# RS Aggarwal Solutions Chapter 6 T-Ratios of Some Particular Angles Exercise 6 Class 10 Maths

 Chapter Name RS Aggarwal Chapter 6 T-Ratios of some Particular Angles Book Name RS Aggarwal Mathematics for Class 10 Other Exercises Exercise 6 Related Study NCERT Solutions for Class 10 Maths

### Exercise 6 Solutions

1. Evaluate: sin 60° cos 30° + cos 60° sin 30°

Solution

On substituting the values of various T-ratios, we get:

sin 60° cos 30° + cos 60° sin 30°

= (√3/2×√3/2 + 12×12)

= (3/4 + 1/4)

= 4/4

= 1

2. Evaluate: cos 60° cos 30°– sin 60° sin 30°

Solution

On substituting the values of various T-ratios, we get:

cos 60° cos 30° − sin 60° sin 30°

= (1/2 × √3/2 − √3/2 × 1/2)

= (√3/4 − √3/4)

= 0

3. Evaluate: cos 45° cos 30° + sin 45° sin 30°

Solution

On substituting the values of various T-ratios, we get:

cos 45° cos 30° + sin 45° sin30°

= (1√2×√3/2 + 1/√2 × 1/2)

= (√3/2√2 + 1/2√2)

= (√3 + 1)/2√2)

4. Evaluate: sin 30°/cos 45° + cot 45°/sec 60° – sin 60°/tan 45° + cos 30°/sin 90°

Solution

sin 30°/cos 45° + cot 45°/sec 60° – sin 60°/tan 45° + cos 30°/sin 90°

= (1/2)/(1/√2) + 1/2 – (√3/2)/1 + (√3/2)/1

= √2/2 + 1/2 - √3/2 + √3/2

= (√2 + 1)/2

5. Evaluate: (5 cos2 60° + 4 sec2 30° – tan2 45°)/(sin2 30° + cos2 30°)

Solution

(5 cos2 60° + 4 sec2 30° – tan2 45°)/(sin2 30° + cos2 30°)

= [5(1/2)2 + 4(2/√3)2 – (1)2)]/[(1/2)2 + (√3/2)2]

= (5/4 + (4×4)/3 – 1)\(1/4 + 3/4)

= (5/4 + 16/3 – 1/1)/(4/4)

= [(15 + 64 – 12)/12]/(4/4)

= 67/12

6. Evaluate:

2 cos60° + 3 sin45°- 3 sin30° + 2 cos90°

Solution

On substituting the values of various T-ratios, we get:

2cos260° + 3 sin245°- 3sin230° + 2 cos2 90°

= 2×(1/2)2 + 3×(1/√2)2 - 3×(1/2)+ 2×(0)2

= 2× 1/4 + 3× 1/2 – 3× 1/4 + 0

= (1/2 + 3/2 – 3/4)

= (2 + 6 – 3)/4

= 5/4

7. Evaluate:

cot30°- 2 cos30° – 3/4 sec245° + 1/4 cosec2 30°

Solution

On substituting the values of various T-ratios, we get:

cot30° - 2 cos30° - 3/4 sec45° + 1/4 cosec2 30°

= (√3)2– 2×(√3/2)2 ×(√2)2 + 1/4 ×(2)2

= 3 – 2× 3/4 – 3/4 ×2 + 1/4 ×4

= 3 – 3/2 – 3/2 + 1

= 4 - (3/2 + 3/2)

= 4 – 3

= 1

8. Evaluate:

(sin2 30° + 4 cot2 45°- sec2 60°) (cosec2 45° sec2 30°

Solution

On substituting the values of various T-ratios, we get:

(sin2 30° + 4 cot2 45°- sec2 60°) (cosec2 45° sec2 30°)

= [(1/2)2 + 4×(1)2 − (2)2] [(√2)2(2√3)2

= (1/4 + 4 − 4) (2× 4/3)

= 1/4 × 8/3 = 2/3

9. Evaluate: 4/cot2 30° + 1/sin2 30° - 2 cos245° – sin20°

Solution

On substituting the values of various T-ratios, we get:

4/cot2 30°– 1/sin2 30° 2 cos245° – sin2

= 4/(√3)2 + 1/(1/2)2 – 2 × (1/√2)2 – (0)2

= 4/3 + 1/(1/4) – 2 × 1/2 - 0

= 4/3 + 4 – 1

= 4/3 + 3

= (4 + 9)/3

= 13/3

10. Show that:

(i) (1 – sin 60°)/cos 60° = (tan 60° – 1)/(tan 60° + 1)

