Class 10 RD Sharma Solutions Chapter 5 Trigonometric Ratios Exercise 5.2 | Set 1 (original) (raw)

Last Updated : 21 Feb, 2021

Evaluate each of the following(1-13)

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)2

= 9/4 - 1/2

= (9 - 2)/4

= 7/4