CHAPTER 11 TELANGANA CLASS X MATHEMATICS - CHAPTER 11: TRIGONOMETRY (EXERCISE 11.1 TO 11.4 FULL SOLUTIONS)

BOARD OF SECONDARY EDUCATION, TELANGANA

CLASS 10 MATHEMATICS - CHAPTER 11: TRIGONOMETRY

Complete Exercise-Wise Questions & Step-by-Step Solutions (Ex 11.1 - 11.4) | Muqeem Tutorial

SUBJECT: MATHEMATICS CHAPTER 11 (FULL EXERCISES) WEIGHTAGE: 8 - 10 MARKS

EXERCISE 11.1 — Trigonometric Ratios in Right Triangles

Q1. In △ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine: (i) sin A, cos A  |  (ii) sin C, cos C.

Solution:
By Pythagoras theorem in △ABC: AC = √(AB² + BC²) = √(24² + 7²) = √(576 + 49) = √625 = 25 cm.
• (i) For Angle A: Opposite = BC = 7, Adjacent = AB = 24, Hypotenuse = AC = 25.
👉 sin A = 7/25,   cos A = 24/25

• (ii) For Angle C: Opposite = AB = 24, Adjacent = BC = 7, Hypotenuse = AC = 25.
👉 sin C = 24/25,   cos C = 7/25

Q2. In a right triangle PQR, right-angled at Q, PQ = 12 cm and PR = 13 cm. Find the value of tan P - cot R.

Solution:
QR = √(PR² - PQ²) = √(13² - 12²) = √(169 - 144) = √25 = 5 cm.
• tan P = Opposite/Adjacent = QR/PQ = 5/12.
• cot R = Adjacent/Opposite = QR/PQ = 5/12.
tan P - cot R = 5/12 - 5/12 = 0.
👉 tan P - cot R = 0

Q3. If sin A = 3/4, calculate cos A and tan A.

Solution:
sin A = Opposite/Hypotenuse = 3/4 ⇒ Opposite = 3k, Hypotenuse = 4k.
Adjacent = √(16k² - 9k²) = k√7.
• cos A = Adjacent/Hypotenuse = (k√7)/(4k) = √7/4.
• tan A = Opposite/Adjacent = 3k/(k√7) = 3/√7.
👉 cos A = √7/4,   tan A = 3/√7

Q4. Given 15 cot A = 8, find sin A and sec A.

Solution:
cot A = 8/15 ⇒ Adjacent = 8k, Opposite = 15k.
Hypotenuse = √[(8k)² + (15k)²] = √(64k² + 225k²) = √289k² = 17k.
• sin A = Opposite/Hypotenuse = 15/17.
• sec A = Hypotenuse/Adjacent = 17/8.
👉 sin A = 15/17,   sec A = 17/8

Q5. Given sec θ = 13/12, calculate all other trigonometric ratios.

Solution:
sec θ = 13/12 ⇒ Hypotenuse = 13k, Adjacent = 12k ⇒ Opposite = √(169k² - 144k²) = 5k.
• sin θ = 5/13
• cos θ = 12/13
• tan θ = 5/12
• cosec θ = 13/5
• cot θ = 12/5

EXERCISE 11.2 — Ratios of Specific Angles (0°, 30°, 45°, 60°, 90°)

Ratio \ Angle	0°	30°	45°	60°	90°
sin θ	0	1/2	1/√2	√3/2	1
cos θ	1	√3/2	1/√2	1/2	0
tan θ	0	1/√3	1	√3	ND

Q1. Evaluate the following expressions:

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

• (ii) 2 tan² 45° + cos² 30° - sin² 60°:
= 2(1)² + (√3/2)² - (√3/2)² = 2 + 3/4 - 3/4 = 2

• (iii) cos 45° / (sec 30° + cosec 30°):
= (1/√2) / (2/√3 + 2) = (1/√2) / [(2 + 2√3)/√3] = √3 / [2√2(1 + √3)] = (3√2 - √6) / 8

• (iv) (sin 30° + tan 45° - cosec 60°) / (sec 30° + cos 60° + cot 45°):
= (1/2 + 1 - 2/√3) / (2/√3 + 1/2 + 1) = (3/2 - 2/√3) / (3/2 + 2/√3) = (43 - 24√3) / 11

Q2. Choose the correct option and justify your choice:

• (i) 2 tan 30° / (1 + tan² 30°) =
= [2(1/√3)] / [1 + 1/3] = (2/√3) / (4/3) = 6 / 4√3 = √3/2 = sin 60°.
👉 Answer: (A) sin 60°

• (ii) (1 - tan² 45°) / (1 + tan² 45°) =
= (1 - 1) / (1 + 1) = 0 / 2 = 0.
👉 Answer: (D) 0

• (iii) 2 tan 30° / (1 - tan² 30°) =
= (2/√3) / (1 - 1/3) = (2/√3) / (2/3) = √3 = tan 60°.
👉 Answer: (C) tan 60°

Q3. If tan (A + B) = √3 and tan (A - B) = 1/√3; 0° < A + B ≤ 90°; A > B, find A and B.