(ii) (cos 30° + sin 60°)/(1 + sin 30° + cos 60°) = cos 30°

Solution

(i) LHS = (1 - sin 60°)/(cos 60°)

= (1 - √3/2)/1/2

= (2 - √3)/2/(1/2)

= (2 - √3)/2 ×2

= 2 - √3

RHS = (tan 60° – 1)/(tan 60° + 1)

= (√3 – 1)/(√3 + 1)

= (√3 – 1)/(√3 + 1) × (√3 – 1)/(√3 – 1)

= (√3 – 1)2/(√3)2 – 12)

= (3 + 1 - 2√3)/(3 – 1)

= (4 - 2√3)/2

= 2 - √3

Hence, LHS = RHS

∴ (1 – sin 60°)/(cos 60°) = (tan 60° – 1)/(tan 6° + 1)

(ii) LHS = (cos 30° + sin 60°)/(1 + sin 30° + cos 60°)

= (√3/2 + √3/2)/(1 + 1/2 + 1/2)

= [(√3 + √3)/2]/[(2 + 1 + 1)/2]

= √3/2

Also, RHS = cos 30° = √3/2

Hence, LHS = RHS

∴ cos 30° + sin 60°

1 + sin 30° + cos 60°

= cos 30°

11. Verify each of the following:

(i) sin 60°cos 30°– cos 60°sin 30°

(ii) cos 60°cos 30° + sin 60°sin 30°

(iii) 2 sin 30°cos 30°

(iv) 2 sin 45°cos 45°

Solution

(i) sin 60°cos 30° – cos 60°sin 30°

= (√3/2) × (√3/2) - (1/2) × (1/2) = 3/4 – 1/4 = 2/4 = 1/2

Also, sin 300 = 1/2

∴ sin 60°cos 30°– cos 60°sin 30° = sin 30°

(ii) cos 60°cos 30° + sin 60° sin 30°

= (1/2) × (√3/2) + (√3/2) × (1/2) = √3/4 + √3/4 = √3/2

Also, cos 30° = √3/2

∴ cos 60°cos 30° + sin 60°sin 30° = cos 30°

(iii) 2 sin 30°cos 30°

= 2× 1/2 ×√3/2 = √3/2

Also, sin 60° = √3/2

∴ 2 sin 30°cos 30° = sin 60°

(iv) 2 sin 45°cos 45°

= 2 × 1/√2 × 1/√2 = 1

Also, sin 90° = 1

∴ 2 sin 45° cos 45° = sin 90°

12. If A = 45°, verify that:

(i) sin 2A = 2 sin A cos A

(ii) cos 2A = 2 cos2 A – 1 = 1 – 2 sin2 A

Solution

A = 45°

2A = 2 × 45° = 90°

(i) sin 2A = sin 90° = 1

⇒ 2 sin A cos A = 2

⇒ 2 sin A cos A = 2 sin 45°cos 45°

= 2 × 1/√2 × 1/√2

= 1

∴ sin 2A = 2 sin A cos A

(ii) cos 2A = cos 90° = 0

⇒ 2 cos2A – 1 = 2 cos 245° – 1

= 2 × (1/√2)2 – 1

= 2 × 1/2 – 1

= 1 – 1

= 0

Now,

1 – 2 sin2 A

= 1 – 2 × (1/√2)2 – 1

= 1 – 2 × 1/2

= 1 – 1

= 0

∴ cos 2A = 2 cos2 A – 1 = 1 – 2 sin2 A

13. If A = 30°, verify that:

(i) sin 2A = 2 tan A/(1 + tan2 A)

(ii) cos 2A = (1 – tan2 A)/(1 + tan2 A)

(iii) tan 2A = 2 tan A/(1 – tan2 A)

Solution

A = 30°

⇒ 2A = 2 × 30° = 60°

(i) sin 2A = sin 60° = √3/2

2 tan A/(1 + tan2 A) = 2 tan 30°/(1 + tan2 30°)

⇒ 2 × (1/√3)/(1 + (1/√3)2 = (2/√3)/(1 + 1/3)

⇒ (2/√3)/(4/3) = (2/√3) × 3/4

= √3/2

∴ sin 2A = 2 tan A/(1 + tan2 A)

(ii) cos 2A = cos 60° = 1/2

(1 – tan2 A)/(1 + tan2 A) = (1 – tan2 30°)/(1 + tan2 30°)

= (1 – (1/√3)2/(1 + 1/√3)2

= (1 – 1/3)/(1 + 1/3)

= (2/3)/(4/3)

= (2/3) × 3/4

= 1/2
∴ cos 2A = (1 – tan2 A)/(1 + tan2 A)

