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Trigonometric Ratios Of Complementary Angles

Class 10th Mathematics RS Aggarwal Solution
Exercise 7
  1. Without using trigonometric tables, evaluate: sin16^circle /cos74^circle…
  2. sec11^circle /cosec79^circle Solve:
  3. tan27^circle /cot63^circle Solve:
  4. cos35^circle /sin55^circle Solve:
  5. cosec42^circle /sec48^circle Solve:
  6. cot38^circle /tan52^circle Solve:
  7. cos 81° - sin 9° = 0 Without using trigonometric tables, prove that:…
  8. tan 71 - cot 19 = 0 Without using trigonometric tables, prove that:…
  9. cosec 80 - sec 10 = 0 Without using trigonometric tables, prove that:…
  10. cosec^2 72 - tan^2 18 = 1 Without using trigonometric tables, prove that:…
  11. cos^2 75 + cos^2 15 =1 Without using trigonometric tables, prove that:…
  12. tan^2 66 - cot^2 24 = 0 Without using trigonometric tables, prove that:…
  13. sin^2 48o + sin^2 42 = 1 Without using trigonometric tables, prove that:…
  14. cos^2 57 - sin^2 33 = 0 Without using trigonometric tables, prove that:…
  15. (sin 65 + cos 25)(sin 65 - cos 25) = 0 Without using trigonometric tables,…
  16. sin 53° cos 37° + cos 53° sin 37° = 1 Without using trigonometric tables, prove…
  17. cos 54° cos 36° - sin 54° sin 36° = 0
  18. sec 70° sin 20° + cos 20° cosec 70° = 2
  19. sin 35° sin 55° - cos 35° cos 55° = 0
  20. (sin 72° + cos 18°)(sin 72° - cos 18°) = 0
  21. tan 48° tan 23° tan 42° tan 67° = 1
  22. sin70^circle /cos20^circle + cosec20^circle /sec70^circle - 2cos70^circle…
  23. Prove that : cos80^circle /sin10^circle + cos59^circle cos31^circle = 2…
  24. 2sin68^circle /cos22^circle - 2cot15^circle /5tan75^circle - 3tan45^circle…
  25. sin18^circle /cos72^circle + root 3 (tan10^circle tan30^circle tan40^circle…
  26. 7cos55^circle /3sin35^circle - 4 (cos70^circle cosec20^circle)/3 (tan5^circle…
  27. sin θ cos (90° - θ) + sin (90° - θ) cos θ = 1 Prove that:
  28. Prove that: sintegrate heta /cos (90 - theta) + costheta /sin (90 - theta) = 2…
  29. sintegrate heta cos (90^circle - theta) costheta /sin (90^circle - theta) +…
  30. cos (90^circle - theta) sec (90^circle - theta) tantheta /cosec (90^circle -…
  31. cos (90^circle - theta)/1+sin (90^circle - theta) + 1+sin (90^circle -…
  32. sec (90 - theta) cosectheta -tan (90 - theta) cos^225^circle + cos^265^circle…
  33. cot θ tan (90° -θ) - sec (90° - θ)cosec θ +√3tan 12°tan 60°tan 78° = 2…
  34. tan 5° tan 25° tan 30° tan 65° tan 85° = 1/√3 Prove that:
  35. cot 12° cot 38° cot 52° cot 60° cot 78° = 1/√3
  36. cos 15° cos 35° cosec 55° cos 60° cosec 75° =1/2
  37. cos 1° cos 2° cos 3° ... cos 180° = 0
  38. Prove: (sin49^circle /cos41^circle)^2 + (cos41^circle /sin49^circle)^2 = 2…
  39. sin (70°+θ) — cos (20° — θ) = 0 Prove that:
  40. tan (55° — θ) — cot (35° + θ) = θ
  41. cosec (67° + θ) — sec (23° — θ) = 0
  42. cosec (65° + θ) — sec (25° —θ) — tan (55° — θ) + cot (35° + θ) =0…
  43. sin (50° +θ) — cos (40° — θ) + tan 1°tan 10°tan 80°tan 89° =1
  44. Express each of the following in terms of T-ratios of angles lying between 0°…
  45. cot 65 + tan 49 Express each of the following in terms of T-ratios of angles…
  46. sec 78 + cosec 56 Express each of the following in terms of T-ratios of angles…
  47. cosec 54 + sin 72 Express each of the following in terms of T-ratios of angles…
  48. If A, B and C are the angles of a delta ABC, prove that tan (c+a/2) = cot (b/2)…
  49. If cos 2 = sin 4 , where 2 and 4 are acute angles, find the value of .…
  50. If sec 2A = cosec (A — 42°), where 2A is an acute angle, find the value of A.…
  51. If sin 3A = cos (A — 26°), where 3A is an acute angle, find the value of A.…
  52. If tan 2A = cot (A — 12°), where 2A is an acute angle, find the value of A.…
  53. If sec 4A = cosec (A — 15°), where 4A is an acute angle, find the value of A.…
  54. 2/3 cosec^2 58°-2/3 cot 58° tan 32° -5/3 tan 13° tan 37° tan 45° tan 53° tan…
Exercise 7
Question 1.

