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Transformation Formulae

Class 11th Mathematics RD Sharma Solution
Exercise 8.1
  1. Express each of the following as the sum or difference of sines and cosines: (i)…
  2. Prove that : i. 2sin 5 pi /12 sin pi /12 = 1/2 ii. 2cos 5 pi /12 cos pi /12 =…
  3. Show that: i. sin50^circle cos85^circle = 1 - root 2 sin35^circle /2 root 2 ii.…
  4. 4cosxcos (pi /3 + x) cos (pi /3 - x) = cos3x Prove that:
  5. cos10^circle cos30^circle cos50^circle cos70^circle = 3/16 Prove that:…
  6. cos40^circle cos80^circle cos160^circle = -1/8 Prove that:
  7. sin20^circle sin40^circle sin80^circle = root 3/8 Prove that:
  8. cos20^circle cos40^circle cos80^circle = 1/8 Prove that:
  9. tan 20° tan 40° tan 60° tan 80° = 3 Prove that:
  10. tan 20° tan 30° tan 40° tan 80° = 1 Prove that:
  11. sin10^circle sin50^circle sin60^circle sin70^circle = root 3/16 Prove that:…
  12. sin20^circle sin40^circle sin60^circle sin80^circle = 3/16 Prove that:…
  13. Show that i. sin A sin (B - C) + sin B sin (C - A) + sin C sin (A - B) = 0 ii.…
  14. tanxtan (pi /3 - x) tan (pi /3 + x) = tan3x Prove that:
  15. alpha + beta = pi /2 , cosalpha cosbeta = 1/2
Exercise 8.2
  1. Express each of the following as the product of sines and cosines: i. sin 12x +…
  2. sin 38° + sin 22°= sin 82° Prove that :
  3. cos 100° + cos 20° = cos 40° Prove that :
  4. sin 50° + sin 10° = cos 20° Prove that :
  5. sin 23° + sin 37° = cos 7° Prove that :
  6. sin 105° + cos 105° = cos 45° Prove that :
  7. sin 40° + sin 20° = cos 10° Prove that :
  8. cos 55° + cos 65° + cos 175° = 0 Prove that :
  9. sin 50° - sin 70° + sin 10° = 0 Prove that :
  10. cos 80° + cos 40° - cos 20° = 0 Prove that :
  11. cos 20° + cos 100° + cos 140° = 0 Prove that :
  12. sin 5 pi /18 - cos 4 pi /9 = root 3 sin pi /9
  13. cos pi /12 - sin pi /12 = 1/root 2 Prove that :
  14. sin 80° - cos 70° = cos 50° Prove that :
  15. sin 51° + cos 81° = cos 21° Prove that :
  16. i. cos (3 pi /4 + x) - cos (3 pi /4 - x) = - root 2 sinx ii. cos (pi /4 + x) +…
  17. i. sin65^circle + cos65^circle = root 2 cos20^circle ii. sin 47° + cos 77° = cos…
  18. cos 3A + cos 5A + cos 7A + cos 15A= 4 cos 4A cos 5A cos 6A Prove that :…
  19. cos A + cos 3A + cos 5A + cos 7A = 4 cos A cos 2A cos 4A Prove that :…
  20. sina+sin2a+sin4a+sin5a = 4sin3acos a/2 cos 3a/2 Prove that :
  21. sin3a+sin2a-sina = 4sinacos a/2 cos 3a/2 Prove that :
  22. cos20^circle cos100^circle + cos100^circle cos140^circle - cos140^circle…
  23. sin x/2 sin 7x/2 + sin 3x/2 sin 11x/2 = sin2xsin5x Prove that :
  24. cosxcos x/2 - cos3xcos 9x/2 = sin4xsin 7x/2 Prove that :
  25. sina+sin3a/cosa-cos3a = cota Prove that:
  26. sin9a-sin7a/cos7a-cos9a = cot8a Prove that:
  27. sina-sinb/cosa+cosb = tan a-b/2 Prove that:
  28. sina+sinb/sina-sinb = tan (a+b/2) cot (a-b/2) Prove that:
  29. cosa+cosb/cosb-cosa = cot (a+b/2) cot (a-b/2) Prove that:
  30. sina+sin3a+sin5a/cosa+cos3a+cos5a = tan3a Prove that:
  31. cos3a+2cos5a+cos7a/cosa+2cos3a+cos5a = cos5a/cos3a Prove that:
  32. cos4a+cos3a+cos2a/sin4a+sin3a+sin2a = cot3a Prove that:
  33. sin3a+sin5a+sin7a+sin9a/cos3a+cos5a+cos7a+cos9a = tan6a Prove that:…
  34. sin5a-sin7a+sin8a-sin4a/cos4a+cos7a-cos5a-cos8a = cot6a Prove that:…
  35. sin5acos2a-sin6acosa/sinasin2a-cos2acos3a = tana Prove that:
