NCERT Solutions For Class 11 Math Chapter – 3 Exercise – 3.3

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Prove that:

1.

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 1

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 2

2.

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 3

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 4

Here

= 1/2 + 4/4

= 1/2 + 1

= 3/2

= RHS

3.

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 5

Solution:

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4.

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 7

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 8

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 9

5. Find the value of:

(i) sin 75o

(ii) tan 15o

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 10

(ii) tan 15°

It can be written as

= tan (45° – 30°)

Using formula

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 11

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 12

Prove the following:

6.

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 13

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 14

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 15

7.

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 16

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 17

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 18

8.

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 19

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 20

9.

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 21

Solution:

Consider

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 22

It can be written as

= sin x cos x (tan x + cot x)

So we get

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 23

10. sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x = cos x

Solution:

LHS = sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 24

11.

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 25

Solution:

Consider

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 26

Using the formula

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 27

12. sin2 6x – sin2 4x = sin 2x sin 10x

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 28

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 29

13. cos2 2x – cos2 6x = sin 4sin 8x

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 30

We get

= [2 cos 4x cos (-2x)] [-2 sin 4x sin (-2x)]

It can be written as

= [2 cos 4x cos 2x] [–2 sin 4(–sin 2x)]

So we get

= (2 sin 4x cos 4x) (2 sin 2x cos 2x)

= sin 8x sin 4x

= RHS

14. sin 2x + 2sin 4x + sin 6x = 4cos2 x sin 4x

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 31

By further simplification

= 2 sin 4x cos (– 2x) + 2 sin 4x

It can be written as

= 2 sin 4x cos 2x + 2 sin 4x

Taking common terms

= 2 sin 4x (cos 2x + 1)

Using the formula

= 2 sin 4x (2 cos2 x – 1 + 1)

We get

= 2 sin 4x (2 cos2 x)

= 4cos2 x sin 4x

= R.H.S.

15. cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)

Solution:

Consider

LHS = cot 4x (sin 5x + sin 3x)

It can be written as

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 32

Using the formula

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 33

= 2 cos 4x cos x

Hence, LHS = RHS.

16.

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 34

Solution:

Consider

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 35

Using the formula

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 36

17.

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 37

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 28

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 39

18.

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 40

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 41

19.

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 42

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 43

20.

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 44

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 45

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 46

21.

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 47

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 48

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 49

22. cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 50

23.

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 51

Solution:

Consider

LHS = tan 4x = tan 2(2x)

By using the formula

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 52

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 53

24. cos 4x = 1 – 8sincosx

Solution:

Consider

LHS = cos 4x

We can write it as

= cos 2(2x)

Using the formula cos 2A = 1 – 2 sin2 A

= 1 – 2 sin2 2x

Again by using the formula sin2A = 2sin A cos A

= 1 – 2(2 sin x cos x2

So we get

= 1 – 8 sin2x cos2x

= R.H.S.

25. cos 6x = 32 cos6 x – 48 cos4 x + 18 cos2 – 1

Solution:

Consider

L.H.S. = cos 6x

It can be written as

= cos 3(2x)

Using the formula cos 3A = 4 cos3 A – 3 cos A

= 4 cos3 2x – 3 cos 2x

Again by using formula cos 2x = 2 cos2 – 1

= 4 [(2 cos2 – 1)3 – 3 (2 cos2 x – 1)

By further simplification

= 4 [(2 cos2 x3 – (1)3 – 3 (2 cos2 x2 + 3 (2 cos2 x)] – 6cos2 x + 3

We get

= 4 [8cos6x – 1 – 12 cos4x + 6 cos2x] – 6 cos2x + 3

By multiplication

= 32 cos6x – 4 – 48 cos4x + 24 cos2 x – 6 cos2x + 3

On further calculation

= 32 cos6– 48 cos4x + 18 cos2x – 1

= R.H.S.