sec θ) = 2

💡 Solution:

Step 1:** Convert all ratios into sin θ and cos θ

L.H.S. = (1 + (cos θ)/(sin θ) – (1)/(sin θ))(1 + (sin θ)/(cos θ) + (1)/(cos θ))

  • Step 2: Take LCM in each bracket

= ((sin θ + cos θ – 1)/(sin θ))((cos θ + sin θ + 1)/(cos θ))

  • Step 3: Apply formula (a – b)(a + b) = a2 – b2 where a = (sin θ + cos θ) and b = 1

= ((sin θ + cos θ)2 – 12)/(sin θ cos θ) = (sin ^2θ + cos ^2θ + 2sin θcos θ – 1)/(sin θ cos θ)

  • Step 4: Substitute sin ^2θ + cos ^2θ = 1

= (1 + 2sin θcos θ – 1)/(sin θ cos θ) = (2sin θcos θ)/(sin θcos θ) = 2 = R.H.S.

Question 6

1 cos θ1 cos θ 1 cos θ1…


📝 Question:

(1 + cos θ)/(1 – cos θ) – (1 – cos θ)/(1 + cos θ) = 4cot θ cosecθ

💡 Solution:

Step 1:** Take LCM of denominators

L.H.S. = ((1 + cos θ)2 – (1 – cos θ)2)/((1 – cos θ)(1 + cos θ))

  • Step 2: Expand numerators using (a+b)2 – (a-b)2 = 4ab

= (4cos θ)/(1 – cos ^2θ)

  • Step 3: Substitute 1 – cos ^2θ = sin ^2θ

= (4cos θ)/(sin ^2θ) = 4 · (cos θ)/(sin θ) · (1)/(sin θ) = 4cot θ cosecθ = R.H.S.

Question 7

sin θ1 cos θ 1 cos θsin θ…


📝 Question:

(sin θ)/(1 + cos θ) + (1 + cos θ)/(sin θ) = 2cosecθ

💡 Solution:

Step 1:** Take LCM of denominators

L.H.S. = (sin ^2θ + (1 + cos θ)2)/(sin θ(1 + cos θ))

  • Step 2: Expand (1 + cos θ)2 = 1 + cos ^2θ + 2cos θ

= (sin ^2θ + 1 + cos ^2θ + 2cos θ)/(sin θ(1 + cos θ))

  • Step 3: Substitute sin ^2θ + cos ^2θ = 1

= (1 + 1 + 2cos θ)/(sin θ(1 + cos θ)) = (2 + 2cos θ)/(sin θ(1 + cos θ)) = (2(1 + cos θ))/(sin θ(1 + cos θ))

  • Step 4: Cancel common factor (1 + cos θ)

= (2)/(sin θ) = 2cosecθ = R.H.S.

Question 8

If cos θ sin θ 2sin θ then…


📝 Question:

If cos θ – sin θ = √2sin θ, then prove that cos θ + sin θ = √2cos θ

💡 Solution:

cos θ – sin θ = √2sin θ

  • Step 1: Rearrange terms to group sin θ

cos θ = sin θ + √2sin θ = (1 + √2)sin θ

  • Step 2: Express sin θ in terms of cos θ

sin θ = (cos θ)/(√2 + 1)

  • Step 3: Rationalize the denominator

sin θ = (cos θ (√2 – 1))/((√2 + 1)(√2 – 1)) = ((√2 – 1)cos θ)/(2 – 1) = √2cos θ – cos θ

  • Step 4: Transpose -cos θ to Left Hand Side

cos θ + sin θ = √2cos θ

Question 9

If tan θ ntan φ and sin θ…


📝 Question:

If tan θ = ntan φ and sin θ = msin φ, then prove that cos ^2θ = (m2 – 1)/(n2 – 1)

💡 Solution:

Step 1:** From sin θ = msin φ, express cosecφ

sin φ = (sin θ)/(m) ⇒ cosecφ = (m)/(sin θ)

  • Step 2: From tan θ = ntan φ, express cot φ

tan φ = (tan θ)/(n) ⇒ cot φ = (n)/(tan θ)

  • Step 3: Use standard identity cosec2φ – cot ^2φ = 1

((m)/(sin θ))2 – ((n)/(tan θ))2 = 1

(m2)/(sin ^2θ) – (n2)/((sin ^2θ)/(cos ^2θ)) = 1 ⇒ (m2 – n2cos ^2θ)/(sin ^2θ) = 1

  • Step 4: Substitute sin ^2θ = 1 – cos ^2θ

m2 – n2cos ^2θ = 1 – cos ^2θ

m2 – 1 = n2cos ^2θ – cos ^2θ = (n2 – 1)cos ^2θ

cos ^2θ = (m2 – 1)/(n2 – 1)

