RS Aggarwal Class 10 Solutions Chapter 7 Trigonometric Ratios of Complementary Angles Ratios Exercise 7

RS Aggarwal Solutions Chapter 7 Trigonometric Ratios of Complementary Angles Exercise 7 Class 10 Maths

Chapter Name	RS Aggarwal Chapter 7 Trigonometric Ratios of Complementary Angles
Book Name	RS Aggarwal Mathematics for Class 10
Other Exercises	Exercise 7
Related Study	NCERT Solutions for Class 10 Maths

Exercise 7 Solutions

1. Without using trigonometric tables, evaluate:

(i) sin 16°/cos 74°

(ii) sec 11°/cosec 79°

(iii) tan 27°/cot 63°

(iv) cos 35°/sin 55°

(v) cosec 42°/sec 48°

(vi) cot 30°/tan 52°

Solution

(i) sin 16°/cos 74°

= sin (90° - 74°)/cos 74°

= cos 74°/cos 74° [∵ sin (90° - θ) = cos θ] 

= 1 

(ii) sec 11°/cosec 70°

= sec (90° - 79°)/cosec 79°

= cosec 79°/cosec 79° [∵ sec (90 – θ) = cosec θ]

= 1

(iii) tan 27°/cot 63°

= tan (90° - 63°)/cos 30°

= cos 63°/cos 63° [∵ tan (90 – θ) = cot θ]

= 1

(iv) cos 35°/sin 55°

= cos(90° - 55°)/sin 55°

= sin 55°/sin 55° [∵ tan (90 – θ) = cot θ]

= 1

(v) cosec 42°/sec 48°

= cosec (90° - 48°)/sec 40°

= sec 48°/sec 48° [∵ sec (90 – θ) = cosec θ]

= 1

(vi) cot 38°/tan 52°

= cot (90° - 52°)/tan 52°

= tan 52°/tan 52° [∵ sec (90 – θ) = cosec θ]

= 1

2. Without using trigonometric tables, prove that: 

(i) cos 81° - sin 9°

(ii) tan 71° - cot 19° = 0

(iii) cosec 80° - sec 10 = 0

(iv) cosec² 72° - tan² 18° = 1

(v) cos² 75° + cos² 15° = 1 

(vii) sin² 48° + sin² 42° = 1 

(viii) cos² 57°. sin² 33° = 0

(ix) (sin 65° + cos 25°) (sin 65° - cos 25°) = 0

Solution

(i) LHS = cos 81° – sin 9°

= cos (90° - 9°) – sin 9°

= sin 9° - sin 9°

= 0

= RHS

(ii) LHS = tan 71° - cot 19°

= tan (90° - 19°) – cot 19°

= cot 19° - cot 19°

= 0

= RHS

(iii) LHS = cosec 80° - sec 10°

= cosec (90° - 10°) – sec 10°

= sec 10° - sec 10°

= 0

= RHS

(iv) LHS = cosec² 72° - tan² 18°

= cosec² (90° - 18°) – tan² 18°

= sec² 18° - tan² 18°

= 1

= RHS

(v) LHS = cos² 75° + cos² 15°

= cos² (90° - 15°) + cos² 15°

= sin² 15° + cos² 15°

= 1

(vi) LHS = tan² 66° - cot² 24°

= tan² (90° - 24°) – cot² 24°

= cot² 24° - cot² 24°

= 0

= RHS

(vii) LHS = sin² 48° + sin² 42°

= sin²(90° - 42°) + sin² 42°

= cos² 42° + sin² 42°

= 1

(viii) LHS = cos² 57° - sin² 33°

= cos² (90° - 33°) – sin² 33°

= sin² 33° - sin² 33°

= 0

= RHS

(ix) LHS = (sin 65° + cos 25°) (sin 65° - cos 25°)