Without using trigonometric tables, evaluate:
Answer:

= = = 1

(∵ cos (90-θ) = sin θ and (90-θ) lies in the first quadrant where


all the angles are taken as positive.)
Question 2.

Solve:
Answer:

= = = = = 1
Question 3.

Solve:
Answer:

= = = 1 (∵ cot (90-θ) = tan θ)
Question 4.

Solve:
Answer:

= = = 1
Question 5.

Solve:
Answer:

= = = = = 1
Question 6.

Solve:
Answer:

= = = 1 (∵ tan (90-θ) = cot θ)
Question 7.

Without using trigonometric tables, prove that:

cos 81° - sin 9° = 0
Answer:

Consider cos 81° - sin 9° = cos 81° - sin (90 - 81)°

= cos 81° - cos 81°


= 0


Hence, proved.
Question 8.

Without using trigonometric tables, prove that:
tan 71° - cot 19° = 0
Answer:

Consider tan 71° - cot 19° = tan 71° - cot (90 - 71)°

= tan 71° - tan 71°


= 0


Hence, proved.
Question 9.

Without using trigonometric tables, prove that:
cosec 80° - sec 10° = 0
Answer:

Consider cosec 80° - sec 10° = cosec 80° - sec (90 - 10)°

= cosec 80° - cosec 80°


= 0


Hence, proved.
Question 10.

Without using trigonometric tables, prove that:
cosec2 72° - tan2 18° = 1
Answer:

Consider cosec2 72° - tan2 18° = cosec2 72° - tan2 (90 - 72)°

= cosec2 72° - cot2 72°


= 1


(∵1 + cot2 θ = cosec2 θ ⇒ cosec2 θ - cot2 θ = 1 )


Hence, proved.
Question 11.

Without using trigonometric tables, prove that:
cos275° + cos215° =1
Answer:

Consider cos2 75° + cos2 15° = cos2 75° + cos2 (90 - 75)°

= cos2 75° + sin2 75°


= 1


(∵cos2θ + sin2 θ = 1)


Hence, proved.
Question 12.

Without using trigonometric tables, prove that:
tan266° - cot224° = 0
Answer:

Consider tan2 66° - cot2 24° = tan2 66° - cot2 (90 - 66)°

= tan2 66° - tan2 66°


= 0


Hence, proved.
Question 13.

Without using trigonometric tables, prove that:
sin248o + sin242° = 1
Answer:

Consider sin2 48° + sin2 42° = sin2 48° + sin2 (90 - 48)°

= sin2 48° + cos2 48°


= 1


(∵cos2θ + sin2 θ = 1 )


Hence, proved.
Question 14.