Question 10

If x acosecθ and y bcot θ then…


📝 Question:

If x = acosecθ and y = bcot θ, then prove that x2/a2 – y2/b2 = 1

💡 Solution:

x = acosecθ ⇒ x/a = cosecθ and y = bcot θ ⇒ y/b = cot θ

  • Step 1: Square both ratios

x2/a2 = cosec2θ and y2/b2 = cot ^2θ

  • Step 2: Subtracting second equation from first

L.H.S. = x2/a2 – y2/b2 = cosec2θ – cot ^2θ

  • Step 3: Using identity cosec2θ – cot ^2θ = 1

= 1 = R.H.S.

Question 11

If x rsin A cos C y rsin…


📝 Question:

If x = rsin A cos C, y = rsin A sin C, and z = rcos A, then prove that r2 = x2 + y2 + z2

💡 Solution:

Step 1:** Calculate R.H.S. x2 + y2 + z2

x2 = r2sin ^2 A cos ^2 C

y2 = r2sin ^2 A sin ^2 C

z2 = r2cos ^2 A

  • Step 2: Add x2 + y2 + z2

x2 + y2 + z2 = r2sin ^2 A cos ^2 C + r2sin ^2 A sin ^2 C + r2cos ^2 A

  • Step 3: Factor out r2sin ^2 A from the first two terms

= r2sin ^2 A(cos ^2 C + sin ^2 C) + r2cos ^2 A

  • Step 4: Substitute cos ^2 C + sin ^2 C = 1

= r2sin ^2 A(1) + r2cos ^2 A = r2(sin ^2 A + cos ^2 A)

  • Step 5: Substitute sin ^2 A + cos ^2 A = 1

= r2(1) = r2 = L.H.S.

Question 1

i 2cos 2θ 3cos θ 0


📝 Question:

(i):** 2cos ^2θ – √3cos θ = 0

💡 Solution:

Step 1:** Factor out cos θ

cos θ (2cos θ – √3) = 0

  • Case I: cos θ = 0 ⇒ θ = 90° Case II: 2cos θ – √3 = 0 ⇒ cos θ = √3/2 ⇒ θ = 30° Final Answer: θ = 30°, 90°

Question 1

ii 2sin 2θ cos θ 1


📝 Question:

(ii):** 2sin ^2θ – cos θ = 1

💡 Solution:

Step 1:** Convert sin ^2θ into 1 – cos ^2θ

2(1 – cos ^2θ) – cos θ = 1 ⇒ 2 – 2cos ^2θ – cos θ = 1

2cos ^2θ + cos θ – 1 = 0

  • Step 2: Factorize quadratic equation in cos θ

2cos ^2θ + 2cos θ – cos θ – 1 = 0

2cos θ(cos θ + 1) – 1(cos θ + 1) = 0

(2cos θ – 1)(cos θ + 1) = 0

  • Case I: 2cos θ – 1 = 0 ⇒ cos θ = 1/2 ⇒ θ = 60°* Case II: cos θ + 1 = 0 ⇒ cos θ = -1, which gives θ = 180°(rejected as θ ≤ 90°).
  • Final Answer: θ = 60°

Question 1

iii 3tan 2θ 2sec 2θ 1


📝 Question:

(iii):** 3tan ^2θ = 2sec ^2θ + 1

💡 Solution:

Step 1:** Substitute sec ^2θ = 1 + tan ^2θ

3tan ^2θ = 2(1 + tan ^2θ) + 1 ⇒ 3tan ^2θ = 2 + 2tan ^2θ + 1

  • Step 2: Rearrange terms

3tan ^2θ – 2tan ^2θ = 3 ⇒ tan ^2θ = 3 ⇒ tan θ = √3

  • Step 3: Find angle θ

tan θ = tan 60° ⇒ θ = 60°* Final Answer: θ = 60°

Question 1

iv cos 2θ 3cos θ 2 sin 2θ


📝 Question:

(iv):** cos ^2θ – 3cos θ + 2 = sin ^2θ

💡 Solution:

Step 1:** Substitute sin ^2θ = 1 – cos ^2θ

cos ^2θ – 3cos θ + 2 = 1 – cos ^2θ

2cos ^2θ – 3cos θ + 1 = 0

  • Step 2: Factorize

2cos ^2θ – 2cos θ – cos θ + 1 = 0

2cos θ(cos θ – 1) – 1(cos θ – 1) = 0

(2cos θ – 1)(cos θ – 1) = 0

  • Case I: 2cos θ – 1 = 0 ⇒ cos θ = 1/2 ⇒ θ = 60° Case II: cos θ – 1 = 0 ⇒ cos θ = 1 ⇒ θ = 0° Final Answer: θ = 0°, 60°

