# Class 10 R S AGGARWAL AND V AGGARWAL Solutions Maths Chapter 12 - Trigonometric Ratios of Complementary Angles

## Trigonometric Ratios of Complementary Angles Exercise MCQ

### Solution 1

**Correct
Option: **D

### Solution 2

**Correct
Option: **C

### Solution 3

tan 10˚ tan 15˚ tan 75˚ tan 80˚

= tan 10˚ tan (90˚- 10˚) tan 15˚ tan (90˚- 15˚)

= (tan 10˚ cot 10˚)( tan 15˚ cot 15˚) ……∵ tan (90˚- θ) = cot θ

= (tan 10˚ × 1/tan 10˚)( tan 15˚× 1/tan 15˚) …∵ cot θ = 1/ tan θ

= 1

**Correct
Option: **B

### Solution 4

tan 5˚ tan 25˚ tan 30˚ tan 65˚ tan 85˚

= tan 5˚tan(90˚- 5˚) tan 25˚ tan(90˚- 25˚) tan 30˚

= (tan 5˚cot 5˚)( tan 25˚cot25˚) tan 30˚… ∵ tan (90˚- θ) = cot θ

= (tan 5˚× 1/tan 5˚)( tan 25˚× 1/tan 25˚) ×

…… ∵ cot θ = 1/ tan θ

=

**Correct
Option: **C

### Solution 5

cos 1˚ cos 2˚ cos 3˚ …… cos 180˚

= cos 1˚ cos 2˚ cos 3˚ …… cos 90˚ ….. cos 180˚

= 0 ….. ∵ cos 90˚= 0

**Correct
Option: **A

### Solution 6

sin 43˚ cos 47˚ + cos 43˚ sin 47˚

= sin 43˚ cos(90˚ - 43˚) + cos 43˚ sin(90˚ - 43˚)

= sin 43˚ sin 43˚+ cos 43˚ cos 43˚ … ∵ cos(90˚- θ) = sin θ

= sin^{2}
43˚+ cos^{2} 43˚ …… ∵ sin^{2} θ + cos^{2} θ = 1

= 1

**Correct
Option: **B

### Solution 7

sec 70˚ sin 20˚ + cos 20˚ cosec 70˚

= sec (90˚- 20˚) sin 20˚ + cos 20˚ cosec (90˚- 20)

= cosec 20˚sin 20˚ + cos 20˚sec 20˚

…. ∵ sec (90˚- θ) = cosec θ, cosec (90˚- θ) = sec θ

= cosec 20˚× 1/cosec 20˚ + cos 20˚× 1/cos 20˚

…. ∵ sin θ = 1/ cosec θ, sec θ = 1/ cos θ

= 1 + 1

= 2

**Correct
Option: **C

### Solution 8

cosec^{2}
57˚- tan^{2} 33˚

= cosec^{2}
57˚ - tan^{2} (90˚ - 57˚)

= cosec^{2}
57˚ - cot^{2} 57˚

= 1 …. ∵ cosec^{2}
θ - cot^{2} θ = 1

**Correct
Option: **A

### Solution 9

sec^{2}
10˚- cot^{2} 80˚

= sec^{2}
10˚ - cot^{2} (90˚ - 10˚)

= sec^{2}
10˚ - tan^{2} 10˚

= 1 …. ∵ sec^{2}
θ - tan^{2} θ = 1

**Correct
Option: **B

### Solution 10

**Correct
Option: **D

### Solution 11

sin 38˚ - cos 52˚

= sin 38˚ - cos (90˚ - 38˚)

= sin 38˚- sin 38˚ …. ∵ cos (90˚ - θ) = sin θ

= 0

**Correct
Option: **A

### Solution 12

**Correct
Option: **B

### Solution 13

**Correct
Option: **C

### Solution 14

**Correct
Option: B**

### Solution 15

sin (60˚ + θ) - cos (30˚- θ)

= cos [90˚ - (60˚ + θ)] - cos (30˚- θ) …..∵ sin (90˚ + θ) = cos θ

= cos (30˚- θ) - cos (30˚- θ)

= 0

**Correct
Option: **C

### Solution 16

sin A = cos B

⇒ sin A = sin (90˚- B)

∴ A = 90˚- B ⇒ A + B = 90˚

**Correct
Option: **D

### Solution 17

cos(α + β) = 0

⇒ cos(α + β) = cos 90˚

∴ α + β = 90˚ ⇒ α = 90˚- β

∴ α - β = 90˚- β - β = 90˚- 2β

⇒ sin(α - β) = sin(90˚- 2β) = cos 2β … ∵ cos(90˚- θ) = sin θ

**Correct
Option: **B

### Solution 18

sin(45˚+ θ) - cos(45˚- θ)