Without using trigonometric tables, prove that:
cos257° - sin233° = 0
Answer:

Consider cos2 57° - sin2 33° = cos2 57° - sin2 (90 - 57)°

= cos2 57° - cos2 57°


= 0


Hence, proved.
Question 15.

Without using trigonometric tables, prove that:
(sin 65° + cos 25°)(sin 65° - cos 25°) = 0
Answer:

Consider (sin 65° + cos 25°)(sin 65° - cos 25°)

= sin2 65° - cos2 25°


= sin2 65° - cos2 (90 - 65)°


= sin2 65° - sin2 65°


= 0


Hence, proved.
Question 16.

Without using trigonometric tables, prove that:

sin 53° cos 37° + cos 53° sin 37° = 1
Answer:

Consider (sin 53° cos 37°) + (cos 53° sin 37°)

= (sin 53° cos (90-53)°) + (cos 53° sin (90-53)°)


= sin2 53° + cos2 53°


= 1


Hence, proved.
Question 17.

cos 54° cos 36° - sin 54° sin 36° = 0
Answer:

Consider (cos 54° cos 36°) - (sin 54° sin 36°)

= (cos 54° cos (90-54)°) - (sin 54° sin (90-54)°)


= (cos 54° sin54°) - (sin 54° cos 54°)


= 0


Hence, proved.
Question 18.

sec 70° sin 20° + cos 20° cosec 70° = 2
Answer:

Consider L.H.S.

= (sec 70° sin 20°) + (cos 20° cosec 70°)


= (sec (90-20)° sin 20°) + (cos 20° cosec (90-20)°)


= (cosec 70° sin70°) + (cos 20° sec 20°)


= 1 + 1 (∵ cosec θ = 1/sinθ and sec θ = 1/cos θ )


= 2 = R.H.S.


Hence, proved.
Question 19.

sin 35° sin 55° - cos 35° cos 55° = 0
Answer:

Consider L.H.S. = (sin 35° sin 55°) - (cos 35° cos 55°)

= (sin 35° sin (90-35)°) - (cos 35° cos (90-35)°)


= (sin 35° cos 35°) - (cos 35° sin 35°)


= 0 = R.H.S.


Hence, proved.
Question 20.

(sin 72° + cos 18°)(sin 72° - cos 18°) = 0
Answer:

Consider (sin 72° + cos 18°)(sin 72° - cos 18°)

= sin2 72° - cos2 18°


= sin2 72° - cos2 (90-72)°


= sin2 72° - sin2 72°

[since, cos(90 - θ) = sinθ]

= 0


Hence, proved.
Question 21.

tan 48° tan 23° tan 42° tan 67° = 1
Answer:

Consider L.H.S.

= tan 48° tan 23° tan 42° tan 67°


= tan 48° tan 42° tan 23° tan 67°


= tan 48° tan (90-48)° tan 23° tan (90-23)°


= (tan 48° cot 48°) (tan 23° cot 23°)


= (1) × (1) = 1 = R.H.S


Hence, proved.
Question 22.

Prove that:
Answer:

Consider L.H.S

= + - 2 cos 70° cosec 20°


= + - 2 cos 70° cosec (90-70)°


= + - 2 cos 70° sec 70°


= 1 + 1 + - 2 = 0 = R.H.S.


Hence, proved.
Question 23.

Prove that :



Answer:

Consider L.H.S.

= + cos 59° cosec 31°


= + cos 59° cosec (90-59)°


= + cos 59° sec 59°


= 1 + 1 = 2 = R.H.S.


Hence, proved.
Question 24.

Prove that:
Answer:

Consider L.H.S.

=


=


=


= 2 –(2/5) – 3(1 × 1 × 1)/5


= 2 – (2/5) – (3/5)


= (10 – 2 - 3)/5


= 5/5


= 1 = R.H.S.


Hence, proved.
Question 25.

Prove that:
Answer:

To prove:
Proof:
Consider L.H.S.