= sin² 65° - cos² 25°

= cos² 25° - cos² 25°

= 0

= RHS

3. Without using trigonometric tables, prove that:

(i) sin 53° cos 37° + cos 53°sin 37° = 1

(ii) cos 54° cos 36° - sin 54° sin 36° = 0

(iii) sec 70° sin 20° + cos 20° cosec 70° = 2

(iv) sin 35° sin 55° - cos 35°cos 55° = 0

(v) (sin 72° + cos 18°) (sin 72° - cos 18°) = 0

(vi) tan 48° tan 23° tan 42° tan 67° = 1

Solution

(i) LHS = sin 53° cos 37° + cos 53° sin37°

= sin (90°- 37°) cos 37° + cos (90° - 37°) sin 37°

= cos² 37° cos 37° + sin 37° sin 37°

= cos² 37° + sin² 37° 

= 1 

= RHS

(ii) LHS = cos 54° cos 36°- sin 54°sin36°

= cos (90° - 36) cos 36° sin (90° - 36°) sin 36°

= sin 36° cos 36° - cos 36° sin 36°

= 0

= RHS

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

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

= cosec 20°. 1/cosec 20° + 1/sec 20°. sec 20°

 = 1 + 1 

= RHS

(iv) LHS = sin 35° sin 55° - cos 35° cos 55° 

= sin 35° cos(90° - 55°) - cos 35° sin (90° - 55°)

= sin 35° cos 35° - cos 35° sin 35° 

= 0

= RHS

(v) LHS = (sin 72° + cos 18°) (sin 72° - cos 18°) 

= (sin 72° + cos 18°) (cos(90° – 72°) - cos 18°]

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

= (sin 72° + cos 18°) (0) 

= RHS

(vi) LHS = tan 48° tan 23° tan 42° tan 67°

= cot (90°- 48°) cot (90° - 23) tan 42° tan 67°

= cot 42° cot 67° tan 42° tan 67°

= 1/tan 42° × 1/tan 67° × tan 42° × tan 67°

= 1

= RHS

4. Without using trigonometric tables, prove that: 

(i) sin 70°/cos 20° + cosec 20°/sec 70° - 2 cos 70° cosec 20° = 0

(ii) cos 80°/sin 10° + cos 59° cosec 31° = 2

(iii) 2 sin 68°/cos 22° - 2 cot 15°/5 tan 75° - (3 tan 45° tan 20° tan 40° tan 50° tan 70°)/5 = 1

(iv) sin 18°/cos 72° + √3 (tan 10° tan 30° tan 40° tan 50° tan 80°) = 2

(v) 7 cos 55°/3 sin 35° = 4(cos 70° cosec 20°)/3(tan 5° tan 25° tan 45° tan 65° tan 85°) = 1

Solution

(i) LHS = sin 70°/cos 20° + cosec 20°/sec 70° - 2 cos 70° cosec 20° 

= sin 70°/sin(90° - 20°) + sec(90° - 20°)/sec 70° - 2 cos 70° sec(90° - 20°)

= sin 70°/sin 70° + sec 70°/sec 70° - 2 cos 70° sec 70°

= 1 + 1 – 2 × cos 70° × 1/cos 70°

= 2 – 2

= 0

= RHS

(ii) LHS = cos 80°/sin 10° + cos 59° cosec 31°

= cos 80°/sin(90° - 10°) + sin(90° - 59°) cosec 31°

= cos 80°/cos 80° + sin 31° cosec 31°

= 1 + sin 31° × 1/sin 31°

= 1 + 1

= 2

= RHS

(iii) LHS = 2 sin 68°/cos 22° - 2 cot 15°/5 tan 75° - (3 tan 45° tan 20° tan 40° tan 50° tan 70°)/5

(iv) LHS = sin 18°/cos 72° + √3 tan (10° tan 30° tan 40° tan 50° tan 80°)

= sin 18°/sin (90° - 72°) + √3[cot (90° - 10°) × 1/√3 × cot (90° - 40°) × tan 50° × tan 80°]

= sin 18°/sin 18° + √3(cot 80° × cot 50° × tan 50° × tan 80°)/√3

= 1 + (1/tan 80° × 1/tan 50° × tan 50° × tan 80°)

