NCERT Grade 11-Trigonometric Functions-Answers

NCERT Solutions for Class 11 Maths

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1.      

2.     

3.    Number of revolutions in 1 minute = 360

∴ Number of revolution in 60 seconds = 360

Number of revolutions in 1 second = 360/60 = 6 revolutions

∴ Angle made by wheel in 6 revolutions = 360 × 6 = 2160°

4.    Here r = 100 cm and l = 22 cm

5.    Given: Diameter AB = 40 cm, Radius OA = 20 cm and Chord AC = 20 cm

∴ ΔAOC is an equilateral triangle.

6.    Let r1 and r2 be radii of two circles in which arcs of same length l subtend angles θ1 = 60° and θ2 = 75° respectively.

7.    (i) Given: length of pendulum (r) = 75 cm and length of arc (l) = 10 cm

(i) Given: length of pendulum (r) = 75 cm and length of arc (l) = 10 cm

(i) Given: length of pendulum (r) = 75 cm and length of arc (l) = 10 cm

8.    

9.    

10.   

11.   

12.   

13.   Here sin 765° = sin (2 × 360° + 45°) = sin 45° 

14.   Here cos ec (-1410)° = cos ec (-4 × 360° + 30°) = cos ec 30° = 2

15.   Here  = tan 1140 = tan (3 × 360° + 60°) = tan 60° = √3

16.   Here 

17.   Here 

18.   Taking L.H.S

19.   Taking L.H.S

20.   Taking L.H.S

21.   

22.   (i) sin 75° = sin(45° + 30°) = sin 45° cos 30° + cos 45° sin 30°

23.   Taking L.H.S

24.   Taking L.H.S

25.   Taking L.H.S

26.   Taking L.H.S

27.   Taking L.H.S

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

= cos [(n + 1)x – (n + 2)x]

= cos [nx + x – nx – 2x]

= cos (-x) = cos x = R.H.S.

28.   Taking L.H.S

29.   L.H.S. = sin2 6x – sin2 4x

= sin (6x + 4x). sin (6x – 4x)

[∵ sin2 x – sin2 y = sin (x + y) sin (x – y)]

= sin 10x sin 4x = R.H.S.

30.   L.H.S. = cos2 2x – cos2 6x

= sin (2x + 6x).sin(6x – 2x)

[∵ cos2 y – cos2 x = sin (x + y) sin (x – y)]

= sin 8x sin 4x = R.H.S.

31.   L.H.S. = sin 2x + 2 sin 4x + sin 6x

32.   L.H.S. = cot 4x (sin 5x + sin 3x)

33.   

34.   

35.   

36.   

37.   

38.   

39.   We know that cot 3x = cot (2x + x)

⇒ cot 3x (cot 2x + cot 2x) = cot 2x cot x – 1

⇒ cot 3x cot 2x + cot 3x cot x = cot 2x cot x – 1

⇒ cot 3x cot 2x + cot 3x cot x – cot 2x cot x + 1 = 0

⇒ cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1

40.   

41.   L.H.S. = cos 4x = 1 – 2 sin2 2x

= 1 – 2 (2sin x cos x)2

= 1 – 2(4 sin2x cos2 x)

= 1 – 8 sin2 x cos2 x = R.H.S.

42.   L.H.S. = cos 6x = 2 cos2 3x – 1

= 2 [4cos3 x – 3cos x]2 – 1

= 2 [16cos2 x + 9cos2 x – 24cos4 x] – 1

= 32cos6 x + 18cos2 x – 48cos4 x – 1

= 32cos6 x – 48 cos4 x + 18 cos2 x – 1 = R.H.S.

43.   Given: tan x = √3 Here x lies in first or third quadrant.

∴ tan x = tan 60° or tan  x = tan (180° + 60°)

⇒ tan x = tan 60° or tan x = tan 240°

⇒ tan x = tan   or tan x = tan 

Therefore, the principal solutions are 

Now, tan x = tan 

⇒ x = nπ +  where n ∈ z.

44.   Given: sec x = 2 ⇒ cos x =  Here x lies in first or fourth quadrant.

∴ cos x = cos 60° or cos x = cos (360° – 60°)

⇒ cos x = cos 60° or cos x = cos 300°

Therefore, the principal solutions are 

Now, cos x = cos 

⇒ x = 2nπ ± where n ∈ z

45.   Given:  Here x lies in second or fourth quadrant.

∴ tan x = -tan 30° = tan (180° – 30°) or tan x = tan (360° – 60°)

⇒ tan x = tan 150° or tan x = tan 330°

Therefore, the principal solutions are 

46.   Given:  Here x lies in third or fourth quadrant.

∴ sin x = -sin 30° = sin (180° + 30°) or in x = sin(360° – 30°)

⇒ sin x = sin 210° or sin x = sin 330°

Therefore, the principal solutions are 

47.   Given: cos 4x = cos 2x

⇒ 4x = 2nπ ± 2x, n ∈ z

⇒ 4x – 2x = 2nπ or 4x + 2 = 2nπ, n ∈ z

⇒ 2x = 2nπ or 6x = 2nπ, n ∈ z

Therefore, the principal solutions are 

48.   Given: cos 3x + cos x – cos 2x = 0

49.   Given: sin 2x + cos x = 0

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