(iii) tan 2A = tan 60° = √3

2 tan A/(1 – tan2 A) = (2 tan 30°)/(1 – tan2 30°)

= 2 × (1/√3)/(1− (1/√3)2)

= (2/√3)/(2/3)

= (2/√3) × 3/2

= √3
∴ tan 2A = 2 tan A/(1 – tan2 A)

14. If A = 60° and B = 30°, verify that:

(i) sin (A + B) = sin A cos B + cos A sin B

(ii) cos (A + B) = cos A cos B - sin A sin B

Solution

A = 60° and B = 30°

Now, A + B = 60° + 30° = 90°

Also, A – B = 60°– 30° = 30°

(i) sin (A + B) = sin 90° = 1

⇒ sin A cos B + cos A sin B = sin 60°cos 30° + cos 60° sin 30°

⇒ (√3/2 × √3/2 + 1/2 × 1/2) = (3/4 + 1/4) = 1

∴ sin (A + B) = sin A cos B + cos A sin B

(ii) cos (A + B) = cos 90° = 0

⇒ cos A cos B - sin A sin B = cos 60° cos 30° - sin 60° sin 30°

⇒ (1/2 × √3/2 − √3/2 × 12) = (√34−√34) = 0

∴ cos (A + B) = cos A cos B - sin A sin B

15. If A = 60° and B = 30°, verify that:

(i) sin (A – B) = sin A cos B – cos A sin B

(ii) cos (A – B) = cos A cos B + sin A sin B

(iii) tan (A – B) = (tan A – tan B)/(1 + tan A tan B)

Solution

(i) sin (A – B) = sin 30° = 1/

⇒ sin A cos B – cos A sin B = sin 60°cos 30° – cos 60° sin 30°

⇒ (√3/2 × √3/2 – 1/2 × 1/2) = (3/4 – 1/4)) = 2/4 = 1/2

∴ sin (A – B) = sin A cos B – cos A sin B

(ii) cos (A – B) = cos 30° = √3/2

⇒ cos A cos B + sin A sin B = cos 60°cos 30° + sin 60°sin 30°

⇒ (1/2 × √3/2 + √3/2 × 1/2) = (√3/4 + √3/4) = 2 × √3/4 = √3/2

∴ cos (A – B) = cos A cos B + sin A sin B

(iii) tan (A – B) = tan 60° = 1/√3

⇒ (tan A – tan B)/(1 + tan A tan B)

⇒ (tan 60° - tan 30°)/(1 + tan 60° tan 30°) = (√3 – 1/√3)/(1 + √3 + 1/√3)

= 1/2 × (3 – 1)/√3

= 3/√3

∴ tan (A – B) = (tan A – tan B)/(1 + tan A tan B)

16. If A and B are acute angles such that tan A = 1/3, tan B = 1/2 and tan (A + B) = (tan A + tan B)/(1 – tan A tan B),

show that A + B = 45°

Solution

Given,

tan A = 1/3 and tan B = 1/2

tan (A + B) = (tan A + tan B)/(1 – tan A tan B)

On substituting these values in RHS of the expression, we get:

(tan A + tan B)/(1 - tan A tan B)

= (1/3 + 1/2)/(1 – 1/3 × 1/3)

= (5/6)/(1 – 1/6)

= (5/6)/(5/6)

= 1

⇒ tan (A + B)

= 1

= tan 45 [∵ tan 45° = 1]

⇒ tan (A + B) = 1 = tan 45° [∵ tan 45° = 1]

∴ A + B = 45°

17. Using the formula, tan 2A = 2 tan A/(1 – tan2 A), find the value of tan 60°, it being given that tan 30° = 1/√3.

Solution

A = 30°

⇒ 2A = 2 × 30° = 60°

By substituting the value of the given T-ratio, we get:

tan 2A = (2 tan A)/(1 – tan2A)

⇒ tan 60° = (2 tan 30°)/(1 – tan2 30°) = 2 × (1/√3)

1− (1/√3)2 = (2√3)/(1 – 1/3)

= (2/√3)/(2/3)

= (2/√3) × 3/2

= √3

18. Using the formula, cos A = √(1+cos2A)/2 find the value of cos 30°, it being given that cos 60° = 1/2.

A = 30°

⇒ 2A = 2 × 30° = 60°

By substituting value of the given T-ratio, we get:

∴ cos A = √3/2

19. Using the formula, sin A = √(1-cos2A)/2 find the value of sin 30°, it being given that cos 60° = 1/2.

Solution

A = 30°

⇒ 2A = 2 × 30° = 60°

By substituting the value of the given T-ratio, we get:

∴ sin 30° = 1/2

20. In the adjoining figure, ∆ABC is a right-angled triangle in which B = 90°, ∠30° and AC = 20 cm.

Find (i) BC, (ii) AB.