Question 1

v cos θ1 sin θ cos θ1 sin…


📝 Question:

(v):** (cos θ)/(1 – sin θ) + (cos θ)/(1 + sin θ) = 4

💡 Solution:

Step 1:** Factor out cos θ and take LCM

cos θ [((1 + sin θ) + (1 – sin θ))/((1 – sin θ)(1 + sin θ))] = 4

  • Step 2: Simplify numerator and denominator

cos θ [(2)/(1 – sin ^2θ)] = 4 ⇒ cos θ [(2)/(cos ^2θ)] = 4

  • Step 3: Cancel cos θ

(2)/(cos θ) = 4 ⇒ cos θ = 2/4 = 1/2

cos θ = cos 60° ⇒ θ = 60°* Final Answer: θ = 60°

Question 1

Express the following as trigonometric ratios of angles…


📝 Question:

Express the following as trigonometric ratios of angles which lie between 0°and 45°

💡 Solution:

(i) sin 56°:**

sin 56°= sin (90°- 34°) = cos 34°

  • (ii) tan 81°:

tan 81°= tan (90°- 9°) = cot 9°

  • (iii) sec 73°:

sec 73°= sec (90°- 17°) = cosec 17°

Question 2

Find the value of the following


📝 Question:

Find the value of the following

💡 Solution:

(i) (cos 80°)/(sin 10°):**

(cos (90°- 10°))/(sin 10°) = (sin 10°)/(sin 10°) = 1

  • (ii) (sin 37°)/(2cos 53°):

(sin 37°)/(2cos (90°- 37°)) = (sin 37°)/(2sin 37°) = 1/2

  • (iii) 3sin 17°sec 73°:

3sin 17°sec (90°- 17°) = 3sin 17°cosec 17°= 3 · 1 = 3

Question 3

Find the value of the following


📝 Question:

Find the value of the following

💡 Solution:

(i) sin 64°- cos 26°:**

sin (90°- 26°) – cos 26°= cos 26°- cos 26°= 0

  • (ii) 3cos 80°cosec 10°+ 2cos 59°cosec 31°:

3cos (90°- 10°)cosec 10°+ 2cos (90°- 31°)cosec 31°= 3sin 10°cosec 10°+ 2sin 31°cosec 31°= 3(1) + 2(1) = 5

  • (iii) (2cos 67°)/(sin 23°) – (tan 40°)/(cot 50°) + cos 0°:

(2cos (90°- 23°))/(sin 23°) – (tan (90°- 50°))/(cot 50°) + 1 = (2sin 23°)/(sin 23°) – (cot 50°)/(cot 50°) + 1 = 2 – 1 + 1 = 2

  • (iv) (sin 35°)/(cos 55°) + (cos 55°)/(sin 35°) – 2cos 60°:

(sin (90°- 55°))/(cos 55°) + (cos (90°- 35°))/(sin 35°) – 2(1/2) = (cos 55°)/(cos 55°) + (sin 35°)/(sin 35°) – 1 = 1 + 1 – 1 = 1

  • (v) (3tan 15°)/(cot 75°) + (sin 31°)/(2cos 59°) + (sec 70°)/(cosec 20°):

(3tan (90°- 75°))/(cot 75°) + (sin (90°- 59°))/(2cos 59°) + (sec (90°- 20°))/(cosec 20°)

= (3cot 75°)/(cot 75°) + (cos 59°)/(2cos 59°) + (cosec 20°)/(cosec 20°) = 3 + 1/2 + 1 = 9/2

Question 4

Prove that tan 15tan 36tan 45tan 54tan 75…


📝 Question:

Prove that tan 15°tan 36°tan 45°tan 54°tan 75°= 1

💡 Solution:

Step 1:** Group complementary angle pairs and substitute tan 45°= 1

L.H.S. = tan 15° · tan 75° · tan 36° · tan 54° · tan 45°* Step 2: Convert first angle of each complementary pair using tan (90°- θ) = cot θ

= tan (90°- 75°) · tan 75° · tan (90°- 54°) · tan 54° · (1)

= (cot 75° · tan 75°) · (cot 54° · tan 54°) · 1

  • Step 3: Using cot θ · tan θ = 1

= 1 · 1 · 1 = 1 = R.H.S.

