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

Last Updated : 21 Feb, 2021

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

Solution:

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

Putting the values of sin 45Â° = cos 45Â°= 1/âˆš2, sin 30Â° = 1/2, cos 30Â° = âˆš3/2 in eq(1)

= (1/âˆš2)(1/2) + (1/âˆš2})(âˆš3/2)

= (1/2âˆš2) + (âˆš3/2âˆš2)

= (1 + âˆš3)/2âˆš2

### Question 2. sin60Â°cos30Â° + cos60Â°sin30Â°

Solution:

Given: sin60Â°cos30Â° + cos60Â°sin30Â°          -(1)

Putting the values of sin 60Â° = cos 30Â° = âˆš3/2, sin 30Â° = cos 60Â° = 1/2 in eq(1)

= (âˆš3/2)(âˆš3/2) + (1/2)(1/2)

= 3/4 + 1/4

= 1

### Question 3. cos60Â°cos45Â° âˆ’ sin60Â°sin45Â°

Solution:

Given: cos60Â°cos45Â° âˆ’ sin60Â°sin45Â°         -(1)

Putting the values of sin 45Â° = cos 45Â° = 1/âˆš2, sin 60Â° = âˆš3/2, cos 60Â° = 1/2 in eq(1)

= (1/2)(1/âˆš2) – (âˆš3/2)(1/âˆš2)

= (1/2âˆš2) – (âˆš3/2âˆš2)

= (1 -âˆš3)/2âˆš2

### Question 4. sin230Â° + sin245Â° + sin260Â° + sin290Â°

Solution:

Given: sin230Â° + sin245Â° + sin260Â° + sin290Â°         -(1)

Putting the values of sin 45Â° = 1/âˆš2, sin 30Â° = 1/2, sin 60Â° = âˆš3/2, sin 90Â° = 1 in eq(1)

= (1/2)2 + (1/âˆš2)2 + (âˆš3/2)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. cos230Â° + cos245Â° + cos260Â° + cos290Â°

Solution:

Given: cos230Â° + cos245Â° + cos260Â° + cos290Â°          -(1)

Putting the values of cos 45Â° = 1/âˆš2, cos 60Â° = 1/2, cos 30Â° = âˆš3/2, cos 90Â° = 0 in eq(1)

= (âˆš3/2)2 +(1/âˆš2)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. tan230Â° + tan260Â° + tan245Â°

Solution:

Given: tan230Â° + tan260Â° + tan245Â°           -(1)

Putting the values of tan 45Â° = 1, tan 30Â° = 1/âˆš3, tan 60Â° = âˆš3 in eq(1)

= (1/âˆš3)2 + (âˆš3)2 + (1)2

= 1/3 + 3 + 1

= (1 + 12)/3

= 13/3

### Question 7. 2sin230Â° – 3cos245Â° + tan260Â°

Solution:

Given: 2sin230Â° – 3cos245Â° + tan260Â°          -(1)

Putting the values of tan 60Â° = âˆš3, cos 45Â° = 1/âˆš2, sin 30Â° = 1/2 in eq(1)

= 2(1/2)2 -3 (1/âˆš2)2 + (âˆš3)2

= 2/4 – 3/2 + 3

= 1/2 – 3/2 + (3Ã—2)/2

= 1/2 – 3/2 + 6/2

= 4/2

= 2

### Question 8. sin230Â°cos245Â° + 4tan230Â° + (1/2)sin290Â° – 2cos230Â° + (1/24)cos20Â°

Solution:

Given: sin230Â°cos245Â° + 4tan230Â° + (1/2)sin290Â° – 2cos230Â° + (1/24)cos20Â°

= (1/2)2(1/âˆš2)2 + 4(1/âˆš3)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(sin460Â° + cos430Â°) âˆ’ 3(tan260Â° âˆ’ tan245Â°) + 5cos245Â°

Solution:

Given: 4(sin460Â° + cos430Â°) âˆ’ 3(tan260Â° âˆ’ tan245Â°) + 5cos245Â°

= 4(âˆš3/2)4 + (âˆš3/2)4) – 3((âˆš3)2 – (1)2) + 5(1/âˆš2)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. (cosec245Â°sec230Â°)(sin230Â° + 4cot245Â° – sec260Â°)

Solution:

Given: (cosec245Â°sec230Â°)(sin230Â° + 4cot245Â° – sec260Â°)

= ((âˆš2)2(2/âˆš3)2((1/2)2 + 4(1)2 – (2)2)

= (8/3) Ã— (1/4) + 4 – 4

= (8/3) Ã— (1/4)

= 2/3

### Question 11. cosec330Â°cos60Â°tan345Â°sin290Â°sec245Â°cot30Â°

Solution:

Given: cosec330Â°cos60Â°tan345Â°sin290Â°sec245Â°cot30Â°

= (2)3(1/2)(1)3(1)2(âˆš2)2(âˆš3)

= (8)(1/2)(2)(âˆš3)

= 8âˆš3

### Question 12. cot230Â° – 2cos260Â° – (3/4)sec245Â° – 4sec230Â°

Solution:

Given: cot230Â° – 2cos260Â° – (3/4)sec245Â° – 4sec230Â°

= (âˆš3)2 – 2(1/2)2 – (3/4)(âˆš2)2 – 4(2/âˆš3)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. (cos0Â° + sin45Â° + sin30Â°)(sin90Â° – cos45Â° + cos60Â°)

Solution:

Given: (cos0Â° + sin45Â° + sin30Â°)(sin90Â° – cos45Â° + cos60Â°)

= (1 + 1/âˆš2 + 1/2)(1 – 1/âˆš2 + 1/2)

= (3/2 + 1/âˆš2)(3/2 – 1/âˆš2)

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

= (3/2)2 – (1/âˆš2)

= 9/4 – 1/2

= (9 – 2)/4

= 7/4

Previous
Next