  36. sin11asina+sin7asin3a/cos11asina+cos7asin3a = tan8a Prove that:
  37. sin3acos4a-sinacos2a/sin4asina+cos6acosa = tan2a Prove that:
  38. sinasin2a+sin3asin6a/sinacos2a+sin3acos6a = tan5a Prove that:
  39. sina+2sin3a+sin5a/sin3a+2sin5a+sin7a = sin3a/sin5a Prove that:
  40. sin (theta + phi) - 2sintegrate heta +sin (theta - phi)/cos (theta + phi) -…
  41. i. sinalpha +sinbeta +singamma -sin (alpha + beta + gamma) = 4sin alpha + beta…
  42. cosa+cosb = 1/2 sina+sinb = 1/4 prove that tan (a+b/2) = 1/2
  43. coseca+seca = cosecb+secb, : tanatanb = cot a+b/2
  44. If sin 2A = λ sin 2B, prove that: tan (a+b)/tan (a-b) = lambda +1/lambda -1…
  45. i. cos (a+b+c) + cos (-a+b+c) + cos (a-b+c) + cos (a+b-c)/sin (a+b+c) + sin…
  46. cos (a-b)/cos (a+b) + cos (c+d)/cos (c-d) = 0 prove that tan A tan B tan C tan…
  47. If cos (α + β) sin (γ + δ) = cos (α - β) sin (γ - δ), prove that cot α cot β…
  48. If y sin ϕ = x sin (2θ + ϕ), prove that (x + y) cot (θ + ϕ) = (y -x) cot θ…
  49. If cos (A + B) sin (C - D) = cos (A - B) sin (C + D), prove that tan A tan B…
  50. xcostheta = ycos (theta + 2 pi /3) = zcos (theta + 4 pi /3) , xy+yz+zx = 0…
  51. msintegrate heta = nsin (theta +2 alpha) , tan (theta + alpha) cotalpha =…
Very Short Answer
  1. If (cos α + cos β)2 + (sin α + sin β)2 = lambda cos^{2} ( { alpha - beta }/{2} )…
  2. Write the value of sin { pi }/{12} sin frac { 5 pi }/{12}
  3. If sin A + sin B = α and cos A + cos B = β, then write the value of tan ( {a+b}/{2}…
  4. If cos A = m cos B, then write the value of cot {a+b}/{2} cot frac {a-b}/{2}…
  5. Write the value of the expression { 1-4sin10^{degree }sin70^{circ} }/{…
  6. If a+b = { pi }/{3} and cos A + cos B = 1, then find the value of cos…
  7. Write the value of sin { pi }/{15} sin frac { 4 pi }/{15} sin frac { 3 pi }/{10}…
  8. If sin 2A = λ sin 2B, then write the value of { lambda +1 }/{ lambda-1 }…
  9. Write the value of {sina+sin3a}/{cosa+cos3a}
  10. If cos (A + B) sin (C – D) = cos (A – B) sin (C + D), then write the value tan A tan B…
Mcq
  1. cos 40o + cos 80o + cos 160o + cos 240o = Mark the Correct alternative in the…
  2. sin 163o cos 347o + sin 73o sin 167o = Mark the Correct alternative in the following:…
  3. If sin2theta +sin2phi = {1}/{2} and cos2theta +cos2phi = {3}/{2} then…
  4. The value of cos 52o + cos 68o + cos 172o is Mark the Correct alternative in the…
  5. The value of sin 78o – sin 66o – sin 42o + sin 6o is Mark the Correct alternative in…
  6. If sin α + sin β = a and cos α – cos β = b, then tan { alpha - beta }/{2} = Mark…
  7. cos 35o + cos 85o + cos 155o = Mark the Correct alternative in the following:…
  8. The value of sin 50o – sin 70o + sin 10o is equal to Mark the Correct alternative in…
  9. sin 47o + sin 61o – sin 11o – sin 25o is equal to Mark the Correct alternative in the…
  10. If cos A = m cos B, then cot {a+b}/{2} cot frac {b-a}/{2} = Mark the Correct…
  11. If A, B, C are in A.P., then {sina-sinc}/{cosc-cosa} = Mark the Correct…
  12. If sin (B + C – A), sin (C + A – B), sin (A + B – C) are in A.P., then cot A, cot B,…
  13. If sinx+siny = root {3} (cosy-cosx) then sin 3x + sin 3y = Mark the Correct…
  14. If tanalpha = {x}/{x+1} and tanbeta = {1}/{2x+1} thenα + β is equal to…
Exercise 8.1
Question 1.