= cos [90˚ - (45˚+ θ)] - cos(45˚- θ)

= cos(45˚- θ) - cos(45˚- θ) …. ∵ cos(90˚- θ) = sin θ

= 0

**Correct
Option: **A

### Solution 19

sec 4A = cosec (A - 10˚)

⇒ cosec (90˚ - 4A) = cosec (A - 10˚) ….∵ cosec (90˚ - θ) = sec θ

∴ 90˚ - 4A = A - 10˚

⇒ 5A = 100˚

⇒ A = 20˚

**Correct
Option: **A

### Solution 20

sin 3A = cos(A - 10˚)

⇒ cos(90˚ - 3A) = cos(A - 10˚) …. ∵ cos(90˚ - θ) = sin θ

∴ 90˚ - 3A = A - 10˚

⇒ 4A = 100˚

⇒ A = 25˚

**Correct
Option: **C

### Solution 21(i)

cot 34˚- tan 56˚

= cot (90˚- 56˚) - tan 56˚

= tan 56˚- tan 56˚ …. ∵ cot (90˚- θ) = tan θ

= 0

∴ cot 34˚- tan 56˚ = 0

### Solution 21(ii)

cosec 31˚- sec 59˚

= cosec (90˚- 59˚) - sec 59˚

= sec 59˚- sec 59˚ …. ∵ cosec (90˚- θ) = sec θ

= 0

∴ cosec 31˚- sec 59˚ = 0

### Solution 21(iii)

cos^{2}
67˚+ cos^{2} 23˚

= cos^{2}
(90˚- 23˚) + cos^{2} 23˚

= sin^{2}
23˚ + ^{ }cos^{2} 23˚

= 1 ….. ∵ sin^{2}
θ + ^{ }cos^{2} θ = 1

∴ cos^{2}
67˚+ cos^{2} 23˚ = 1

### Solution 21(iv)

cosec^{2}
54˚- tan^{2} 36˚

= cosec^{2}
(90˚- 23˚) - tan^{2} 36˚

= sin^{2}
23˚ + cos^{2} 23˚ …. ∵ cosec (90˚- θ) = sin θ

= 1 ….. ∵ sin^{2}
θ + ^{ }cos^{2} θ = 1

∴ cos^{2}
67˚+ cos^{2} 23˚ = 1

### Solution 21(v)

sec^{2}
40˚- cot^{2} 50˚

= sec^{2}
(90˚- 50˚) - cot^{2} 50˚

= cosec^{2}
50˚ - cot^{2} 50˚

= 1 …. ∵ cosec^{2}
50˚ - cot^{2} 50˚ = 1

∴ sec^{2}
40˚- cot^{2} 50˚ = 1

## Trigonometric Ratios of Complementary Angles Exercise Ex. 12

### Solution 1(i)

### Solution 1(ii)

### Solution 1(iii)

### Solution 1(iv)

### Solution 1(v)

### Solution 1(vi)

### Solution 2(i)

L.H.S = cos81˚ - sin9˚

= cos(90˚- 9˚) - sin9˚

= sin9˚ - sin9˚ ….. ∵ cos(90˚- θ) = sin θ

= 0

= R.H.S

Hence proved.

### Solution 2(ii)

L.H.S = tan71˚ - cot19˚

= tan(90˚- 19˚) - cot19˚

= cot19˚ - cot19˚ ….. ∵ tan(90˚- θ) = cot θ

= 0

= R.H.S

Hence proved.

### Solution 2(iii)

L.H.S = cosec60˚ - sec30˚

= cosec(90˚- 30˚) - sec30˚

= sec30˚ - sec30˚ ….. ∵ cosec(90˚- θ) = sec θ

= 0

= R.H.S

Hence proved.

### Solution 2(iv)

L.H.S = cot34˚ - tan56˚

= cot(90˚- 56˚) - tan56˚

= tan56˚ - tan56˚ ….. ∵ cot(90˚- θ) = tan θ

= 0

= R.H.S

Hence proved.

### Solution 2(v)

L.H.S = sin^{2 }48˚
+ sin^{2 }42˚

= sin^{2}
(90˚- 42˚) + sin^{2 }42˚

= cos^{2
}42˚ + sin^{2 }42˚ ….. ∵ sin(90˚- θ) = cos θ

=
1 ….. ∵ sin^{2 }θ+ cos^{2}θ= 1

= R.H.S

Hence proved.

### Solution 2(vi)

L.H.S = cos^{2 }72˚
+ cos^{2 }18˚

= cos^{2}
(90˚- 18˚) + cos^{2 }18˚

= sin^{2
}18˚ + cos^{2 }18˚ ….. ∵ cos(90˚- θ) = sin θ

=
1 ….. ∵ sin^{2 }θ+ cos^{2}θ= 1

= R.H.S

Hence proved.