= 1 + 1

= 2

= RHS

(v) LHS = (7 cos 55°)/3 sin 35° - 4(cos 70° cosec 20°)/3(tan 5° tan 25° tan 45° tan 65° tan 85°)

= (7 cos 55°)/3 cos (90° - 35°) – 4 [sin(90° - 70°) cosec 20°)/3[cot(90° - 5°) × cot(90° - 25°) × 1 × tan 65° × tan 85°

= 7 cos 55°/3 cos 55° - 4(sin 20° cosec 20°)/3(cot 85° cot 65° tan 65° tan 85°)

= 7/3 – 4 (sin 20° × 1/sin 20°)/[3(1/tan 85° × 1/tan 60° × tan 65° × tan 85°)

= 7/3 – 4/3

= 3/3

= 1

= RHS

5. Prove that:

(i) sin θ cos (90° - θ) + sin (90° - θ) cos θ = 1

(ii) sin θ/cos(90° - θ) + cos θ/sin(90° - θ) = 2

(iii) sin θ cos(90° - θ) cos θ/sin(90°- θ) + cos θ sin (90° - θ) sin θ/cos(90° - θ) = 1

(iv) cos (90° - θ) sec(90° - θ)tan θ/cosec(90° – θ) sin(90° - θ) cot(90° - θ) + tan(90° - θ)/cot θ = 2

(v) cos (90° - θ)/(1 + sin(90° - θ)/cos(90° - θ) = 2 cosec θ

(vi) sec(90°- θ) cosec θ – tan(90°- θ) cot θ + cos² 25° + cos² 65°)/3 tan 27° tan 63° = 2/3

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

Solution

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

= sin θ sin θ + cos θ cos θ

= sin² θ + cos² θ

= 1

= RHS

Hence proved.

(ii) LHS = sin θ/cos (90° - θ) + cos θ/(sin(90° - θ)

sin θ/sin θ + cos θ/cos θ

= 1 + 1

= 2

= RHS

Hence proved.

(iii) LHS = sin θ cos(90° - θ) cos θ/sin(90° - θ) + cos θ sin(90°- θ) sin θ/cos (90° - θ)

= (sin θ sin θ cos θ)/cos θ + (cos θ cos θ sin θ)/sin θ

= sin² θ + cos² θ

= 1

= RHS

Hence proved.

(iv) LHS = {(cos 90° - θ) sec(90° - θ) tan θ}/cosec(90° - θ) sin(90° - θ)° cot(90° – θ) + tan(90° - θ)/cot θ

= (sin θ cosec θ tan θ)/(sec θ cos θ tan θ) + cot θ /cot θ

= 1 + 1

= 2

= RHS

Hence proved.

(v) LHS = cos (90° - θ)/(1 + sin(90° - θ) + 1 + sin(90° - θ)/cos(90° - θ)

= sin θ/(1 + cos θ) + (1 + cos θ)/sin θ

= sin² θ + (1 + cos θ)²/(1 + cos θ) sin θ

= (sin² θ + 1 + cos² θ + 2 cos θ)/(1 + cos )sin θ

= (1 + 1 + 2 cos θ)/(1 + cos θ) sin θ

= (2 + 2 cos θ)/(1 + cos θ) sin θ

= 2(1 + cos θ)/(1 + cos θ) sin θ

= 2.(1/sin θ)

= 2 cosec θ  

= RHS

(vi) LHS = sec(90° - θ) cosec θ – tan(90° - θ) cot θ  + cos² 25°  + cos² 65°)/(3 tan 27° tan 63°)

= cosec θ cosec θ - cot θ cot θ + sin² (90° - 25°) + cos² 65°]/3 tan 27° cot (90° - 63°)

= (cosec² θ – cot² θ + sin² 65° cos² 65°)/3 tan 27° cot 27°

= (1 + 1)/(3×tan 27°) × 1/(tan 27°)

= 2/3

= RHS

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

= cot θ cot θ – cosec θ cosec θ + √3 tan 12° × √3 × cot (90° - 78°)

