ML Aggarwal Trigonometrical Ratios of Standards Angles Exe-18.2 Class 9 ICSE Maths Solutions

ML Aggarwal Trigonometrical Ratios of Standards Angles Exe-18.2 Class 9 ICSE Maths Solutions Ch-18. Step by Step Solutions of Exe-18.2 questions on Trigonometrical Ratios of of Standards Angles for ML Aggarwal ICSE Class 9th Mathematics. Visit official website CISCE for detail information about ICSE Board Class-9.

ML Aggarwal Trigonometrical Ratios Exe-18.2 Class 9 ICSE Maths Solutions

 Board ICSE Subject Maths Class 9th Chapter-18 Trigonometrical Ratios of Standards Angles Topics Solution of Exe-18.2 Questions Academic Session 2024-2025

Solution of Exe-18.2 Questions on Trigonometrical Ratios of Standards Angles

ML Aggarwal Class 9 ICSE Maths Solutions Ch-18

Question 1.

(i) cos 18°/sin 72°

(ii) tan 41°/cot 49°

(iii) cosec 17°30’/sec 72°30’

(i) cos 18°/sin 72°

= cos 18°/sin (90° – 18°)

= cos 18°/cos 18°

= 1

(ii) tan 41°/cot49°

= tan 41°/cot (90° – 41°)

= tan 41°/tan 41°

= 1

(iii) cosec 17°30’/sec 72°30’

= cosec 17°30’/sec (90° – 17°30’)

= cosec 17°30’/cosec 17°30’

= 1

Question 2.

(i) cot 40°/tan 50° – ½ (cos 35°/sin 55°)

(ii) (sin 49°/cos 41°)2 + (cos 41°/sin 49°)2

(iii) sin 72°/cos 18° – sec 32°/cosec 58°

(iv) cos 75°/sin 15° + sin 12°/cos 78°– cos 18°/sin 72°

(v) sin 25°/sec 65° + cos 25°/ cosec 65°.

(i)cot 40°/tan 50° – ½ (cos 35°/sin 55°)

It can be written as

=1/2

= 12 + 12

= 1 + 1

= 2

= 1 – 1

= 0

= 1 + 1 – 1

= 1

(v) sin 25°/sec 65° + cos 25°/ cosec 65°

= (sin 25° × cos 65°) + (cos 25° × sin 65°)

= [sin 25° × cos (90°– 25°)] + [cos 25° × sin (90° – 25°)]

= (sin 25°× sin 25°) + (cos 25° × cos 25°)

= sin2 25° + cos2 25°

= 1

Question 3.

(i) sin 62° – cos 28°

(ii) cosec 35° – sec 55°.

(i)sin 62° – cos 28°

= sin (90° – 28°) – cos 28°

= cos 28° – cos 28°

= 0

(ii) cosec 35° – sec 55°

= cosec 35° – sec (90° – 35°)

= cosec 35° – cosec 35°

= 0

Question 4.

(i) cos2 26° + cos 64° sin 26° + tan 36°/ cot 54°

(ii) sec 17°/cosec 73° + tan 68°/cot 22° + cos2 44° + cos2 46°.

(i)cos2 26° + cos 64° sin 26° + tan 36°/ cot 54°

= cos2 26° + cos (90° – 26°) sin 26° + tan 36°/cot (90° – 36°)

cos (90° – θ) = sin θ

cot (90° – θ) = tan θ

sin2 θ + cos2 θ = 1

= cos2 26° + sin2 26° + tan 36°/tan 36°

= 1 + 1

= 2

(ii)sec 17°/cosec 73° + tan 68°/cot 22° + cos2 44° + cos2 46°

= sec (90° – 73°)/cosec 73° + tan (90° – 22°)/cot 22° + cos2 (90° – 46°) + cos2 46°

= cosec 73°/cosec 73° + cot 22°/cot 22° + sin2 46° + cos2 46°

sin2 θ + cos2 θ = 1

= 1 + 1 + 1

= 3

Question 5.

(i) cos 65°/sin 25° + cos 32°/sin 58° – sin 28° sec 62° + cosec2 30°

(ii) sec 29°/ cosec 61° + 2 cot 8° cot 17° cot 45° cot 73° cot 82° – 3 (sin2 38° + sin2 52°).