### Solution 3(i)

L.H.S = sin^{2 }12˚
+ sin^{2 }78˚

= sin^{2}
(90˚- 78˚) + sin^{2 }78˚

= cos^{2
}78˚ + sin^{2 }78˚ ….. ∵ sin(90˚- θ) = cos θ

=
1 ….. ∵ sin^{2 }θ+ cos^{2}θ= 1

= R.H.S

Hence proved.

### Solution 3(ii)

L.H.S = sec^{2 }29˚-
cot^{2 }61˚

= sec^{2
}29˚ - cot^{2} (90˚- 29˚)

= sec^{2
}29˚ - tan^{2}29˚ ….. ∵ cot(90˚- θ) = tan θ

=
1 ….. ∵ sec^{2 }θ - tan^{2}θ=
1

= R.H.S

Hence proved.

### Solution 3(iii)

L.H.S = tan^{2 }56˚-
cot^{2 }34˚

= tan^{2
}56˚- cot^{2} (90˚- 56˚)

= tan^{2
}56˚- tan^{2 }56˚ ….. ∵ cot(90˚- θ) = tan θ

= 0

= R.H.S

### Solution 3(iv)

L.H.S = cos^{2 }57˚
- sin^{2 }33˚

= cos^{2
}57˚ - sin^{2}(90˚- 57˚)

= cos^{2
}57˚ - cos^{2 }57˚ ….. ∵ sin(90˚- θ) = cos θ

= 0

= R.H.S

Hence proved.

### Solution 3(v)

L.H.S = sec^{2 }50˚-
cot^{2 }40˚

= sec^{2
}50˚ - cot^{2} (90˚- 50˚)

= sec^{2
}50˚ - tan^{2}50˚ ….. ∵ cot(90˚- θ) = tan θ

=
1 ….. ∵ sec^{2 }θ - tan^{2}θ=
1

= R.H.S

Hence proved.

### Solution 3(vi)

L.H.S = cosec^{2
}72˚ - tan^{2 }18˚

= cosec^{2
}72˚ - tan^{2} (90˚- 72˚)

= cosec^{2
}72˚ - cot^{2} 72˚ ….. ∵ tan(90˚- θ) = cot θ

=
1 ….. ∵ cosec^{2 }θ - cot^{2}θ=
1

= R.H.S

Hence proved.

### Solution 4(iv)

L.H.S = tan^{}15˚
tan^{}60˚ tan^{}75˚

= (tan^{}15˚ tan^{}75˚) tan^{}60˚

= [tan
15˚tan(90˚- 15˚)] tan^{}60˚

= (tan
15˚cot 15˚) tan^{}60˚ ……∵ tan(90˚- θ) = cot θ

= (tan
15˚×1/tan15˚) tan^{}60˚ …… ∵ cot θ = 1/tan θ

= 1 × √3

= √3

= R.H.S

Hence proved.

### Solution 4(v)

L.H.S = tan^{}48˚
tan^{}23˚ tan^{}42˚ tan^{}67˚ tan^{}45˚

= (tan^{}48˚ tan^{}42˚) (tan^{}23˚ tan^{}67˚) tan^{}45˚

= [tan
48˚tan(90˚- 48˚)] [tan 23˚tan(90˚- 23˚)] tan^{}45˚

= (tan
48˚cot 48˚) (tan 23˚cot 23˚) tan^{}45˚

……∵ tan(90˚- θ) = cot θ

= (tan
48˚×1/tan 48˚) (tan 23˚×1/tan 23˚) tan^{}45˚

…… ∵ cot θ = 1/tan θ

= 1×1×1

= 1

= R.H.S

Hence proved.

### Solution 4(vii)

L.H.S = cosec
39˚cos51˚ + tan21˚ cot 69˚- sec^{2 }21˚

= cosec 39˚cos(90˚-
39˚) + tan 21˚cot(90˚-
21˚) - sec^{2 }21˚

= cosec 39˚sin
39˚+ tan 21˚tan 21˚- sec^{2 }21˚

….. ∵ cot(90˚- θ) = tan θ and cos(90˚- θ) = sin θ

= 1 + tan^{2}21˚-
sec^{2 }21˚

= 1 - (sec^{2
}21˚- tan^{2}21˚)

= 1 - 1 …….
∵ sec^{2 }θ
- tan^{2}θ = 1

= 0

= R.H.S

Hence proved.