= cot² θ – cosec² θ + 3 tan 12° cot 12°

= - 1 + 3 × tan 12° × 1/tan 12°

= - 1 + 3

= 2

= RHS

6. Without using trigonometric tables, prove that:

(i) tan 5° tan 25° tan 30° tan 65° tan 85° = 1

(ii) cot 12° cot 38° cot 52° cot 60° cot 78° = 1/√3

(iii) cos 15° cos 35° cosec 55° cos 60° cosec 75° = 1/2

(iv) cos 1° cos 2° cos 3°..... cos 180° = 0

(v) (sin 49°/cos 41°)² + (cos 41°/sin 49°) = 2

Solution

(i) LHS = tan 5° tan 25° tan 30° tan 65° tan 85°

= tan (90° - 85°) tan(90° - 65°)× 1/√3 × 1/cot 60° 1/cot 85°

= cot 85° cot 65° 1/√3 1/cot 60°. 1/cot 85°

= 1/√3

= RHS

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

= tan (90° - 12°) × tan(90°-38°)×cot 52°× 1/√3 × cot 78°

= 1/√3 × tan 78°× an 52°×cot 52°×cot 78°

= 1/√3 × tan 78° × tan 52° × 1/tan 52° × 1/tan 78°

= 1/√3

= RHS

(iii) LHS = cos 15° cos 35° cosec 55° cos 60° cosec 75°

= cos (90° - 75°) cos (90° - 55°) 1/sin 55° × ½ × 1/sin 75°

= sin 75° sin 55°. 1/sin 55° × 1/2 × 1/sin 75°

= 1/2

= RHS

(iv) LHS = cos 1° cos 2° cos 3° .... cos 180°

= cos 1° × cos 2° × cos 3° × .....× cos 90° × ....× in cos 180°

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

= 0

= RHS

(v) LHS = (sin 49°/cos 41°)² + (cos 41°/sin 49°)

= (cos(90° - 45°)/cos 41°)² + (cos 41°/cos(90° - 49°)²

= (cos 41°/cos 41°)² + (cos(41°/cos 41°)²

= 1² + 1²

= 1 + 1

= 2

= RHS

Disclaimer: The RHS of (v) given in textbook is incorrect. There should be 2 instead 1.

The same has been corrected in the solution here. 

7. Prove that:

(i) sin (70° + θ) – cos(20° - θ) = 0

(ii) tan (55° - θ) – cot (35° + θ) = 0

(iii) cosec (67° - θ) + sec (20° - θ) = 0

(iv) cosec (65° + θ) – sec (25°- θ) – tan(55° - θ) + cot (35° + θ) = 0

(v) sin (50° + θ) – cos (40° - θ) + tan 1° tan 10° tan 80° tan 89° = 1

Solution

(i) LHS = sin (70° + θ) – cos (20° - θ)

= sin {(90° - (20° - θ)} – cos (20° - θ)

= cos (20° - θ) – cos (20° - θ)

= 0

= RHS

(ii) LHS = tan(55° - θ) – cot(35° + θ)

= tan{90° - (35° - θ)} – cot θ(35° - θ)

= cot (35° - θ) – cot(35° - θ)

= 0

= RHS

(iii) LHS = cosec(67°- θ) – sec(23° - θ)

= cosec {90° - 23° - θ)} – sec (23°- θ)

= sec (20° - θ) – sec (23° - θ)

= 0

= RHS

(iv) LHS = cosec (65° - θ) – sec(25°- θ) – tan(55° - θ) + cot(35° + θ)

= cosec{90° - (25° - θ)} – sec(25° - θ) – tan(55° - θ) + cot{90° - (55° - θ)}

= sec(25° - θ) – sec(25° - θ) – tan(55° - θ) + tan(55° - θ)

= 0

= RHS

(v) LHS = sin(50° - θ) – cos (40° - θ) + tan 1° tan 10° tan 80° tan 89°

= sin{90° - (40° - θ)} – cos (40° - θ) + {tan 1° tan(90° - 1°)} {tan 10° tan(90° - 10)}

