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

Page / Exercise

Chapter 12 - Trigonometric Ratios of Complementary Angles 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

= sin2 43˚+ cos2 43˚ …… ∵ sin2 θ + cos2 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

cosec2 57˚- tan2 33˚

= cosec2 57˚ - tan2 (90˚ - 57˚)

= cosec2 57˚ - cot2 57˚

= 1 …. ∵ cosec2 θ - cot2 θ = 1

Correct Option: A

Solution 9

sec2 10˚- cot2 80˚

= sec2 10˚ - cot2 (90˚ - 10˚)

= sec2 10˚ - tan2 10˚

= 1 …. ∵ sec2 θ - tan2 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)

cos2 67˚+ cos2 23˚

= cos2 (90˚- 23˚) + cos2 23˚

= sin2 23˚ + cos2 23˚

= 1 ….. sin2 θ + cos2 θ = 1

∴ cos2 67˚+ cos2 23˚ = 1

Solution 21(iv)

cosec2 54˚- tan2 36˚

= cosec2 (90˚- 23˚) - tan2 36˚

= sin2 23˚ + cos2 23˚ …. ∵ cosec (90˚- θ) = sin θ

= 1 ….. ∵ sin2 θ + cos2 θ = 1

∴ cos2 67˚+ cos2 23˚ = 1

Solution 21(v)

sec2 40˚- cot2 50˚

= sec2 (90˚- 50˚) - cot2 50˚

= cosec2 50˚ - cot2 50˚

= 1 …. ∵ cosec2 50˚ - cot2 50˚ = 1

∴ sec2 40˚- cot2 50˚ = 1

Chapter 12 - Trigonometric Ratios of Complementary Angles 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 = sin2 48˚ + sin2 42˚

= sin2 (90˚- 42˚) + sin2 42˚

= cos2 42˚ + sin2 42˚ ….. ∵ sin(90˚- θ) = cos θ

= 1 ….. ∵ sin2 θ+ cos2θ= 1

= R.H.S

Hence proved.

Solution 2(vi)

L.H.S = cos2 72˚ + cos2 18˚

= cos2 (90˚- 18˚) + cos2 18˚

= sin2 18˚ + cos2 18˚ ….. ∵ cos(90˚- θ) = sin θ

= 1 ….. ∵ sin2 θ+ cos2θ= 1

= R.H.S

Hence proved.

Solution 3(i)

L.H.S = sin2 12˚ + sin2 78˚

= sin2 (90˚- 78˚) + sin2 78˚

= cos2 78˚ + sin2 78˚ ….. ∵ sin(90˚- θ) = cos θ

= 1 ….. ∵ sin2 θ+ cos2θ= 1

= R.H.S

Hence proved.

Solution 3(ii)

L.H.S = sec2 29˚- cot2 61˚

= sec2 29˚ - cot2 (90˚- 29˚)

= sec2 29˚ - tan229˚ ….. ∵ cot(90˚- θ) = tan θ

= 1 ….. ∵ sec2 θ - tan2θ= 1

= R.H.S

Hence proved.

Solution 3(iii)

L.H.S = tan2 56˚- cot2 34˚

= tan2 56˚- cot2 (90˚- 56˚)

= tan2 56˚- tan2 56˚ ….. ∵ cot(90˚- θ) = tan θ

= 0

= R.H.S

Solution 3(iv)

L.H.S = cos2 57˚ - sin2 33˚

= cos2 57˚ - sin2(90˚- 57˚)

= cos2 57˚ - cos2 57˚ ….. ∵ sin(90˚- θ) = cos θ

= 0

= R.H.S

Hence proved.

Solution 3(v)

L.H.S = sec2 50˚- cot2 40˚

= sec2 50˚ - cot2 (90˚- 50˚)

= sec2 50˚ - tan250˚ ….. ∵ cot(90˚- θ) = tan θ

= 1 ….. ∵ sec2 θ - tan2θ= 1

= R.H.S

Hence proved.

Solution 3(vi)

L.H.S = cosec2 72˚ - tan2 18˚

= cosec2 72˚ - tan2 (90˚- 72˚)

= cosec2 72˚ - cot2 72˚ ….. ∵ tan(90˚- θ) = cot θ

= 1 ….. ∵ cosec2 θ - cot2θ= 1

= R.H.S

Hence proved.

Solution 4(iv)

L.H.S = tan15˚ tan60˚ tan75˚

= (tan15˚ tan75˚) tan60˚

= [tan 15˚tan(90˚- 15˚)] tan60˚

= (tan 15˚cot 15˚) tan60˚ ……∵ tan(90˚- θ) = cot θ

= (tan 15˚×1/tan15˚) tan60˚ …… ∵ cot θ = 1/tan θ

= 1 × √3

= √3

= R.H.S

Hence proved.

Solution 4(v)

L.H.S = tan48˚ tan23˚ tan42˚ tan67˚ tan45˚

= (tan48˚ tan42˚) (tan23˚ tan67˚) tan45˚

= [tan 48˚tan(90˚- 48˚)] [tan 23˚tan(90˚- 23˚)] tan45˚

= (tan 48˚cot 48˚) (tan 23˚cot 23˚) tan45˚

……∵ tan(90˚- θ) = cot θ

= (tan 48˚×1/tan 48˚) (tan 23˚×1/tan 23˚) tan45˚

…… ∵ 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˚- sec2 21˚

= cosec 39˚cos(90˚- 39˚) + tan 21˚cot(90˚- 21˚) - sec2 21˚

= cosec 39˚sin 39˚+ tan 21˚tan 21˚- sec2 21˚

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

= 1 + tan221˚- sec2 21˚

= 1 - (sec2 21˚- tan221˚)

= 1 - 1 ……. ∵ sec2 θ - tan2θ = 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 + sin2 65˚+ sin2 25˚+ √3 tan 5˚ tan 45˚ tan 85˚

= cot θ cot θ - cosec θ cosec θ + sin2 (90˚- 25˚)+ sin2 25˚+ √3 (tan 5˚ tan 85˚) tan 45˚

= (cot2θ - cosec2θ) + (cos2 25˚+ sin2 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