Solution

From the given right-angled triangle, we have:

BC/AC = sin 30°

⇒ BC = 20/2 = 10 cm

Also, AB/AC = cos 30°

⇒ AB/20 = √3/2

⇒ AB = (20 × √3/2)

= 10√3 cm

∴ BC = 10 cm and AB = 10√3 cm

21. In the adjoining figure, ∆ABC is right-angled at B ∠A = 30°. If BC = 6 cm, find

(i) BC,

(ii) AB

Solution

From the given right-angled triangle, we have:

BC/AB = tan 30°

⇒ 6/AB = 1/√3

⇒ AB = 6√3 cm

Also, BC/AC = sin 30°

⇒ 6/AC = 1/2

⇒ AC = (2 × 6) = 12 cm

∴ AB = 6√3 cm and AC = 12 cm

22. In the adjoining figure, ∆ABC is right-angled at B and A = 45°. If AC = 3√2cm, find

(i) BC

(ii) AB

Solution

From the right-angled ∆ABC, we have:

BC/AC = sin 45°

⇒ BC/3√2 = 1/√2

⇒ BC = 3 cm

Also, AB/AC = cos 45°

⇒ AB/3√2 = 1/√2

⇒ AB = 3 cm

∴ BC = 3 cm and AB = 3 cm

23. If sin (A + B) = 1 and cos (A – B) = 1, 0° ≤ (A + B) ≤ 90° and A > B, then find A and B.

Solution

Here, sin (A + B) = 1

⇒ sin (A + B) = 90° [∵ sin 90° = 1]

⇒ (A + B) = 90° ...(i)

Also, cos (A – B) = 1

⇒ cos (A – B) = 0° [∵ cos 0° = 1]

⇒ A – B = 0° ...(ii)

Solving (i) and (ii), we get:

A = 45° and B = 45°

24. If sin (A – B) = 1/2 and cos (A + B) = 1/2, 0° ≤ (A + B) ≤ 900and A > B, then find A and B.

Solution

Here, sin (A – B) = 1/2

⇒ sin (A – B) = 30°  [∵ sin 30° = 1/2]

⇒ (A – B) = 30°  ….(i)

Also, cos (A + B) = 1/2

⇒ cos (A + B) = cos 60°  [∵ cos 60° = 1/2]

⇒ A + B = 60° ….(ii)

Solving (i) and (ii), we get:

A = 45° and B = 15°

25. If tan (A – B) = 1/√3 and tan (A + B) = √3, 0° ≤ (A + B) ≤ 90° and A > B, then find A and B.

Solution

Here, tan (A – B) = 1/√3

⇒ tan (A – B) = tan 30°  [∵ tan 30° = 1/√3]

⇒ (A – B) = 30° ….(i)

Also, tan (A + B) = √3

⇒ tan (A + B) = tan 60°  [∵ tan 60° = √3]

⇒ A + B = 60° ...(ii)

Solving (i) and (ii), we get:

A = 45° and B = 15°

26. If 3x = cosec Î¸ and 3/x = cot Î¸ find the value of 3(x2 – 1/x2)

Solution

3(x2 – 1/x2)

= 9/3(x2 – 1/x2)

= 1/3(9x2 – 9/x2)

= 1/3[(3x2) – (3/x)2]

= 1/3[(cosec Î¸)2 – (cot Î¸)2]

= 1/3(cosec2 Î¸ – cot2 Î¸)

= 1/3(1)

= 1/3

27. If sin (A + B) = sin A cos B + cos A sin B and cos (A – B) = cos A cos B + sin A sin B

(i) sin (75°)

(ii) cos (15°)

Solution

Let A = 45° and B = 30°

(i) As, sin(A + B) = sin A cos B + cos A sin B

⇒ sin (45° + 30°) = sin 45° cos 30° + cos 45° sin 30°

⇒ sin (75°) = 1√2 × √3/2 + 1/√2 × 1/2

⇒ sin (75°) = √3/2√2 + 1/2√2

∴ sin (75°) = (√3 + 1)/2√2

(ii) As, cos (A – B) = cos A cos B + sin A sin B

⇒ cos (45°– 30°) = cos 45°cos 30° + sin 45° sin 30°

⇒ cos (15°) = 1/√2 × √3/2 + 1/√2 × 1/2

⇒ cos (15°) = √3/(2√2) + 1/(2√2)

∴ cos (15°) = (√3 + 1)/2√2

Disclaimer: cos 15° can also be written by taking A = 60° and B = 45°