# 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 30Â° = 1/2, tan 30Â° = 1/âˆš3, tan 60Â° = âˆš3, sin 90Â° = cos 0Â° = 1 in eq(1)

= 3/2

### Question 15.

Solution:

Given:            -(1)

= tan230Â° + cosec260Â° âˆ’ cos245Â°

Putting the values of cosec60Â° = 2/âˆš3, cos45Â° = 1/âˆš2, tan30Â° = 1/âˆš3 in eq(1)

= 4/3 + 4/3 – 1/2

= (4 + 4)/3 – 1/2

= 8/3 – 1/2

= 13/6

### Question 16. 4(sin430Â° + cos260Â°) – 3(cos245Â° – sin290Â°) – sin260Â°

Solution:

Given: 4(sin430Â° + cos260Â°) – 3(cos245Â° – sin290Â°) – sin260Â°       -(1)

Putting the values of cos45Â° = 1/âˆš2, sin30Â° = cos30Â° = 1/2, sin60Â° = âˆš3/2, sin90Â° = 1 in eq(1)

= 4((1/2)4 + (1/2)2) – 3((1/âˆš2)2 – (1)2) – (âˆš3/2)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

cosecÎ¸ = 1/sinÎ¸, cotÎ¸ = 1/tanÎ¸

Now,

= tan45Â°sin30Â° + sec60Â°tan45Â° –

= 1 – 1/2 + 2 – 5/2

= (1 – 5)/2 + 2

= -2 + 2

= 0

### Question 20. 2sin3x = âˆš3

Solution:

Given,

2sin3x = âˆš3

sin3x = âˆš3/2

As we know that,

sin60Â° = âˆš3/2

So, sin3x = sin60Â°

Now,

3x = 60Â°

x = 60Â°/3

x = 20Â°

### Question 21. 2sin x/2

Solution:

Given

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

We know sin30Â° = 1/2

So,

sinx/2 = sin30Â°

x/2 = 30Â°

x = 60Â°

### Question 22. âˆš3sinx = cosx

Solution:

Given,

âˆš3sinx = cosx or sinx/cosx = 1/âˆš3

tan x = 1/âˆš3

We know tan30Â° = 1/âˆš3

So,

tan x = tan30Â°

x = 30Â°

### Question 23. tan x = sin45Â°cos45Â° + sin30Â°

Solution:

Given: tan x = sin45Â°cos45Â° + sin30Â°

Putting the values of sin 45Â° = cos 45Â° = 1/âˆš2, sin 30Â° = 1/2

tan x = ((1/âˆš2 Ã— 1/âˆš2) + 1/2)

tan x = (1/âˆš2)2 + 1/2

tan x = 1/2 + 1/2

tan x = 1

tan x = tan45Â°

So, x = 45Â°

### Question 24. âˆš3tan 2x = cos60Â° + sin45Â°cos45Â°

Solution:

Putting the values of sin 45Â° = cos 45Â° = 1/âˆš2, cos 60Â° = 1/2 in the given equation

âˆš3tan 2x = 1/2 + 1/âˆš2 Ã— 1/âˆš2

âˆš3tan 2x = 1/2 + (1/âˆš2)2

âˆš3tan 2x = 1/2 + 1/2

tan 2x = 1/âˆš3

tan 2x = tan30Â°

So, x = 15Â°

### Question 25. cos 2x = cos60Â°cos30Â° + sin60Â°sin30Â°

Solution:

Given: cos 2x = cos60Â°cos30Â° + sin60Â°sin30Â°

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

cos 2x = 1/2 Ã— âˆš3/2 + âˆš3/2 Ã— 1/2

cos 2x = âˆš3/4 + âˆš3/4

cos 2x = âˆš3/2

cos 2x = cos30Â°

So, x = 15Â°

### (i) tan2Î¸ =

Solution:

Putting the value of given Î¸ in the above equation

We get

âˆš3 =

âˆš3 =

âˆš3 = âˆš3

Hence Proved

### (ii) tan2Î¸ =

Solution:

Putting the value of given Î¸ in the above equation

We get

âˆš3 =

âˆš3 =

âˆš3 = âˆš3

Hence Proved

### (iii) cos2Î¸ =

Solution:

Putting the value of given Î¸ in the above equation

We get

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

1/2 = 2/4 or 1/2

Hence Proved

### (iv) cos3Î¸ = 4cos3Î¸ âˆ’ 3cosÎ¸

Solution:

Putting the value of given Î¸ in the above equation

We get

cos3(30Â°) = 4cos3 30Â° âˆ’ 3cos 30Â°

cos 90Â° = 4cos330Â° âˆ’ 3cos 30Â°

0 = 4(3âˆš3/8) – 4(3âˆš3/8)

0 = 3(âˆš3/2) – 3(âˆš3/2)

0 = 0

Hence Proved

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