Express each of the following as the sum or difference of sines and cosines:

(i) 2 sin 3x cos x

(ii) 2 cos 3x sin 2x

(iii) 2 sin 4x sin 3x

(iv) 2 cos 7x cos 3x
Answer:

(i) We know, 2 sin A cos B = sin(A + B) + sin(A – B)


2 sin 3x cos x = sin (3x + x) + sin (3x – x)


= sin (4x) + sin (2x)


= sin 4x + sin 2x


(ii) We know, 2 cos A sin B = sin (A + B) – sin (A – B)


2 cos 3x sin 2x = sin (3x + 2x) – sin (3x – 2x)


= sin (5x) – sin (x)


= sin 5x – sin x


(iii) We know, 2 sin A sin B = cos (A – B) – cos (A + B)


2 sin 4x sin 3x = cos (4x – 3x) – cos (4x + 3x)


= cos (x) – cos (7x)


= cos x – cos 7x


(iv) We know, 2 cos A cos B = cos (A + B) + cos (A – B)


2 sin 3x cos x = cos (7x + 3x) + cos (7x – 3x)


= cos (10x) + cos (4x)


= cos 10x + cos 4x
Question 2.

Prove that :

i.

ii.

iii.
Answer:

i. We know, 2 sin A sin B = cos (A – B) – cos (A + B)






= cos 60° - cos 90°




Hence Proved


ii. We know, 2 cos A cos B = cos (A + B) + cos (A – B)






= cos 90° + cos 60°




Hence Proved


iii. We know, 2 sin A cos B = sin (A + B) + sin (A – B)





= sin 90° + sin 60°




Hence Proved
Question 3.

Show that:

i.

ii.


Answer:

i. We know, 2 sin A cos B = sin (A + B) + sin (A – B)






{sin (-x) = - sin x}



{sin (180° - x) = sin x}






Hence Proved


ii. We know, 2 sin A cos B = sin (A + B) + sin (A – B)



Take L.H.S




{sin (-x) = - sin x}




= R.H.S.


Hence Proved
Question 4.

Prove that:
Answer:

Take L.H.S



{2 cos A cos B = cos (A + B) + cos (A – B)}




= 2 cos x {cos 120° + cos 2x}


= 2 cos x {cos (180° - 60°) + cos 2x}


{cos (180° - A) = - cos A}


= 2 cos x (cos 2x - cos 60°)


= 2 cos 2x cos x – 2 cos x cos 60°



= cos 3x + cos x – cos x


= cos 3x = R.H.S.


Hence Proved
Question 5.

Prove that:
Answer:

Take L.H.S




Multiplying & Dividing by 2:




{∵ 2 cos A cos B = cos (A + B) + cos (A – B)}





{∵ cos (-A) = cos A}




Multiplying & Dividing by 2:



{∵ 2 cos A cos B = cos (A + B) + cos (A – B)}





{∵ cos (180° - A) = - cos A}





= R.H.S


Hence Proved
Question 6.

Prove that:
Answer:

Take L.H.S


cos 40° cos 80° cos 160°


Multiplying & Dividing by 2:



{∵ 2 cos A cos B = cos (A + B) + cos (A – B)}






{∵ cos (180° - A) = - cos A & cos (180° + A) = - cos A}






Multiplying & Dividing by 2:



{∵ 2 cos A cos B = cos (A + B) + cos (A – B)}





{∵ cos (180° - A) = - cos A}





= R.H.S


Hence Proved
Question 7.

Prove that:
Answer:

Take L.H.S


sin 20° sin 40° sin 80°


Multiplying & Dividing by 2:



{∵ 2 sin A sin B = cos (A – B) – cos (A + B)}





{∵ cos (-A) = cos A}




Multiplying & Dividing by 2:



{∵ 2 cos A sin B = sin (A + B) – sin (A – B)}




{∵ sin (-A) = - sin A}



{∵ sin (180° - A) = sin A}





= R.H.S


Hence Proved
Question 8.

Prove that:
Answer:

Take L.H.S


cos 20° cos 40° cos 80°


Multiplying & Dividing by 2:



{∵ 2 cos A cos B = cos (A + B) + cos (A – B)}








Multiplying & Dividing by 2:



{∵ 2 cos A cos B = cos (A + B) + cos (A – B)}





{∵ cos (180° - A) = - cos A}





= R.H.S


Hence Proved
Question 9.

Prove that:

tan 20° tan 40° tan 60° tan 80° = 3
Answer:

Take L.H.S


tan 20° tan 40° tan 60° tan 80°



Multiplying & Dividing by 2:



{∵ 2 sin A sin B = cos (A – B) – cos (A + B) &


2 cos A cos B = cos (A + B) + cos (A – B)}





{∵ cos (-A) = cos A}





{∵ 2 cos A sin B = sin (A + B) – sin (A – B) &


2 cos A cos B = cos (A + B) + cos (A – B)}




{∵ sin (-A) = - sin A}




{∵ sin (180° - A) = sin A & cos (180° - A) = - cos A}




= 3


= R.H.S


Hence Proved
Question 10.

Prove that:

tan 20° tan 30° tan 40° tan 80° = 1
Answer:

Take L.H.S


tan 20° tan 40° tan 30° tan 80°



Multiplying & Dividing by 2:



{∵ 2 sin A sin B = cos (A – B) – cos (A + B) &


2 cos A cos B = cos (A + B) + cos (A – B)}





{∵ cos (-A) = cos A}





{∵ 2 cos A sin B = sin (A + B) – sin (A – B) &


2 cos A cos B = cos (A + B) + cos (A – B)}




{∵ sin (-A) = - sin A}




{∵ sin (180° - A) = sin A & cos (180° - A) = - cos A}




= 1


= R.H.S


Hence Proved
Question 11.

Prove that:
Answer:

Take L.H.S



Multiplying & Dividing by 2:



{∵ 2 sin A sin B = cos (A – B) – cos (A + B)}





{∵ cos (-A) = cos A}




Multiplying & Dividing by 2:



{∵ 2 cos A sin B = sin (A + B) – sin (A – B)}




{∵ sin (-A) = - sin A}



{∵ sin (180° - A) = sin A}





= R.H.S


Hence Proved
Question 12.

Prove that:
Answer:

Take L.H.S



Multiplying & Dividing by 2:



{∵ 2 sin A sin B = cos (A – B) – cos (A + B)}





{∵ cos (-A) = cos A}




Multiplying & Dividing by 2:



{∵ 2 cos A sin B = sin (A + B) – sin (A – B)}




{∵ sin (-A) = - sin A}



{∵ sin (180° - A) = sin A}





= R.H.S


Hence Proved
Question 13.