(i) cos 65°/sin 25° + cos 32°/sin 58° – sin 28° sec 62° + cosec2 30°

= cos 65°/sin (90° – 65°) + cos 32°/sin (90°–32°) – sin 28° sec (90°–28°) + cosec2 30°

= cos 65°/cos 65° + cos 32°/cos 32° – sin 28° cosec 28° + cosec2 30°

cosec 30° = 2

= 1 + 1 – 1 + 4

= 5

(ii) sec 29°/cosec 61° + 2 cot8° cot17° cot45° cot73° cot82° – 3 (sin2 38° + sin2 52°)

= sec 29°/cosec (90° – 29°) + 2 cot8° cot17° cot45° cot (90°–17°) cot (90°–8°) – 3 [sin238° + sin2 (90°–38°)]

= sec 29°/sec 29° + (2 cot8° cot17° × 1 × tan17° tan8°) – 3 (sin2 38° + cos2 38°)

= 1 + (2 cot 8° tan 8° cot 17° tan 17° × 1) – (3 × 1)

cosec (90° – θ) = sec θ

⇒ cot (90° – θ) = tan θ

⇒ sin2 θ + cos2 θ = 1

= 1 + (2 × 1 × 1 × 1) – 3

= 1 + 2 – 3

= 0

Question 6. Express each of the following in terms of trigonometric ratios of angles between 0° to 45°:

(i) tan 81° + cos 72°

(ii) cot 49° + cosec 87°.

(i)tan 81° + cos 72°

= tan (90° – 9°) + cos (90° – 18°)

= cot 9° + sin 18°

(ii) cot 49° + cosec 87°

= cot (90° – 41°) + cosec (90° – 3°)

= tan 41° + sec 3°

Without using trigonometric tables, prove that (7 to 11):

Question 7.

(i) sin2 28° – cos2 62° = 0

(ii) cos2 25° + cos2 65° = 1

(iii) cosec2 67° – tan2 23° = 1

(iv) sec2 22° – cot2 68° = 1.

(i) sin2 28° – cos2 62° = 0

LHS = sin2 28° – cos2 62°

= sin2 28° – cos2 (90° – 28°)

= sin2 28° – sin2 28°

= 0

= RHS

(ii) cos2 25° + cos2 65° = 1

LHS = cos2 25° + cos2 65°

= cos2 25° + cos2 (90° – 25°)

sin2 θ + cos2 θ = 1

= cos2 25° + sin2 25°

= 1

(iii) cosec2 67° – tan2 23° = 1

LHS = cosec2 67° – tan2 23°

= cosec2 67° – tan2 (90° – 67°)

cosec2 θ – cot2 θ = 1

= cosec2 67° – cot2 67°

= 1

(iv) sec2 22° – cot2 68° = 1

LHS = sec2 22° – cot2 68°

= sec2 22° – cot2 (90° – 22°)

sec2 θ – tan2 θ = 1

= sec2 22° – tan2 22°

= 1

Question 8.

(i) sin 63° cos 27° + cos 63° sin 27° = 1

(ii) sec 31° sin 59° + cos 31° cosec 59° = 2.

(i)sin 63° cos 27° + cos 63° sin 27° = 1

LHS = sin 63° cos 27° + cos 63° sin 27°

= sin 63° cos (90° – 63°) + cos 63° sin (90° – 63°)

= sin 63° sin 63° + cos 63° cos 63°

sin2 θ + cos2 θ = 1

= sin2 63° + cos2 63°

= 1

(ii) sec 31° sin 59° + cos 31° cosec 59° = 2

LHS = sec 31° sin 59° + cos 31° cosec 59°

= 1/cos 31° × sin 59° + (cos 31° × 1/sin 59°)

= sin 59°/cos (90° – 59°) + cos 31°/sin (90° – 31°)

= sin 59°/sin 59° + cos 31°/cos 31°

= 1 + 1

= 2

= RHS

Question 9.

(i) sec 70° sin 20° – cos 20° cosec 70° = 0

(ii) sin2 20° + sin2 70° – tan2 45° = 0.