= cos (40° - θ) - cos (40° - θ) + (tan 1° cot 1°) (tan 10° cot 10°)

= 1(1/cot 1° × cot 1°)(tan 10° × 1/tan 10°)

= 1 × 1

= 1

= RHS

8. Express each of the following in terms of trigonometric ratios of lying between 0° and 45°.

(i) sin 67° + cos 45°

 (ii) cot 65° + tan 49°

(iii) sec 78° + cosec 56°

 (iv) cosec 54° + sin 72°

Solution

(i) sin 67° + cos 75°

= cos (90° - 67°) + sin (90° - 75°)

= cos 23° + sin 15°

(ii) cot 65° + tan 49°

= cos(90° - 65°) + cot(90° - 49°)

= cos 25° + cot 41°

(iii) sec 78° + cosec 56°

= sec (90° - 12°) + cosec (90° - 34°)

= cosec 12° + sec 34°

(iv) cosec 54° + sin 72°

= sec (90° - 54°) + cos(90° - 72°)

= sec 36° + cos 18°

9. If A, B, C are the angles of a ∆ABC, prove that tan (C + A)/2 = cot π/2.

Solution

In ∆ABC,

A + B + C = 180°

⇒ A + C = 180° - B ...(i)

Now,

LHS = tan (C + A)/2

= tan (180° - θ)/2 [Using (i)]

= tan (90° - B/2)

= cot B/2

= RHS

10. If cos 2θ = sin 4θ and 2θ is acute, then find the value of θ.

Solution

We have:

cos 2θ = sin 4θ

⇒ sin (90° - 2θ) = sin 4θ

Comparing both sides, we get

90° - 2θ = 4θ

⇒ 2θ - 4θ = 90°

⇒ 6θ = 90°

⇒ θ = 90°/6

∴ θ = 15°

Hence, the value of θ is 15°

11. If sec 2A = cosec (A - 12°), where 2A is an acute angle, find the value A,

Solution

We have:

sec 2A = cosec (A - 42°)

⇒ cosec (90° - 2A) = cosec (A - 42°)

Comparing both sides, we get

90° - 2A = A - 42°

⇒ 2A + A = 90° + 42°

⇒ 3A = 132°

⇒ A = 132°/3

∴ A = 40°

Hence, the value of A is 44°.

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

Solution

sin 3A = cos (A - 26°)

⇒ cos (90° - 3A) = cos (A - 26°) [∵ sin θ = cos (90° - θ)]

⇒ 90° - 3A = A - 26°

⇒ 116° = 4A

⇒ A = 116°/4 = 29°

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

Solution

tan 2A = cot(A - 12°)

⇒ cot (90° - 2A) = cot (A - 12°) [∵ tan θ = (cot (90° - θ)]

⇒ (90° - 2A) = (A - 12°)

⇒ 102° = 3A

⇒ A = 102°/3

= 34°

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

Solution

sec 4A = cosec (A - 15°)

⇒ cosec (90° - 4A) = cosec (A - 15°) [∵ secθ = cosec (90° - θ)]

⇒ (90° - 4A) = (A - 15°)

⇒ 105° = 5A

⇒ A = 105°/5°

= 21°

15. Without using trigonometric tables, evaluate the following:

Solution

2/3 cosec² 58° - 2/3 cot 58° tan 32° - 5/3 tan 13° tan 37° tan 45° tan 53° tan 77°

= 2/3 (cosec² 58° - cot 58° tan 32°) – 5/3 tan 13° tan(90° - 13°) tan 37° tan(90° - 37°) (tan 45°)

= 2/3{cosec² 58° - cot 58° tan (90° - 58°)} – 5/3 tan 13° cot 13° tan 37° cot 37° (1)

= 2/3(cosec² 58° - cot 58° tan 58°) - 5/3 tan 13° 1/tan 13° tan 37° 1/tan 37°

= 2/3(cosec² 58° - cot² 58°) – 5/3

= 2/3 – 5/3

= - 1

Hence proved.