Show that

i. sin A sin (B – C) + sin B sin (C – A) + sin C sin (A – B) = 0

ii. sin (B – C) cos (A – D) + sin (C – A) cos(B – D) + sin(A – B) cos(C – D) = 0
Answer:

i. Take L.H.S


sin A sin (B – C) + sin B sin (C – A) + sin C sin (A – B)


Multiplying & Dividing by 2:



{∵ 2 sin A sin B = cos (A – B) – cos (A + B)}





= 0


= R.H.S


Hence Proved


ii. sin (B – C) cos (A – D) + sin (C – A) cos(B – D) + sin(A – B) cos(C – D) = 0


Answer:


Take L.H.S


sin (B – C) cos(A – D) + sin (C – A) cos(B – D) + sin(A – B) cos(C – D)


Multiplying & Dividing by 2:



{∵ 2 sin A cos B = sin (A + B) + sin (A – B)}






= 0


= R.H.S


Hence Proved
Question 14.

Prove that:
Answer:

Take L.H.S.




= tan x tan (60° - x) tan (60° + x)



Multiplying & Dividing by 2:



{∵ 2 sin A sin B = cos (A – B) – cos (A + B) &


2 cos A cos B = cos (A + B) + cos (A – B)}




{cos (-A) = cos A}



{cos (180° - A) = - cos A}




Multiplying & Dividing by 2:



{∵ 2 cos A sin B = sin (A + B) – sin (A – B) &


2 cos A cos B = cos (A + B) + cos (A – B)}







= tan 3x


= R.H.S.


Hence Proved
Question 15.



Answer:

Take L.H.S:





cos α cos β


Multiplying & Dividing by 2:



{2 cos A cos B = cos (A + B) + cos (A – B)}







We know,
Exercise 8.2
Question 1.

Express each of the following as the product of sines and cosines:

i. sin 12x + sin 4x

ii. sin 5x – sin x

iii. cos 12x + cos 8x

iv. sin 2x + cos 4x
Answer:

i.


sin 12x + sin 4x




= 2 sin 8x cos 4x


ii.


sin 5x – sin x




= 2 cos 3x sin 2x


iii.


cos 12x + cos 8x




= 2 cos 10x cos 2x


iv. sin 2x + cos 4x


= sin 2x + sin (90° - 4x)




= 2 sin (45° - x) cos (3x – 45°)


= 2 sin (45° - x) cos {-(45° - 3x)}


{cos (-x) = cos x}


= 2 sin (45° - x) cos (45° - 3x)
Question 2.

Prove that :

sin 38° + sin 22°= sin 82°
Answer:

Take L.H.S:


sin 38° + sin 22°





= 2 sin 30° cos 8°



= cos 8°


= cos (90° - 82°)


{cos (90° - A) = sin A}


= sin 82°


= R.H.S


Hence Proved
Question 3.

Prove that :

cos 100° + cos 20° = cos 40°
Answer:

Take L.H.S:


cos 100° + cos 20°





= 2 cos 60° cos 40°



= cos 40°


= R.H.S


Hence Proved
Question 4.

Prove that :

sin 50° + sin 10° = cos 20°
Answer:

Take L.H.S:


sin 50° + sin 10°





= 2 sin 30° cos 20°



= cos 20°


= R.H.S


Hence Proved
Question 5.

Prove that :

sin 23° + sin 37° = cos 7°
Answer:

Take L.H.S:


sin 23° + sin 37°





= 2 sin 30° cos -7°


{cos (-A) = cos A}



= cos 7°


= R.H.S


Hence Proved
Question 6.

Prove that :

sin 105° + cos 105° = cos 45°
Answer:

Take L.H.S:


sin 105° + cos 105°


= sin 105° + sin (90° - 105°)


{sin (90° - A) = cos A}


= sin 105° + sin (- 15°)


{sin(-A) = - sin A}


= sin 105° - sin (15°)





= 2 cos 60° sin 45°




= cos 45°


= R.H.S


Hence Proved
Question 7.

Prove that :

sin 40° + sin 20° = cos 10°
Answer:

Take L.H.S:


sin 40° + sin 20°





= 2 sin 30° cos 10°



= cos 10°


= R.H.S


Hence Proved
Question 8.

Prove that :

cos 55° + cos 65° + cos 175° = 0
Answer:

Take L.H.S:


cos 55° + cos 65° + cos 175°





{cos (180° - A) = - cos A}


= 2 cos 60° cos (-5°) – cos 5°


{cos (-A) = cos A}



= cos 5° – cos 5°


= 0


= R.H.S


Hence Proved
Question 9.

Prove that :

sin 50° – sin 70° + sin 10° = 0
Answer:

Take LHS:


sin 50° – sin 70° + sin 10°





= 2 cos 60° sin (-10°) + sin 10°


{sin (-A) = -sin (A)}



= - sin 10° + sin 10°


= 0


= R.H.S


Hence Proved
Question 10.

Prove that :

cos 80° + cos 40° – cos 20° = 0
Answer:

Take L.H.S:


cos 80° + cos 40° – cos 20°





= 2 cos 60° cos 20° – cos 20°



= cos 20° – cos 20°


= 0


= R.H.S


Hence Proved
Question 11.

