# Class 10 RD Sharma Solutions – Chapter 5 Trigonometric Ratios – Exercise 5.2 | Set 2

• Last Updated : 21 Feb, 2021

### Question 14.

Solution:

Given:           -(1)

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

= 3/2

### Question 15.

Solution:

Given:            -(1)

= tan230° + cosec260° − cos245°

Putting the values of cosec60° = 2/√3, cos45° = 1/√2, tan30° = 1/√3 in eq(1)

= 4/3 + 4/3 – 1/2

= (4 + 4)/3 – 1/2

= 8/3 – 1/2

= 13/6

### Question 16. 4(sin430° + cos260°) – 3(cos245° – sin290°) – sin260°

Solution:

Given: 4(sin430° + cos260°) – 3(cos245° – sin290°) – sin260°       -(1)

Putting the values of cos45° = 1/√2, sin30° = cos30° = 1/2, sin60° = √3/2, sin90° = 1 in eq(1)

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

= 4((1/16) + (1/4)) – 3((1/2) – 1) – (3/4)

= 4(5/16) – 3(-1/2) – 3/4

= 5/4 + 3/2 – 3/4

= (5 + 6 – 3)/4

= 8/4

= 2

### Question 17.

Solution:

Given:

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

= 3 + 2 + 4

= 9

Solution:

Given:

### Question 19.

Solution:

Given:

As we know that

cosecθ = 1/sinθ, cotθ = 1/tanθ

Now,

= tan45°sin30° + sec60°tan45° –

= 1 – 1/2 + 2 – 5/2

= (1 – 5)/2 + 2

= -2 + 2

= 0

### Question 20. 2sin3x = √3

Solution:

Given,

2sin3x = √3

sin3x = √3/2

As we know that,

sin60° = √3/2

So, sin3x = sin60°

Now,

3x = 60°

x = 60°/3

x = 20°

### Question 21. 2sin x/2

Solution:

Given

2sinx/2 = 1 or sin x/2 = 1/2

We know sin30° = 1/2

So,

sinx/2 = sin30°

x/2 = 30°

x = 60°

### Question 22. √3sinx = cosx

Solution:

Given,

√3sinx = cosx or sinx/cosx = 1/√3

tan x = 1/√3

We know tan30° = 1/√3

So,

tan x = tan30°

x = 30°

### Question 23. tan x = sin45°cos45° + sin30°

Solution:

Given: tan x = sin45°cos45° + sin30°

Putting the values of sin 45° = cos 45° = 1/√2, sin 30° = 1/2

tan x = ((1/√2 × 1/√2) + 1/2)

tan x = (1/√2)2 + 1/2

tan x = 1/2 + 1/2

tan x = 1

tan x = tan45°

So, x = 45°

### Question 24. √3tan 2x = cos60° + sin45°cos45°

Solution:

Putting the values of sin 45° = cos 45° = 1/√2, cos 60° = 1/2 in the given equation

√3tan 2x = 1/2 + 1/√2 × 1/√2

√3tan 2x = 1/2 + (1/√2)2

√3tan 2x = 1/2 + 1/2

tan 2x = 1/√3

tan 2x = tan30°

So, x = 15°

### Question 25. cos 2x = cos60°cos30° + sin60°sin30°

Solution:

Given: cos 2x = cos60°cos30° + sin60°sin30°

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

cos 2x = 1/2 × √3/2 + √3/2 × 1/2

cos 2x = √3/4 + √3/4

cos 2x = √3/2

cos 2x = cos30°

So, x = 15°

### (i) tan2θ =

Solution:

Putting the value of given θ in the above equation

We get

√3 =

√3 =

√3 = √3

Hence Proved

### (ii) tan2θ =

Solution:

Putting the value of given θ in the above equation

We get

√3 =

√3 =

√3 = √3

Hence Proved

### (iii) cos2θ =

Solution:

Putting the value of given θ in the above equation

We get

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

1/2 = 2/4 or 1/2

Hence Proved

### (iv) cos3θ = 4cos3θ − 3cosθ

Solution:

Putting the value of given θ in the above equation

We get

cos3(30°) = 4cos3 30° − 3cos 30°

cos 90° = 4cos330° − 3cos 30°

0 = 4(3√3/8) – 4(3√3/8)

0 = 3(√3/2) – 3(√3/2)

0 = 0

Hence Proved

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