=
Now look for the pairs whose angles give sum of 90°.
Here the sum of angles of tan 10° and tan 80° gives 90°.
Also the sum of angles of tan 40° and tan 50° gives 90°.
So, now change tan 80° into tan(90-10°) and tan 50° into tan(90-40°)


=
We know tan(90-θ) = cot θ
tan 30° = 1/√3


=


= 1 +
Since tan θ = 1/cot θ


= 1 +


= 1 + 1 = 2 = R.H.S.


Hence, proved.
Note: In such questions take the pairs whose angles give sum of 90° ,change one of them in the form of 90-θ and substitute remaining known values.
Like in this case tan 50° is changed into tan(90-40)° so that it will can be written as cot 50° and
tan 50° × cot 50° gives us value 1.
Question 26.

Prove that:
Answer:

Consider L.H.S.

=


=


=


=


= (7/3) – (4/3) = 43/3 = 1 = R.H.S.


Hence, proved.
Question 27.

Prove that:

sin θ cos (90° - θ) + sin (90° - θ ) cos θ = 1
Answer:

Consider L.H.S.

= sin θ cos (90° - θ) + sin (90° - θ) cos θ

= sin θ sin θ + cos θ cos θ

= sin2 θ + cos2 θ

= 1 = R.H.S.

Hence, proved.
Question 28.

Prove that:



Answer:

Consider L.H.S.

=


=


= 1 +1 = 2 = R.H.S.


Hence, proved.
Question 29.

Prove that:
Answer:

Consider L.H.S.

=


=


=


=


=


=


Hence, proved.
Question 30.

Prove that:
Answer:

Consider L.H.S.

=


=


= = 1+1 = 2 = R.H.S.


Hence, proved.
Question 31.

Prove that:
Answer:

Consider L.H.S.

=


=


= =


=


= 2 cosec θ = R.H.S.


Hence, proved.
Question 32.

Prove that:
Answer:

Consider L.H.S.

=


=


=


= = 2/3 = R.H.S.


Hence, proved.
Question 33.

cot θ tan (90° -θ) - sec (90° - θ)cosec θ +√3tan 12°tan 60°tan 78° = 2
Answer:

Consider L.H.S.

= cot θ tan (90° -θ) - sec (90° - θ)cosec θ +√3tan 12°tan 60°tan 78°


= cot θ cot θ - cosec θ cosec θ +√3 tan 60° tan 12° tan 78°


= cot2 θ – cosec2 θ +√3 tan 60° tan 12° tan(90-12)°


= - (cosec2 θ - cot2 θ) +√3 tan 60° tan 12° cot 12°


= - 1 + √3(√3 × tan 12° cot 12°)


= - 1 + √3(√3 × 1)


= - 1 + 3


= 2 = R.H.S.


Hence, proved.
Question 34.

Prove that:

tan 5° tan 25° tan 30° tan 65° tan 85° = 1/√3
Answer:

Consider L.H.S.

= tan 5° tan 25° tan 30° tan 65° tan 85°


= tan 5° tan 85° tan 25° tan 65° tan 30°


= tan 5° tan (90-5)° tan 25° tan (90-25)° tan 30°


= (tan 5° cot 5°) (tan 25° cot 25°) tan 30°


= 1 × 1 × (1/√3)


= 1/√3 = R.H.S.


Hence, proved.
Question 35.

cot 12° cot 38° cot 52° cot 60° cot 78° = 1/√3
Answer:

Consider L.H.S.

= cot 12° cot 38° cot 52° cot 60° cot 78°


= (cot 12° cot 78°) (cot 38° cot 52°) cot 60°


= (cot 12° cot (90-12)°) (cot 38° cot (90-38)°) cot 60°


= (cot 12° tan 12°) (cot 38° tan 38°) cot 60°


= 1 × 1 × (1/√3)


= 1/√3 = R.H.S.