(i)sec 70° sin 20° – cos 20° cosec 70° = 0

LHS = sec 70° sin 20° – cos 20° cosec 70°

= sin 20°/cos 70° – cos 20°/sin 70°

= sin 20°/cos (90° – 20°) – cos 20°/sin (90° – 20°)

= sin 20°/sin 20° – cos 20°/cos 20°

= 1 – 1

= 0

= RHS

(ii)sin2 20° + sin2 70° – tan2 45° = 0

LHS = sin2 20° + sin2 70° – tan2 45°

= sin2 20° + sin2 (90° – 20°) – tan2 45°

= sin2 20° + cos2 20° – tan2 45°

sin2 θ + cos2 θ = 1 and tan 45° = 1

= 1 – 1

= 0

= RHS

Question 10.

(i) cot 54°/tan 36° + tan 20°/cot 70° – 2 = 0

(ii) sin 50°/cos 40° + cosec 40°/sec 50° – 4 cos 50° cosec 40° + 2 = 0.

(i) cot 54°/tan 36° + tan 20°/cot 70° – 2 = 0

LHS = cot 54°/tan 36° + tan 20°/cot 70° – 2

= cot 54°/tan (90° – 54°) + tan 20°/cot (90° – 20°) – 2

= cot 54°/ cot 54° + tan 20°/ tan 20° – 2

= 1 + 1 – 2

= 0

= RHS

(ii)sin 50°/cos 40° + cosec 40°/sec 50° – 4 cos 50° cosec 40° + 2 = 0

LHS = sin 50°/cos 40° + cosec 40°/sec 50° – 4 cos 50° cosec 40° + 2

= sin 50°/cos (90° – 50°) + cosec 40°/sec (90° – 40°) – 4 cos 50° cosec (90° – 50°) + 2

= sin 50°/sin 50° + cosec 40°/cosec 40° – cos 50° sec 50° + 2

= 1 + 1 – (4 cos50°/cos 50°) + 2

= 1 + 1 – 4 + 2

= 4 – 4

= 0

= RHS

Question 11.

(i) cos 70°/sin 20° + cos 59°/sin 31° – 8 sin2 30° = 0

(ii) cos 80°/sin 10° + cos 59° cosec 31° = 2.

(i) cos 70°/sin 20° + cos 59°/sin 31° – 8 sin2 30° = 0

LHS = cos 70°/sin 20° + cos 59°/sin 31° – 8 sin2 30°

= cos 70°/sin (90° – 70°) + cos 59°/sin (90° – 59°) – 8 sin2 30°

sin 30° = ½

= cos 70°/cos 70° + cos 59°/ cos 59° – 8 (1/2)2

= 1 + 1 – (8 × ¼)

= 2 – 2

= 0

= RHS

(ii)cos 80°/sin 10° + cos 59° cosec 31° = 2

LHS = cos 80°/sin 10° + cos 59° cosec 31°

= cos 80°/sin (90° – 80°) + cos 59°/ sin 31°

= cos 80°/ cos 80° + cos 59°/ sin (90° – 59°)

= 1 + cos 59°/ cos 59°

= 1 + 1

= 2

= RHS

Trigonometrical Ratios of Standards Angles Exercise-18.2

ML Aggarwal Class 9 ICSE Maths Solutions

Page 437

Question 12. without using trigonometrical tables, evaluate:

(i)…………..

(ii)………….

(iii) sin2 34° + sin2 56° + 2 tan 18° tan 72° – cot2 30°.

(i)

= 2 + 1 – 3

= 0

(ii)

sin2 θ + cos2 θ = 1 and cosec2 θ – cot2 θ = 1

= 1/1

= 1

(iii)sin2 34° + sin2 56° + 2 tan 18° tan 72° – cot2 30°

= sin2 34° + [sin (90° – 34°)]2 + 2 tan 18° tan (90° – 18°) – cot2 30°

= sin2 34° + cos2 34° + 2 tan 18° cot 18° – (√3)2

= 1 + (2 tan 18°× 1/tan 18°) – 3

= 1 + 2 – 3

= 0

……………………

= cos θ/cos θ + sin θ/sin θ

= 1 + 1

= 2

= RHS

(ii) cos θ sin (90° – θ) + sin θ cos (90° – θ) = 1

LHS = cos θ sin (90° – θ) + sin θ cos (90° – θ)