Question 5

Prove that sin 35sin 55 cos 35cos 55…


📝 Question:

Prove that sin 35°sin 55°- cos 35°cos 55°= 0

💡 Solution:

Step 1:** Convert sin 35°and cos 35°using complementary angle relations

sin 35°= sin (90°- 55°) = cos 55°cos 35°= cos (90°- 55°) = sin 55°* Step 2: Substitute in L.H.S.

L.H.S. = (cos 55° · sin 55°) – (sin 55° · cos 55°) = 0 = R.H.S.

Question 6

Prove that tan 90 θcosecsup2supθ tan θ cos…


📝 Question:

Prove that (tan (90°- θ))/(cosec2θ · tan θ) = cos ^2θ

💡 Solution:

Step 1:** Use tan (90°- θ) = cot θ

L.H.S. = (cot θ)/(cosec2θ · tan θ)

  • Step 2: Substitute cot θ / tan θ = cot ^2θ

= (cot ^2θ)/(cosec2θ) = ((cos ^2θ)/(sin ^2θ))/((1)/(sin ^2θ))

  • Step 3: Cancel sin ^2θ

= cos ^2θ = R.H.S.

Question 7

Prove that tan 90 θcosecθ tan θ sin…


📝 Question:

Prove that (tan (90°- θ))/(cosecθ · tan θ) = sin θ cos θ

💡 Solution:

Step 1:** Substitute tan (90°- θ) = cot θ

L.H.S. = (cot θ)/(cosecθ · tan θ) = (cot ^2θ)/(cosecθ)

  • Step 2: Express in terms of sin θ and cos θ

= ((cos ^2θ)/(sin ^2θ))/((1)/(sin θ)) = (cos ^2θ)/(sin θ) · sin θ = cos ^2θ = R.H.S.

Question 8

If sin A cos B then prove that…


📝 Question:

If sin A = cos B, then prove that A + B = 90°

💡 Solution:

sin A = cos B

  • Step 1: Use identity cos B = sin (90°- B)

sin A = sin (90°- B)

  • Step 2: Equate angles

A = 90°- B ⇒ A + B = 90°

Question 9

If cosec 2A sec A 36 where 2A…


📝 Question:

If cosec 2A = sec (A – 36°), where 2A is an acute angle, then find the value of A

💡 Solution:

cosec 2A = sec (A – 36°)

  • Step 1: Use identity cosec 2A = sec (90°- 2A)

sec (90°- 2A) = sec (A – 36°)

  • Step 2: Equate angles

90°- 2A = A – 36° ⇒ 90°+ 36°= A + 2A

126°= 3A ⇒ A = (126°)/(3) = 42°* Final Answer: A = 42°

Question 10

If A B 90 sec A a cot…


📝 Question:

If A + B = 90°, sec A = a, cot B = b, then prove that a2 – b2 = 1

💡 Solution:

A + B = 90° ⇒ B = 90°- A

  • Step 1: Express b in terms of angle A

b = cot B = cot (90°- A) = tan A

  • Step 2: Calculate a2 – b2

L.H.S. = a2 – b2 = sec ^2 A – tan ^2 A

  • Step 3: Using identity sec ^2 A – tan ^2 A = 1

= 1 = R.H.S.

Question 11

If A B and C are interior angles…


📝 Question:

If A, B and C are interior angles of a triangle ABC, then prove that tan ((B + C)/(2)) = cot (A/2)

💡 Solution:

Step 1:** Sum of interior angles of Δ ABC is 180°

A + B + C = 180° ⇒ B + C = 180°- A

  • Step 2: Divide both sides by 2

(B + C)/(2) = 90°- A/2

  • Step 3: Take tangent (tan ) on both sides

tan ((B + C)/(2)) = tan (90°- A/2)

  • Step 4: Using tan (90°- θ) = cot θ

tan ((B + C)/(2)) = cot (A/2)

Question 12

If sec 34 x then find the value…


📝 Question:

If sec 34°= x, then find the value of cot ^2 56°+ cosec 56°

💡 Solution:

Step 1:** Convert 56°angles to 34°using complementary relations

cot 56°= cot (90°- 34°) = tan 34°cosec 56°= cosec(90°- 34°) = sec 34°= x

  • Step 2: Express tan ^2 34°in terms of sec 34°

tan ^2 34°= sec ^2 34°- 1 = x2 – 1

  • Step 3: Substitute back into the expression

cot ^2 56°+ cosec 56°= tan ^2 34°+ sec 34°= (x2 – 1) + x = x2 + x – 1

  • Final Answer: x2 + x – 1

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