Hence, proved.
Question 36.

cos 15° cos 35° cosec 55° cos 60° cosec 75° =1/2
Answer:

Consider L.H.S. = cos 15° cos 35° cosec 55° cos 60° cosec 75°

= cos 15° cosec 75° cos 35° cosec 55° cos 60°


= cos 15° cosec (90-15)° cos 35° cosec (90-35)° cos 60°


= (cos 15° sec 15°) × (cos 35° sec 35°) × cos 60°


= (1) × (1) × (1/2)


= 1/2 = R.H.S.


Hence, proved.
Question 37.

cos 1° cos 2° cos 3° ... cos 180° = 0
Answer:

Consider L.H.S. = cos 1° cos 2° cos 3° ... cos 180°

= cos 1° cos 2° cos 3° …× cos 90° × … cos 180°


= cos 1° cos 2° cos 3° ... × 0 × cos 180°


= 0 (∵ cos 90° = 0)


Hence, proved.
Question 38.

Prove:
Answer:

Consider L.H.S.

=


=


=


= 12 + 12


1 + 1 = 2 = R.H.S.


Hence, proved.
Question 39.

Prove that:

sin (70°+θ) — cos (20° — θ) = 0
Answer:

Consider L.H.S.

= sin (70°+θ) — cos (20° — θ)


= sin (70°+θ) — cos [90°-(70° + θ)]


= sin (70°+θ) — sin (70° + θ)


= 0 = R.H.S.


Hence, proved.
Question 40.

tan (55° — θ) — cot (35° + θ) = θ
Answer:

Consider L.H.S.

= tan (55° — θ) — cot (35° + θ)


= tan (90°-(35° +θ)) — cot (35° + θ)


= cot (35° + θ) - cot (35° + θ)


= 0


Hence, proved.
Question 41.

cosec (67° + θ) — sec (23° — θ) = 0
Answer:

Consider L.H.S.

= cosec (67° + θ) — sec (23° — θ)


= cosec (67° + θ) — sec (90° —(23° + θ))


= cosec (67° + θ) - cosec (67° + θ)


= 0


Hence, proved.
Question 42.

cosec (65° + θ) — sec (25° —θ) — tan (55° — θ) + cot (35° + θ) =0
Answer:

Consider L.H.S.

= cosec (65° + θ) — sec (25° —θ) — tan (55° — θ) + cot (35° + θ)


= cosec (65° + θ) — sec (90° - (65° + θ)) — tan (90° - (35° + θ)) + cot (35° + θ)


= cosec (65° + θ) — cosec (65° + θ)) — cot (35° + θ) + cot (35° + θ)


= 0 = R.H.S.


Hence, proved.
Question 43.

sin (50° +θ) — cos (40° — θ) + tan 1°tan 10°tan 80°tan 89° =1
Answer:

Consider L.H.S.

= sin (50° +θ) — cos (40° — θ) + tan 1° tan 10° tan 80° tan 89°


= sin ((90°-(40° - θ)) — cos (40° — θ) + (tan 1° tan 89°)(tan10° tan 80°)


= cos (40° - θ) - cos (40° - θ) + [tan 1° tan (90°-1°)][tan10° tan(90°-10°)]


= 0 + [(tan 1° cot 1°] [tan10° cot 10°]


= 0 + [1] × [1]


= 0 + 1 = 1 = R.H.S.


Hence, proved.
Question 44.

Express each of the following in terms of T-ratios of angles lying between 0° and 45°:

sin 67° + cos 75°
Answer:

Consider sin 67° + cos 75° = sin (90-23)° + cos (90-15)°

= cos 23° + sin 15°
Question 45.

Express each of the following in terms of T-ratios of angles lying between 0° and 45°:
cot 65° + tan 49°
Answer:

Consider cot 65° + tan 49° = cot (90-25)° + tan (90-41)°

= tan 25° + cot 41°
Question 46.

