R S AGGARWAL AND V AGGARWAL Solutions for Class 10 Maths Chapter 13 - Trigonometric Identities

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Chapter 13 - Trigonometric Identities Ex. 13A

Solution 1

L.H.S = sin4θ - cos4θ

= (sin2θ)2 - (cos2θ)2

= (sin2θ - cos2θ) (sin2θ + cos2θ) … ∵ a2 - b2 = (a - b)(a + b)

= 1 (sin2θ - cos2θ) …. ∵ sin2θ + cos2θ = 1

= 1 - cos2θ - cos2θ …. ∵ sin2θ = 1 - cos2θ

= 1 - 2 cos2θ

= R.H.S

Hence proved.

Solution 2

Hence proved.

Solution 3

Hence proved.

Solution 4

Hence proved.

Solution 5

Hence proved.

Solution 6

Hence proved.

Solution 7

L.H.S = sec4 θ - tan4 θ

= (sec2 θ)2 - (tan2 θ)2

= (sec2 θ - tan2 θ) (sec2 θ + tan2 θ)

= 1 × (sec2 θ + tan2 θ) …. ∵ sec2 θ - tan2 θ = 1

= 1 + tan2 θ + tan2 θ …. ∵ sec2 θ = 1 + tan2 θ

= 1 + 2 tan2 θ

= R.H.S

Hence proved.

Solution 8

Hence proved.

Solution 9

L.H.S = R.H.S

Hence proved.

Solution 10

L.H.S = R.H.S

sin2 θ tan θ + cos2 θ cot θ + 2 sin θ cos θ = tan θ + cot θ

Solution 11

Hence proved.

Solution 12

Hence proved.

Solution 13

L.H.S = R.H.S

Solution 14

 

Hence proved.

Solution 16

∵ L.H.S = R.H.S

Solution 17

L.H.S = R.H.S

Solution 18

Hence proved.

Solution 19

Hence proved.

Solution 20

Hence proved.

Solution 21

Hence proved.

Solution 22

From (1) and (2), we get

L.H.S =

= 1 - sin θ cos θ + 1 + sin θ cos θ

= 2

= R.H.S

Hence proved.

Solution 23

Hence proved.

Solution 24

Hence proved.

Solution 25

Hence proved.

Solution 26

Hence proved.

Solution 27

Solution 15

LHS = RHS

Solution 28

Solution 29

Solution 30

Solution 33

Solution 34

Solution 35

Solution 31(i)

Solution 31(ii)

Solution 31(iii)

Solution 37

Chapter 13 - Trigonometric Identities Ex. 13B

Solution 5

Hence proved.

Solution 9

We have,

x = (sec A + sin A) and y = (sec A - sin A)

⇒ x + y = 2 sec A ⇒ sec A =  ⇒ cos A =  …. (1)

And x - y = 2 sin A sin A =  ….. (2)

Solution 10

m = (cos θ - sin θ) and n = (cos θ + sin θ)

m = (cos θ - sin θ) and n = (cos θ + sin θ) … given

Hence proved.

Solution 11

Hence proved.

Solution 12

Hence proved.

Solution 13

We know that,

sec2 θ - tan2 θ = 1

⇒ (sec θ + tan θ) (sec θ - tan θ) = 1 … a2 - b2 = (a - b)(a + b)

⇒ p (sec θ - tan θ) = 1 …. ∵ (sec θ + tan θ) = p

⇒ (sec θ - tan θ) =

We have,

(sec θ + tan θ) = p …. (1) and (sec θ - tan θ) =  …. (2)

Adding (1) and (2), we get

2 sec θ =  cos θ =  …. (3)

Subtracting (2) from (1), we get

2 tan θ =

tan θ

Solution 14

Hence proved.

Solution 15

Hence proved.

Solution 1

Solution 2

Solution 3

Solution 4

Solution 6

Solution 7

Solution 8

Chapter 13 - Trigonometric Identities MCQ

Solution 2

Correct Option: C

Solution 4

Correct Option: D

Solution 5

sin θ - cos θ = 0 …. Given

⇒ (sin θ - cos θ)2 = sin2 θ + cos2 θ - 2 sin θ cos θ

⇒ 0 = 1 - 2 sin θ cos θ …. ∵ sin2 θ + cos2 θ = 1

⇒ 2 sin θ cos θ = 1

⇒ sin θ cos θ =

⇒ sin2 θ cos2 θ =

∴ sin4 θ + cos4 θ

= (sin2 θ + cos2 θ)2 - 2 sin2 θ cos2 θ

=

Correct Option: B

Solution 6

cos 9α = sin α

⇒ sin(90˚- 9α) = sin α

∴ 90˚- 9α = α ⇒ 10α = 90˚ ⇒ α = 9˚

∴ tan 5α = tan 5(9˚) = tan 45˚ = 1

Correct Option: C

Solution 7

Correct Option: D

Solution 10

Correct Option: B

Solution 14

cosec2 θ = 1 + cot2 θ =

Correct Option: A

Solution 15

Correct Option: C

Solution 16

In a ∆ABC, ∠C = 90˚ ⇒ ∠A + ∠B = 90˚

∴ cos(A + B) = cos 90˚ = 0

Correct Option: A

Solution 17

We have, cos A + cos2 A = 1 … (1)

cos A = 1 - cos2 A = sin2 A sin4 A = cos2 A

(sin2 A + sin4 A) = cos A + cos2 A = 1 …from (1)

Correct Option: A

Solution 22

Solution 23

Solution 24

Solution 25

Solution 26

Correct option: (b)

sec2 60° - 1 = (2)2 - 1 = 4 - 1 = 3

Solution 27

Solution 28

Solution 29

Solution 30

Solution 31

  

Solution 32

  

 

Solution 33

Solution 21

  

Solution 35

  

Solution 36

Solution 37

Solution 20

Solution 19

Solution 40

  

Solution 11

Solution 13

Solution 43

Solution 44

Solution 18

Solution 46

Solution 1

Solution 48

Solution 8

Solution 9

Solution 12

Solution 52