Ex-5.2 Trigonometric Ratios (Part - 1), Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10 PDF Download

Evaluate each of the following:

 Q 1 . sin 45° sin 30°  + cos 45° cos 30°

Solution:

Sin 45°sin 30°  + cos 45° cos 30°                                                                                            [1]

We know that by trigonometric ratios we have ,

sin45° = 1/√2                      sin30° = 1/2

cos45° = 1/√2                     cos30° = √3/2

Substituting the values in equation 1 , we get

Q 2 . sin 60° cos 30°  + cos 60° sin 30°

Solution:

sin 60° cos 30°  + cos 60° sin 30°                                                                                             [1]

By trigonometric ratios we have ,

sin60° = √3/2                      sin30° = 1/2

cos30° = √3/2                     cos60° = 1/2

Substituting the values in equation 1 , we get

Q 3 . cos 60° cos 45°  – sin 60° sin 45°

Solution:

cos 60° cos 45°  – sin 60° sin 45°                                                                                              [1]

We know that by trigonometric ratios we have ,

cos60° = 1/2                       cos45° = 1/√2

sin60° = √3/2                      sin45° = 1/√2

Substituting the values in equation 1 , we get

Q.4: sin²30° + sin²45° + sin²60° + sin²90°

Solution:

sin²30° + sin²45° + sin²60° + sin²90°                              [1]

We know that by trigonometric ratios we have ,

sin30° = 1/2       sin45° = 1/√2

sin60° = √3/2    sin90° = 1

Substituting the values in equation 1 , we get

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

= 5/2

Q 5. cos²30° + cos²45° + cos²60° + cos²90°

Solution:

cos²30° + cos²45° + cos²60° + cos²90°              [1]

We know that by trigonometric ratios we have ,

cos30° = √3/2                     cos45° = 1/√2

cos60° = 1/2                      cos90° = 0

Substituting the values in equation 1 , we get

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

= 3/2

Q 6 . tan²30° + tan²45° + tan²60°

Solution:

tan²30° + tan²45° + tan²60°                                               [1]

We know that by trigonometric ratios we have ,

tan30° = 1/√3                             tan60° = √3

tan45° = 1

Substituting the values in equation 1 , we get

= 1/3 + 3 + 1

= 13/3

Q 7 . 2sin²30° − 3cos²45° + tan²60°

Solution:

2sin²30° − 3cos²45° + tan²60°                    [1]

We know that by trigonometric ratios we have ,

sin30° = 1/2                                cos45° = 1/√2

tan60° = √3

Substituting the values in equation 1 , we get

= 2

Q8:sin²30°cos²45° + 4tan230° + 1/2sin²90° − 2cos²90° + 1/24cos²0° 

Solution:

sin²30°cos²45° + 4tan²30° + 1/2sin²90° − 2cos²90° + 1/24cos²0°                                          [1]

We know that by trigonometric ratios we have ,

sin30° = 1/2

cos45° = 1/√2

tan30° = 1/√3

sin90° = 1

cos90° = 0

cos0° = 1

Substituting the values in equation 1 , we get

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

= 48/24 = 2

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

Solution:

4(sin⁴60° + cos⁴30°) − 3(tan²60° − tan²45°) + 5cos²45°                                       [1]

We know that by trigonometric ratios we have ,

sin60° = √3/2                      cos45° = 1/√2

tan60° = √3                        cos30° = √3/2

Substituting the values in equation 1 , we get

= 14/2 − 6 = 7 – 6 = 1

Q 10 . (cosec²45°sec²30°)(sin²30° + 4cot²45° − sec²60°)

Solution:

(cosec²45°sec²30°)(sin²30° + 4cot²45° − sec²60°)            [1]

We know that by trigonometric ratios we have ,

cosec45° = √2                            sec30° = 2/√3

sin30° = 1/2                    cot45° = 1

sec60° = 2

Substituting the values in equation 1 , we get

= 3⋅4/3⋅1/4

= 2/3

Q11. cosec³30°cos60°tan³45°sin²90°sec²45°cot30°

Solution:

Given,

= cosec³30°cos60°tan³45°sin²90°sec²45°cot30°

= 2³(1/2)(1³)(1²)(√2²)(√3)

= (2)³×(1/2)×(1³)×(1²)×(√2²)×(√3)

= 8×(1/2)×(1)×(1)×(2)×(√3)

= 8√3

Q12. cot²30° − 2cos²60° − 3/4sec²45° – 4sec²30°

Solution:

Given,

= cot²30° − 2cos²60° − 34sec²45° – 4sec²30°

= 3 − 1/2 − 3/2 − 16/3

= − 13/3

Q13. (cos0° + sin45° + sin30°)(sin90° − cos45° + cos60°)

Solution:

Given,

(cos0° + sin45° + sin30°)(sin90° − cos45° + cos60°)

Q14. 

Solution:

Given,

Q15. 

Solution:

Given,

Q16. 4(sin⁴30° + cos²60°) − 3(cos²45° − sin²90°) − sin²60°

Solution:

Given,

4(sin⁴30° + cos²60°) − 3(cos²45° − sin²90°) − sin²60°

Q17.

Solution:

Given,

= 3 + 2 + 4 = 9

Q18. 

Solution:

Given,

Q19. 

Solution:

Given,

= 5/2 – 5/2 = 0

Q20. 2sin3x = √3

Solution:

Given,

2sin3x = √3

⇒ sin3x = √3/2

⇒ sin3x = sin60°

⇒ 3x = 60°

⇒ x = 20°

Q21) 2sinx/2 = 1,x = ?

Solution:

sinx/2 = 1/2

sinx/2 = sin30°

x/2 = 30°

x = 60°

Q22) √3sinx = cosx

Solution:

√3tanx = 1

tanx = 1/√3

∴tanx = tan45°

x = 45°

Q23) Tan x = sin 45° cos 45°  + sin 30°

Solution:

Tanx = 1/√2.12√ + 1/2       [∵sin45° = 1/√2cos45° = 1/√2sin30° = 1/2]

Tanx = 1/2 + 1/2

Tan x = 1

Tan x = 45°

x = 45°

Q24) √3Tan2x = cos60° + sin45°cos45°

Solution:

√3Tan2x = 1/2 +1/√2.1/√2   [∵cos60° = 12sin45° = cos45° = 1/√2]

√3Tan2x = 1√3  ⇒ tan2x = tan30°

2x = 30°

x = 15⁰

Q25) cos2x = cos60°cos30° + sin60°sin30°

Solution:

cos2x = 1/2.3√2 + √3/2.1/2    [∵cos60° = sin30° = 1/2sin60° = cos30° = √3/2]

cos2x = 2.3√4

cos2x = √3/2

cos2x = cos30°

2x = 30°

x = 150