Express each of the following in terms of T-ratios of angles lying between 0° and 45°:
sec 78° + cosec 56°
Answer:

Consider sec 78° + cosec 56° = sec (90-12)° + cosec (90-34)°

= cosec 12° + sec 34°
Question 47.

Express each of the following in terms of T-ratios of angles lying between 0° and 45°:
cosec 54° + sin 72°
Answer:

Consider cosec 54° + sin 72° = cosec (90-36)° + sin (90-18)°

= sec 36° + cos 18°
Question 48.

If A, B and C are the angles of a ABC, prove that
Answer:

Since A, B and C are the angles of a triangle, therefore sum of the angles equals 180°.

∴ A + B + C = 180° ⇒ C + A = 180° - B


Now, consider L.H.S. =


=


=


= cot = R.H.S.


Hence, proved.
Question 49.

If cos 2 θ = sin 4 θ, where 2θ and 4θ are acute angles, find the value of θ .
Answer:

given: cos 2θ = sin 4θ

To find: the value of θ

Solution:

Consider cos 2θ = sin 4θ,

Since,

sin( 90° - θ) = cos θ

∴We can rewrite it as:

sin (90° - 2θ) = sin 4θ


On comparing both sides, we get,


90° - 2θ = 4θ

⇒ 90° = 4θ + 2θ

⇒ 6θ = 90°

⇒ θ = 15°
Question 50.

If sec 2A = cosec (A — 42°), where 2A is an acute angle, find the value of A.
Answer:

We are given that: sec 2A = cosec (A – 42°)

∴We can rewrite it as: cosec (90° - 2A) = cosec (A – 42°)


On comparing both sides, we get,


90° - 2A = A – 42 °


⇒ A + 2A = 90° + 42°


⇒ 3A = 132°


⇒ A = 44°
Question 51.

If sin 3A = cos (A — 26°), where 3A is an acute angle, find the value of A.
Answer:

We are given that: sin 3A = cos (A - 26°)

∴We can rewrite it as: cos (90° - 3A) = cos (A - 26°)


On comparing both sides, we get,


90° - 3A = A - 26°


⇒ A + 3A = 90° + 26°


⇒ 4A = 116°


⇒ A = 29°
Question 52.

If tan 2A = cot (A — 12°), where 2A is an acute angle, find the value of A.
Answer:

We are given that: tan 2A = cot (A – 12°)

∴We can rewrite it as: cot (90° - 2A) = cot (A – 12°)


On comparing both sides, we get,


90° - 2A = A – 12 °


⇒ A + 2A = 90° + 12°


⇒ 3A = 102°


⇒ A = 34°
Question 53.

If sec 4A = cosec (A — 15°), where 4A is an acute angle, find the value of A.
Answer:

We are given that: sec 4A = cosec (A – 15°)

∴We can rewrite it as: cosec (90° - 4A) = cosec (A – 15°)


On comparing both sides, we get,


90° - 4A = A – 15 °


⇒ A + 4A = 90° + 15°


⇒ 5A = 105°


⇒ A = 21°
Question 54.

Prove that:

2/3 cosec2 58°-2/3 cot 58° tan 32° -5/3 tan 13° tan 37° tan 45° tan 53° tan 77° = —1.
Answer:

Consider L.H.S.

= cosec2 58°- cot 58° tan 32° - tan 13° tan 37° tan 45° tan 53° tan 77°


= cosec2 58°- cot 58° tan (90-58)° - × [tan 13° tan 77°] × [tan 37° tan 53° ] × tan 45°


= cosec2 58°- cot 58° cot 58° - × [tan 13° tan (90-13)°] × [tan 37° tan (90-37)° ] × 1


= cosec2 58°- cot2 58° - × [tan 13° cot 13°] × [tan 37° cot 37° ]×1


= [cosec2 58°- cot2 58°] - × [1] × [1] × 1


= [1] -


= (2/3) – (5/3)


= (2 - 5)/3


= -3/3


= -1 = R.H.S.


Hence, proved.