Prove that :

cos 20° + cos 100° + cos 140° = 0
Answer:

Take L.H.S:


cos 20° + cos 100° + cos 140°





{cos (180° - A) = - cos A}


= 2 cos 60° cos (-40°) – cos 40°


{cos (-A) = cos A}



= cos 40° – cos 40°


= 0


= R.H.S


Hence Proved
Question 12.



Answer:

Take L.H.S:



{cos A = sin (90° - A)}












= R.H.S.


Hence Proved
Question 13.

Prove that :
Answer:

Take L.H.S:



{cos A = sin (90° - A)}









= 2 cos 45° sin 30°




= R.H.S.


Hence Proved
Question 14.

Prove that :

sin 80° – cos 70° = cos 50°
Answer:

sin 80° – cos 70° = cos 50°


⇒ sin 80° = cos 50° + cos 70°


Take RHS:


cos 50° + cos 70°





= 2 cos 60° cos (-10°)


{cos (-A) = cos A}



= cos 10°


= cos (90° - 80°)


{cos (90° - A) = sin A}


= sin 80°


= L.H.S


Hence Proved
Question 15.

Prove that :

sin 51° + cos 81° = cos 21°
Answer:

Take L.H.S:


sin 51° + cos 81°


= sin 51° + sin (90° - 81°)


{sin (90° - A) = cos A}


= sin 51° + sin 9°





= 2 sin 30° cos 21°



= cos 21°


= R.H.S


Hence Proved
Question 16.

Prove that:

i.

ii.


Answer:

Take L.H.S:









{sin (π – A) = sin A}






= R.H.S.


Hence Proved


ii. Take L.H.S:











= R.H.S.


Hence Proved
Question 17.

Prove that:

i.

ii. sin 47° + cos 77° = cos 17°
Answer:

Take L.H.S:


sin 65° + cos 65°


= sin 65° + sin (90° - 65°)


{sin (90° - A) = cos A}


= sin 65° + sin 25°





= 2 sin 45° cos 20°




= R.H.S


Hence Proved


ii. Take L.H.S:


sin 47° + cos 77°


= sin 47° + sin (90° - 77°)


{sin (90° - A) = cos A}


= sin 47° + sin 13°





= 2 sin 30° cos 17°



= cos 17°


= R.H.S


Hence Proved
Question 18.

Prove that :

cos 3A + cos 5A + cos 7A + cos 15A= 4 cos 4A cos 5A cos 6A
Answer:

Take L.H.S.


cos 3A + cos 5A + cos 7A + cos 15A


= (cos 5A + cos 3A) + (cos 15A + cos 7A)






= 2 cos 4A cos A + 2 cos 11A cos 4A


= 2 cos 4A (cos 11A + cos A)






= 4 cos 4A cos 5A cos 6A


= R.H.S.


Hence Proved
Question 19.

Prove that :

cos A + cos 3A + cos 5A + cos 7A = 4 cos A cos 2A cos 4A
Answer:

Take L.H.S.


cos A + cos 3A + cos 5A + cos 7A


= (cos 3A + cos A) + (cos 7A + cos 5A)






= 2 cos 2A cos A + 2 cos 6A cos A


= 2 cos A (cos 6A + cos 2A)






= 4 cos A cos 2A cos 4A


= R.H.S.


Hence Proved
Question 20.

Prove that :
Answer:

Take L.H.S.


sin A + sin 2A + sin 4A + sin 5A


= (sin 2A + sin A) + (sin 5A + sin 4A)














= R.H.S.


Hence Proved
Question 21.

Prove that :
Answer:

Take L.H.S.


sin 3A + sin 2A – sin A


= (sin 3A – sin A) + sin 2A






{sin 2A = 2 sin A cos A}


= 2 cos 2A sin A + 2 sin A cos A


= 2 sin A (cos 2A + cos A)






= R.H.S.


Hence Proved
Question 22.

Prove that :
Answer:

Take L.H.S:


cos 20° cos 100° + cos 100° cos 140° - cos 140° cos 200°


Multiplying & Dividing by 2:



{∵ 2 cos A cos B = cos (A + B) + cos (A – B)}





{cos (180° + A) = - cos A ;


cos (90° + A) = - sin A &


cos (360° - A) = cos A}












= R.H.S.


Hence Proved
Question 23.

Prove that :
Answer:

Take L.H.S.:



Multiplying & Dividing by 2:



{∵ 2 sin A sin B = cos (A – B) – cos (A + B)}















= sin 5x sin 2x


= R.H.S.


Hence Proved
Question 24.

Prove that :
Answer:

Take L.H.S.:



Multiplying & Dividing by 2:



{∵ 2 cos A cos B = cos (A + B) + cos (A – B)}

















= R.H.S.


Hence Proved
Question 25.

Prove that:
Answer:

Take L.H.S.:







{sin (-A) = - sin A}




= cot A


= R.H.S.


Hence Proved
Question 26.

Prove that:
Answer:

Take L.H.S.:







{sin (-A) = - sin A}




= cot 8A


= R.H.S.


Hence Proved
Question 27.

Prove that:
Answer:

Take L.H.S.:








= R.H.S.


Hence Proved
Question 28.

Prove that:
Answer:

Take L.H.S.:








= R.H.S.


Hence Proved
Question 29.

Prove that:
Answer:

Take L.H.S.:







{sin (-x) = - sin x}




= R.H.S.


Hence Proved
Question 30.

Prove that:
Answer:

Take L.H.S.:









= tan 3A


= R.H.S.