### Solution 5(i)

sin 72˚ + cosec 72˚

= sin(90˚- 18˚) + cosec(90˚- 18˚)

= cos 18˚+ sec 18˚

### Solution 5(ii)

cosec 66˚ + tan 66˚

= cosec(90˚- 24˚) + tan(90˚- 24˚)

= sec 24˚+ cot 24˚

### Solution 5(iii)

tan 68˚ + sec 68˚

= tan(90˚- 22˚) + sec(90˚- 22˚)

= cot 22˚+ cosec 22˚

### Solution 5(iv)

cot 59˚ + cosec 59˚

= cot(90˚- 31˚) + cosec(90˚- 31˚)

= tan 31˚+ sec 31˚

### Solution 5(v)

cos 51˚ + cot 49˚ - sec 47˚

= cos(90˚- 39˚) + cot(90˚- 41˚) - sec(90˚- 43˚)

= sin 39˚+ tan 41˚- cosec 43˚

### Solution 5(vi)

sin 67˚ + cos 75˚

= sin(90˚- 23˚) + cos(90˚- 15˚)

= cos 23˚+ sin 15˚

### Solution 6

sin 3A = cos (A - 10˚)

⇒ cos(90˚- 3A) = cos (A - 10˚) = ….. ∵ cos(90˚ - θ) = sin θ

∴ 90˚- 3A = A - 10˚

⇒ 4A = 100˚

⇒ A = 25˚

### Solution 7

tan A = cot (A + 10˚)

⇒ cot(90˚- A) = cot(A + 10˚) = ….. ∵ cot(90˚ - θ) = tan θ

∴ 90˚- A = A + 10˚

⇒ 2A = 80˚

⇒ A = 40˚

### Solution 8

cos 2A = sin (A - 15˚)

⇒ sin(90˚- 2A) = sin (A - 15˚) = ….. ∵ sin(90˚ - θ) = cos θ

∴ 90˚- 2A = A - 15˚

⇒ 3A = 105˚

⇒ A = 35˚

### Solution 10

tan 2θ = cot (θ + 6˚)

⇒ cot (θ + 6˚) = cot(90˚- 2θ) ….. ∵ cot(90˚- θ) = tan θ

∴ θ + 6˚= 90˚- 2θ

⇒ 3θ = 84˚

⇒ θ = 28˚

### Solution 11

sin (θ + 36˚) = cos θ

⇒ sin (θ + 36˚) = sin(90˚- θ) ….. ∵ sin(90˚ - θ) = cos θ

∴ θ + 36˚= 90˚- θ

⇒ 2θ = 54˚

⇒ θ = 27˚

### Solution 12

3 cot 31˚tan 15˚cot 27˚tan 75˚cot 63˚cot 59˚

= 3 (cot 31˚cot 59˚)(tan 15˚tan 75˚)(cot 27˚cot 63˚)

= 3 [cot 31˚cot (90˚- 31˚)] [tan 15˚tan (90˚- 15˚)] [cot 27˚cot (90˚- 27˚)]

= 3 (cot 31˚tan 31˚)(tan 15˚cot 15˚)(cot 27˚tan 27˚)

….. ∵ tan(90˚- θ) = cot θ and cot(90˚- θ) = tan θ

= 3 …..∵ tan θ = 1/ cot θ and cot θ = 1/ tan θ

### Solution 13

### Solution 14

### Solution 15

### Solution 16

### Solution 17

cot θ tan(90˚- θ) -
sec(90˚- θ) cosec θ + sin^{2} 65˚+ sin^{2} 25˚+ √3
tan 5˚ tan 45˚ tan 85˚

= cot θ cot θ -
cosec θ cosec θ + sin^{2} (90˚- 25˚)+ sin^{2} 25˚+ √3
(tan 5˚ tan 85˚) tan 45˚

= (cot^{2}θ
- cosec^{2}θ) + (cos^{2} 25˚+ sin^{2} 25˚) +
√3 [tan 5˚ tan (90˚- 5˚) × 1

= - 1 + 1 + √3 (tan 5˚cot 5˚)

= 0 + √3 (tan 5˚× )

= √3

### Solution 18

### Solution 19

### Solution 20

sin(50˚+ θ) - cos(40˚- θ) + tan1˚tan10˚tan20˚tan70˚tan80˚ tan89˚

= cos [90˚- (50˚- θ)] - cos(40˚- θ) + (tan1˚tan89˚) (tan10˚tan80˚) (tan20˚tan70˚) ….. ∵ sin θ = cos(90˚- θ)

= cos(40˚- θ)] - cos(40˚- θ) + [tan1˚tan(90˚- 1˚)] [tan10˚tan(90˚- 10˚)] [tan20˚tan(90˚- 20˚)]

= 0 + (tan1˚cot1˚)( tan10˚cot10˚)( tan20˚cot20˚) … ∵ tan(90˚- θ)= cot θ

= 1 × 1 × 1 …….∵ tan θ =

= 1

### Solution 4(i)

### Solution 4(ii)

### Solution 4(iii)

### Solution 4(vi)

### Solution 5(vii)

### Solution 9

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