= cos θ. cos θ + sin θ. sin θ

= cos2 θ + sin2 θ

= 1

= RHS

= tan θ/cot θ + cos θ/cos θ

= (tan θ × tan θ) + 1

= tan2 θ + 1

= sec2 θ

= RHS

……………………

= sin A cos A/cot A

= (sin A cos A × sin A)/cos A

= sin2 A

= 1 – cos2 A

= RHS

= cos A/sec A + sin A/cosec A

= (cos A × cos A) + (sin A × sin A)

= cos2 A + sin2 A

= 1

= RHS

…………………..

(i) It can be written as

= cos θ/cos θ + sin θ/cosec θ – 3 tan2 300

= 1 + (sin θ × sin θ) – 3 (1/√3)2

= sin2 θ + 1 – (3 × 1/3)

= sin2 θ + 1 – 1

= sin2 θ

(ii) It can be written as

= (sec θ cos θ tan θ)/(sin θ cosec θ tan θ) + cot θ/cot θ

So we get

= sec θ cos θ/sin θ cosec θ + 1

= 1/1 + 1

= 1 + 1

= 2

Question 16.

cos (90° – θ) = sin θ, tan (90° – θ) = cot θ and tan θ cot θ = 1

= 1/1

= 1

= RHS

Question 17. Find the value of A if

(i) sin 3A = cos (A – 6°), where 3A and A – 6° are acute angles

(ii) tan 2A = cot (A – 18°), where 2A and A – 18° are acute angles

(iii) If sec 2A = cosec (A – 27°) where 2A is an acute angle, find the measure of ∠A.

(i) sin 3A = cos (A – 6°), where 3A and A – 6° are acute angles

sin 3A = cos (A – 6°)

cos (90° – θ) = sin θ

cos (90° – 3A) = cos (A – 6°)

90° – 3A = A – 6°

90° + 6° = A + 3A

⇒ 96° = 4A

A = 96°/4 = 24°

Hence, the value of A is 24°.

(ii)tan 2A = cot (A – 18°)

cot (90° – θ) = tan θ

cot (90° – 2A) = cot (A – 18°)

90° – 2A = A – 18°

90° + 18° = A + 2A

3A = 108°

⇒ A = 108°/3 = 36°

Hence, the value of A is 36°.

(iii)sec 2A = cosec (A – 27°)

cosec (90° – θ) = sec θ

cosec (90° – 2A) = cos (A – 27°)

90° – 2A = A – 27°

90° + 27° = A + 2A

3A = 117°

⇒ A = 117°/3 = 39°

Hence, the value of A is 39°.

Question 18. Find the value of θ (0° < θ < 90°) if:

(i) cos 63° sec (90° – θ) = 1

(ii) tan 35° cot (90° – θ) = 1.

(i)cos 63° sec (90° – θ) = 1

cos 63° = 1/sec(90° – θ)

1/sec θ = cos θ

cos 63° = cos (90° – θ)

90° – θ = 63°

θ = 90° – 63° = 27°

(ii)tan 35° cot (90° – θ) = 1

tan 35° = 1/cot (90° – θ)

1/cot θ = cos θ

tan 35° = tan (90° – θ)

35° = 90° – θ

θ = 90° – 35°

= 55°

Question 19. If A, B and C are the interior angles of a △ ABC, show that

(i) cos (A + B)/2 = sin C/2

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

A, B and C are the interior angles of a △ ABC

∠A + ∠B + ∠C = 180°

Dividing both sides by 2

(∠A + ∠B + ∠C)/2 = 180°/2

⇒ A/2 + B/2 + C/2 = 90°

(i)cos (A + B)/2 = sin C/2

(A + B)/2 = 90° – C/2

cos (90° – C/2) = sin C/2

Here, cos (90° – θ) = sin θ

sin C/2 = sin C/2

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

(A + C)/2 = 90° – B/2

= tan (90° – B/2)

= cot B/2

= RHS

—  : End of ML Aggarwal Trigonometrical Ratios of Standards Angles Exe-18.2 Class 9 ICSE Maths Solutions :–

Return to :-  ML Aggarawal Maths Solutions for ICSE  Class-9

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