Hence Proved
Question 31.

Prove that:
Answer:

Take L.H.S.:










= R.H.S.


Hence Proved
Question 32.

Prove that:
Answer:

Take L.H.S.:









= cot 3A


= R.H.S.


Hence Proved
Question 33.

Prove that:
Answer:

Take L.H.S.:










= tan 6A


= R.H.S.


Hence Proved
Question 34.

Prove that:
Answer:

Take L.H.S.:










= cot 6A


= R.H.S.


Hence Proved
Question 35.

Prove that:
Answer:

Take L.H.S.:



Multiplying & Dividing by 2:



{∵ 2 sin A sin B = cos (A – B) – cos (A + B);


2 sin A cos B = sin (A + B) + sin (A – B) &


2 cos A cos B = cos (A + B) + cos (A – B)}












= tan A


= R.H.S.


Hence Proved
Question 36.

Prove that:
Answer:

Take L.H.S.:



Multiplying & Dividing by 2:



{∵ 2 sin A sin B = cos (A – B) – cos (A + B) &


2 cos A sin B = sin (A + B) – sin (A – B)}










= tan 8A


= R.H.S.


Hence Proved
Question 37.

Prove that:
Answer:

Take L.H.S.:



Multiplying & Dividing by 2:



{∵ 2 sin A sin B = cos (A – B) – cos (A + B);


2 sin A cos B = sin (A + B) + sin (A – B) &


2 cos A cos B = cos (A + B) + cos (A – B)}




{sin (-A) = - sin A}









= tan 2A


= R.H.S.


Hence Proved
Question 38.

Prove that:
Answer:

Take L.H.S.:



Multiplying & Dividing by 2:



{∵ 2 sin A sin B = cos (A – B) – cos (A + B) &


2 sin A cos B = sin (A + B) + sin (A – B)}




{sin (-A) = - sin A}










= tan 5A


= R.H.S.


Hence Proved
Question 39.

Prove that:
Answer:

Take L.H.S.:










= R.H.S.


Hence Proved
Question 40.

Prove that:
Answer:

Take R.H.S:










= tan θ


= R.H.S


Hence Proved
Question 41.

Prove that:

i.

ii. cos (A + B + C) + cos (A – B + C) + cos (A + B – C) + cos (-A + B + C) = 4 cos A cos B cos C
Answer:

Take L.H.S:







{sin (-A) = - sin A}











{sin (-A) = - sin A}




= R.H.S.


Hence Proved


ii. Take L.H.S.:


cos(A + B + C) + cos (A – B + C) + cos(A + B – C) + cos(-A + B + C)


= {cos (A + B + C) + cos (A – B + C)} + {cos (A + B – C) + cos (-A + B + C)}







= 2 cos (A + C) cos B + 2 cos B cos (A – C)


= 2 cos B {cos (A + C) + cos (A – C)}






= 4 cos A cos B cos C


= R.H.S.


Hence Proved
Question 42.

prove that


Answer:

Given,
Question 43.



Answer:

Given,


cosec A + sec A = cosec B + sec B


⇒ sec A – sec B = cosec B – cosec A









{sin (-A) = - sin A}
Question 44.

If sin 2A = λ sin 2B, prove that:
Answer:

Given,


sin 2A = λ sin 2B



Now,
Question 45.

Prove that:

i.

ii. sin(B– C)cos(A – D) + sin(C – A)cos(B – D) + sin(A – B)cos(C – D) = 0
Answer:

Take L.H.S.










{sin (-A) = -sin A & cos (-A) = cos A}









= cot C


= R.H.S.


Hence Proved


ii. Take L.H.S.:


sin(B– C)cos(A – D) + sin(C – A)cos(B – D) + sin(A – B)cos(C – D)


Multiplying & Dividing by 2:



{∵ 2 sin A cos B = sin (A + B) + sin (A – B)}






{sin (-A) = -sin A}




= 0


= R.H.S.


Hence Proved
Question 46.

prove that tan A tan B tan C tan D= -1
Answer:

Given,




Adding 1 both sides:






Now,



Subtracting 1 both sides:







Dividing equation (i) by equation (ii):











⇒ -1 = tan A tan B tan C tan D


This proves that tan A tan B tan C tan D = -1
Question 47.

If cos (α + β) sin (γ + δ) = cos (α - β) sin (γ - δ), prove that cot α cot β cot γ = cot δ
Answer:

Given,


cos (α + β) sin (γ + δ) = cos (α - β) sin (γ – δ)



Adding 1 both sides:





Now,



Subtracting 1 both sides:





Dividing equation (i) by equation (ii):










{sin (-A) = -sin A & cos (-A) = cos A}




⇒ cot α cot β cot γ = cot δ


Hence Proved
Question 48.

If y sin ϕ = x sin (2θ + ϕ), prove that (x + y) cot (θ + ϕ) = (y -x) cot θ
Answer:

Given, y sin ϕ = x sin (2θ + ϕ)



Adding 1 both sides:




Now,



Adding 1 both sides:




Dividing equation (i) by equation (ii):










{sin (-A) = -sin A & cos (-A) = cos A}



⇒ (x + y) cot (θ + ϕ) = (y -x) cot θ


Hence Proved
Question 49.

