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

Trigonometric Identities Exercise 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

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 15

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 28

Solution 29

Solution 30

Solution 31(i)

Solution 31(ii)

Solution 31(iii)

Trigonometric Identities Exercise Ex. 13B

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Hence proved.

Solution 6

Solution 7

Solution 8

Now,

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

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 16

Now,

Trigonometric Identities Exercise MCQ

Solution 1

Solution 2

Correct Option: (c)

Solution 3

Solution 4

Correct Option: (d)

Solution 5

Correct Option: (b)

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

=

Solution 6

Correct Option: (c)

cos 9α = sin α

⇒ sin(90˚- 9α) = sin α

∴ 90˚- 9α = α ⇒ 10α = 90˚ ⇒ α = 9˚

∴ tan 5α = tan 5(9˚) = tan 45˚ = 1

Solution 7

Correct Option: (d)

Solution 8

Correct option: (a)


Solution 9

Solution 10

Correct Option: (b)

Solution 11

Correct option: (a)


Solution 12

Solution 13

Solution 14

Correct Option: (a)

cosec2 θ = 1 + cot2 θ =

Solution 15

Correct Option: (c)

Solution 16

Correct Option: (a)

In a ∆ABC, ∠C = 90˚ ⇒ ∠A + ∠B = 90˚

∴ cos(A + B) = cos 90˚ = 0

Solution 17

Correct Option: (a)

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)

Solution 18

Solution 19

Solution 20

Correct option: (c)


We know that

 

cosec2 x - cot2 x = 1


Solution 21

  

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