Solution:
1. tan (A + B) = √3 = tan 60° ⇒ A + B = 60° --- (Eq 1)
2. tan (A - B) = 1/√3 = tan 30° ⇒ A - B = 30° --- (Eq 2)
Adding equations: 2A = 90° ⇒ A = 45°.
Substituting A = 45°: B = 60° - 45° = 15°.
👉 A = 45°,   B = 15°

EXERCISE 11.3 — Trigonometric Ratios of Complementary Angles

Q1. Evaluate the following:

• (i) tan 36° / cot 54°:
tan 36° = tan(90° - 54°) = cot 54° ⇒ cot 54° / cot 54° = 1

• (ii) cos 12° - sin 78°:
cos 12° = cos(90° - 78°) = sin 78° ⇒ sin 78° - sin 78° = 0

• (iii) cosec 31° - sec 59°:
cosec 31° = cosec(90° - 59°) = sec 59° ⇒ sec 59° - sec 59° = 0

• (iv) sin 15° sec 75°:
sec 75° = sec(90° - 15°) = cosec 15° ⇒ sin 15° cosec 15° = 1

• (v) tan 26° tan 64°:
tan 64° = cot 26° ⇒ tan 26° cot 26° = 1

Q2. Show that tan 48° tan 16° tan 42° tan 74° = 1.

Proof:
LHS = tan 48° tan 16° tan(90° - 48°) tan(90° - 16°)
= tan 48° tan 16° cot 48° cot 16°
= (tan 48° cot 48°) × (tan 16° cot 16°) = 1 × 1 = 1 = RHS.
👉 Hence Proved.

Q3. If tan 2A = cot (A - 18°), where 2A is an acute angle, find the value of A.

Solution:
tan 2A = cot(90° - 2A).
cot (90° - 2A) = cot (A - 18°)
90° - 2A = A - 18° ⇒ 3A = 108° ⇒ A = 36°.
👉 A = 36°

Q4. If A, B and C are interior angles of a triangle ABC, then show that sin [(B + C)/2] = cos (A/2).

Proof:
In △ABC, A + B + C = 180° ⇒ B + C = 180° - A.
Dividing by 2: (B + C)/2 = 90° - A/2.
Taking sine on both sides: sin [(B + C)/2] = sin (90° - A/2) = cos (A/2).
👉 Hence Proved.

EXERCISE 11.4 — Trigonometric Identities

Q1. Evaluate: (1 + tan θ + sec θ)(1 + cot θ - cosec θ).

Solution:
Convert to sin θ and cos θ:
= [1 + sin θ/cos θ + 1/cos θ] [1 + cos θ/sin θ - 1/sin θ]
= [(cos θ + sin θ + 1)/cos θ] [(sin θ + cos θ - 1)/sin θ]
= [(sin θ + cos θ)² - 1²] / [sin θ cos θ]
= [sin² θ + cos² θ + 2 sin θ cos θ - 1] / [sin θ cos θ]
= [1 + 2 sin θ cos θ - 1] / [sin θ cos θ] = 2 sin θ cos θ / sin θ cos θ = 2.
👉 Value = 2

Q2. Prove that (cosec θ - cot θ)² = (1 - cos θ) / (1 + cos θ).

Proof:
LHS = (1/sin θ - cos θ/sin θ)² = [(1 - cos θ) / sin θ]² = (1 - cos θ)² / sin² θ
= (1 - cos θ)² / (1 - cos² θ) = (1 - cos θ)² / [(1 - cos θ)(1 + cos θ)]
= (1 - cos θ) / (1 + cos θ) = RHS.
👉 Hence Proved.

Q3. Prove that √[ (1 + sin A) / (1 - sin A) ] = sec A + tan A.

Proof:
Rationalizing inside the square root:
LHS = √[ (1 + sin A)(1 + sin A) / ((1 - sin A)(1 + sin A)) ]
= √[ (1 + sin A)² / (1 - sin² A) ] = √[ (1 + sin A)² / cos² A ]
= (1 + sin A) / cos A = 1/cos A + sin A/cos A = sec A + tan A = RHS.
👉 Hence Proved.

Q4. Prove that (sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A.

Proof:
LHS = (sin² A + cosec² A + 2 sin A cosec A) + (cos² A + sec² A + 2 cos A sec A)
= (sin² A + cos² A) + 2(1) + 2(1) + cosec² A + sec² A
= 1 + 2 + 2 + (1 + cot² A) + (1 + tan² A)
= 7 + tan² A + cot² A = RHS.
👉 Hence Proved.

Q5. Prove that (sec A + tan A)(1 - sin A) = cos A.

Proof:
LHS = (1/cos A + sin A/cos A)(1 - sin A) = [(1 + sin A)/cos A] (1 - sin A)
= (1 - sin² A) / cos A = cos² A / cos A = cos A = RHS.
👉 Hence Proved.