If cos (A + B) sin (C – D) = cos (A – B) sin (C + D), prove that tan A tan B tan C + tan D = 0
Answer:

Given,


cos (A + B) sin (C – D) = cos (A – B) sin (C + D)



Adding 1 both sides:





Now,



Subtracting 1 both sides:





Dividing equation (i) by equation (ii):











⇒ -tan D = tan A tan B tan C


⇒ tan D = -tan A tan B tan C


⇒ tan A tan B tan C + tan D = 0


Hence Proved
Question 50.



Answer:

Given,



Now,




Now,



Now,




Now,


xy + yz + xz






{∵ cos (A + B) = cos A cos B – sin A sin B}







{sin (180° - A) = sin A ; sin (180° + A) = -sin A;


cos (180° - A) = -cos A & cos (180° + A) = -cos A }







= 0


= R.H.S.


Hence Proved
Question 51.



Answer:

Given, m sin θ = n sin (θ + 2α)




Using Componendo - Dividendo:









Hence Proved
Very Short Answer
Question 1.

If (cos α + cos β)2 + (sin α + sin β)2 write the value of λ.
Answer:

(cos α + cos β)2 + (sin α + sin β)2


= (cos2 α + 2cos α cos β + cos2 β) + (sin2 α + 2sin α sin β + sin2 β)


Rearranging the terms,


= (cos2 α + sin2 α) + (cos2 β + sin2 β) + 2(cos α cos β + sin α sin β)


= 1 + 1 + 2cos (α – β)


= 2{1 + cos (α – β)}


=


= 4cos2


Comparing with question, λ = 4 (Ans)
Question 2.

Write the value of


Answer:





(Ans)
Question 3.

If sin A + sin B = α and cos A + cos B = β, then write the value of


Answer:

sin A + sin B = α


…(I)


cos A + cos B = β


…(II)


Dividing equation (I) by equation (II), we get –



or, (Ans)
Question 4.

If cos A = m cos B, then write the value of


Answer:






=





Ans:



Question 5.

Write the value of the expression


Answer:















= 1 (Ans)
Question 6.

If and cos A + cos B = 1, then find the value of


Answer:

cos A + cos B = 1







(Ans)
Question 7.

Write the value of


Answer:












…(I)


Now,


















Since, sin is positive,



Now,


And,







or,


Putting these values in (I),






(Ans)
Question 8.

If sin 2A = λ sin 2B, then write the value of


Answer:







(Ans)
Question 9.

Write the value of


Answer:




= tan 2A (Ans)
Question 10.

If cos (A + B) sin (C – D) = cos (A – B) sin (C + D), then write the value tan A tan B tan C.
Answer:

cos (A + B) sin (C – D) = cos (A – B) sin (C + D)


(To make the math easier, while expanding this out I’ll use c A instead of cos A and s A instead of sin A)


⇒ (c A c B – s A s B)(s C c D – c C s D) = (c A c B + s A s B)(s C c D + c C s D)


⇒ c A c B s C c D – c A c B c C s D – s A s B s C c D + s A s B c C s D = c A c B s C c D +


c A c B c C s D + s A s B s C c D + s A s B c C s D


⇒ 2s A s B s C c D = -2c A c B c C s D


⇒ sin A sin B sin C cos D = -cos A cos B cos C sin D



⇒ tan A tan B tan C = - tan D (Ans)
Mcq
Question 1.

Mark the Correct alternative in the following:

cos 40o + cos 80o + cos 160o + cos 240o =

A. 0

B. 1

C. 1/2

D. -1/2
Answer:

cos 40° + cos 80° + cos 160° + cos 240°


= (cos 40° + cos 80°) + (cos 160° + cos 240°)


= 2 + 2


= 2cos 60° cos 20° + 2cos 200° cos 40°


= 2⋅ 1/2 ⋅cos 20° + 2cos (180 + 20)° cos 40°


= cos 20° - 2cos 20° cos 40°


= cos 20° - (cos (40° + 20°) + cos (40° - 20°))


= cos 20° - (cos 60° + cos 20°)


= - cos 60°


= - 1/2 (Ans)
Question 2.

Mark the Correct alternative in the following:

sin 163o cos 347o + sin 73o sin 167o =

A. 0

B. 1/2

C. 1

D. None of these
Answer:

sin 163° cos 347° + sin 73° sin 167°


= sin (180° - 17° ) cos (360° - 13°) + sin (90° - 17°) sin (180° - 13°)


= sin 17° cos 13° + cos 17° sin 13°


= sin (17° + 13°)


= sin 30°


= 1/2 (Ans)
Question 3.

Mark the Correct alternative in the following:

If and then

A. 3/8

B. 5/8

C. 3/4

D. 5/4
Answer:

sin 2θ + sin 2ϕ = 1/2


⇒ 2sin (θ + ϕ) cos (θ – ϕ) = 1/2 …(I)


cos 2θ + cos 2ϕ = 3/2


⇒ 2cos (θ + ϕ) cos (θ – ϕ) = 3/2 …(II)


Dividing equation (I) by equation (II) –


tan (θ + ϕ) = 1/3


⇒ tan2 (θ +ϕ) = 1/9


⇒ 1 + tan2 (θ + ϕ) = 1 + 1/9 = 10/9





Substituting this value of cos (θ + ϕ) in (II), we get –





(Ans)
Question 4.

