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• RD Sharma Class 10 Solutions

# Class 10 RD Sharma Solutions – Chapter 5 Trigonometric Ratios – Exercise 5.2 | Set 1

### Question 1. sin 45° sin 30° + cos 45° cos 30°

Solution:

Given: sin 45° sin 30° + cos 45° cos 30°         -(1)

Putting the values of sin 45° = cos 45°= 1/√2, sin 30° = 1/2, cos 30° = √3/2 in eq(1)

= (1/√2)(1/2) + (1/√2})(√3/2)

= (1/2√2) + (√3/2√2)

= (1 + √3)/2√2

### Question 2. sin60°cos30° + cos60°sin30°

Solution:

Given: sin60°cos30° + cos60°sin30°          -(1)

Putting the values of sin 60° = cos 30° = √3/2, sin 30° = cos 60° = 1/2 in eq(1)

= (√3/2)(√3/2) + (1/2)(1/2)

= 3/4 + 1/4

= 1

### Question 3. cos60°cos45° − sin60°sin45°

Solution:

Given: cos60°cos45° − sin60°sin45°         -(1)

Putting the values of sin 45° = cos 45° = 1/√2, sin 60° = √3/2, cos 60° = 1/2 in eq(1)

= (1/2)(1/√2) – (√3/2)(1/√2)

= (1/2√2) – (√3/2√2)

= (1 -√3)/2√2

### Question 4. sin230° + sin245° + sin260° + sin290°

Solution:

Given: sin230° + sin245° + sin260° + sin290°         -(1)

Putting the values of sin 45° = 1/√2, sin 30° = 1/2, sin 60° = √3/2, sin 90° = 1 in eq(1)

= (1/2)2 + (1/√2)2 + (√3/2)2 + 12

= 1/4 + 1/2 + 3/4 + 1

= 1/4 + 2/4 + 3/4 + 4/4

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

= 10/4

= 5/2

### Question 5. cos230° + cos245° + cos260° + cos290°

Solution:

Given: cos230° + cos245° + cos260° + cos290°          -(1)

Putting the values of cos 45° = 1/√2, cos 60° = 1/2, cos 30° = √3/2, cos 90° = 0 in eq(1)

= (√3/2)2 +(1/√2)2 + (1/2)2 + 02

= 3/4 + 1/2 + 1/4

= 3/4 + 2/4 + 1/4

= (1 + 2 + 3)/4

= 6/4

= 3/2

### Question 6. tan230° + tan260° + tan245°

Solution:

Given: tan230° + tan260° + tan245°           -(1)

Putting the values of tan 45° = 1, tan 30° = 1/√3, tan 60° = √3 in eq(1)

= (1/√3)2 + (√3)2 + (1)2

= 1/3 + 3 + 1

= (1 + 12)/3

= 13/3

### Question 7. 2sin230° – 3cos245° + tan260°

Solution:

Given: 2sin230° – 3cos245° + tan260°          -(1)

Putting the values of tan 60° = √3, cos 45° = 1/√2, sin 30° = 1/2 in eq(1)

= 2(1/2)2 -3 (1/√2)2 + (√3)2

= 2/4 – 3/2 + 3

= 1/2 – 3/2 + (3×2)/2

= 1/2 – 3/2 + 6/2

= 4/2

= 2

### Question 8. sin230°cos245° + 4tan230° + (1/2)sin290° – 2cos230° + (1/24)cos20°

Solution:

Given: sin230°cos245° + 4tan230° + (1/2)sin290° – 2cos230° + (1/24)cos20°

= (1/2)2(1/√2)2 + 4(1/√3)2 + (1/2)(1)2 – 2(0)2 + (1/24)(1)2

= 1/8 + 4/3 + 1/2 + 1/24

= 3/24 + 32/24 + 12 + 24 + 1/24

= 48/24

= 2

### Question 9. 4(sin460° + cos430°) − 3(tan260° − tan245°) + 5cos245°

Solution:

Given: 4(sin460° + cos430°) − 3(tan260° − tan245°) + 5cos245°

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

= 4(9/16 + 9/16) – 3(3 – 1) + 5/2

= 4(18/16) – 3(2) + 5/2

= 9/2 – 12/2 + 5/2

= (9 – 12 + 5)/2

= 2/2

= 1

### Question 10. (cosec245°sec230°)(sin230° + 4cot245° – sec260°)

Solution:

Given: (cosec245°sec230°)(sin230° + 4cot245° – sec260°)

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

= (8/3) × (1/4) + 4 – 4

= (8/3) × (1/4)

= 2/3

### Question 11. cosec330°cos60°tan345°sin290°sec245°cot30°

Solution:

Given: cosec330°cos60°tan345°sin290°sec245°cot30°

= (2)3(1/2)(1)3(1)2(√2)2(√3)

= (8)(1/2)(2)(√3)

= 8√3

### Question 12. cot230° – 2cos260° – (3/4)sec245° – 4sec230°

Solution:

Given: cot230° – 2cos260° – (3/4)sec245° – 4sec230°

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

= 3 – 2/4 – 6/4 – 16/3

= 3 – 1/2 – 3/2 – 16/3

= (18 – 3 – 9 – 32)/6

= -26/6

= -13/3

### Question 13. (cos0° + sin45° + sin30°)(sin90° – cos45° + cos60°)

Solution:

Given: (cos0° + sin45° + sin30°)(sin90° – cos45° + cos60°)

= (1 + 1/√2 + 1/2)(1 – 1/√2 + 1/2)

= (3/2 + 1/√2)(3/2 – 1/√2)

Using identity (a + b)(a – b) = a2 – b2

= (3/2)2 – (1/√2)

= 9/4 – 1/2

= (9 – 2)/4

= 7/4

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