Mark the Correct alternative in the following:

The value of cos 52o + cos 68o + cos 172o is

A. 0

B. 1

C. 2

D. 3/2
Answer:

cos 52° + cos 68° + cos 172°


= (cos 52° + cos 68°) + cos 172°


= 2cos 60° cos 8° + cos (180° - 8°)


= 2 ⋅ 1/2 ⋅ cos 8° - cos 8°


= cos 8° - cos 8°


= 0 (Ans)
Question 5.

Mark the Correct alternative in the following:

The value of sin 78o – sin 66o – sin 42o + sin 6o is

A. 1/2

B. -1/2

C. -1

D. None of these
Answer:

sin 78° - sin 66° - sin 42° + sin 6°


= (sin 78° - sin 42°) – (sin 66° - sin 6°)


= 2cos 60° sin 18° - 2cos 36° sin 30°


= 2 ⋅ 1/2 ⋅ sin 18° - 2 ⋅ 1/2 ⋅ cos 36°


= sin 18° - cos 36°


=


= - 1/2 (Ans)
Question 6.

Mark the Correct alternative in the following:

If sin α + sin β = a and cos α – cos β = b, then

A.

B.

C.

D. None of these
Answer:

sin α + sin β = a


⇒ = a …(I)


cos α – cos β = b


⇒ = b …(II)


Dividing equation (II) by equation (I) –



or,


(Ans)
Question 7.

Mark the Correct alternative in the following:

cos 35o + cos 85o + cos 155o =

A. 0

B.

C.

D. cos 275o
Answer:

cos 35° + cos 85° + cos 155°


= 2 + cos 155°


= 2cos 60° cos 25° + cos (180°-25°)


= 2 ⋅ 1/2 ⋅ cos 25° - cos 25°


= cos 25° - cos 25°


= 0 (Ans)
Question 8.

Mark the Correct alternative in the following:

The value of sin 50o – sin 70o + sin 10o is equal to

A. 1

B. 0

C. 1/2

D. 2
Answer:

sin 50° - sin 70° + sin 10°


= (sin 50° + sin 10°) – sin 70°


= 2sin 30° cos 20° - sin (90° - 20°)


= 2 ⋅ 1/2 ⋅ cos 20° - cos 20°


= cos 20° - cos 20°


= 0
Question 9.

Mark the Correct alternative in the following:

sin 47o + sin 61o – sin 11o – sin 25o is equal to

A. sin 36o

B. cos 36o

C. sin 7o

D. cos 7o
Answer:

sin 47° + sin 61° - sin 11° - sin 25°


= (sin 47° - sin 25°) + (sin 61° - sin 11°)


= 2cos 36° sin 11° + 2cos 36° sin 25°


= 2cos 36° (sin 11° + sin 25°)


= 2cos 36° (2sin 18° cos 7°)


= 4cos 36° sin 18° cos 7°




= cos 7°
Question 10.

Mark the Correct alternative in the following:

If cos A = m cos B, then =

A.

B.

C.

D. None of these
Answer:











Question 11.

Mark the Correct alternative in the following:

If A, B, C are in A.P., then

A. tan B

B. cot B

C. tan 2 B

D. None of these
Answer:

Let common difference be d.


Then A = B – d and C = B + d


So,





= cot B (Ans)
Question 12.

Mark the Correct alternative in the following:

If sin (B + C – A), sin (C + A – B), sin (A + B – C) are in A.P., then cot A, cot B, cot C are in

A. GP

B. HP

C. AP

D. None of these
Answer:

sin (B + C – A), sin (C + A – B), sin (A + B – C) are in A.P.


⇒ 2sin (C + A – B) = sin (B + C – A) + sin (A + B – C)


⇒ 2sin (C + A – B) = 2sin {(B + C – A + A + B – C)/2} cos {(B + C – A – A – B + C)/2}


⇒ sin {(C + A) – B} = sin B cos (C – A)


⇒ sin (C + A) cos B – cos (C + A) sin B = sin B cos (C – A)


⇒ {sin C cos A + cos C sin A} cos B – {cos C cos A – sin C sin A} sin B = sin B {cos C cos A + sin C sin A}


⇒ cos A cos B sin C + sin A cos B cos C – cos A sin B cos C + sin A sin B sin C = cos A sin B cos C + sin A sin B sin C


⇒ cos A cos B sin C + sin A cos B cos C = 2cos A sin B cos C


⇒ cos B (cos A sin C + sin A cos C) = 2cos A sin B cos C





∴ cot A, cot B, cot C are in H.P. (Ans)
Question 13.

Mark the Correct alternative in the following:

If then sin 3x + sin 3y =

A. 2 sin 3x

B. 0

C. 1

D. None of these
Answer:

sin x + sin y = √3 (cos y – cos x)






⇒ x + y = 0 or


⇒ x = -y


or


⇒ x = -y


or


Putting these values of x in sin 3x + sin 3y,


For x = -y


sin 3x + sin 3y


= sin 3(-y) + sin 3y


= -sin 3y + sin 3y


= 0


For


sin 3x + sin 3y



= sin (3y + π) + sin 3y


= -sin 3y + sin 3y


= 0


So, sin 3x + sin 3y = 0 (Ans)
Question 14.

Mark the Correct alternative in the following:

If and then
α + β is equal to

A. π/2

B. π/3

C. π/6

D. π/4
Answer:

tan α = and tan β =


tan (α + β) =





⇒ tan (α + β) = 1



(Ans)