# RD Sharma Solutions for Class 11 Maths Chapter 5 Trigonometric Functions

## RD Sharma Solutions Class 11 Maths Chapter 5 – Free PDF Download

RD Sharma Solutions for Class 11 Maths Chapter 5Â – Trigonometric Functions is provided here for students to study and score good marks in their board exams. In earlier classes, we have studied trigonometric ratios for acute angles as the ratio of the sides of a right-angled triangle. In this chapter, we will extend the definitions of trigonometric ratios to any angle in terms of radian measure and study them as trigonometric functions. Our RD Sharma Solution module utilizes various shortcut tips and practical examples to explain all the exercise questions in a simple and easily understandable language. Experts at BYJUâ€™S have designed RD Sharma Class 11 Solutions in a very lucid and clear manner that helps students solve problems in the most efficient possible ways. Pdfs can be downloaded easily from the links provided below.

RD Sharma Solutions for Class 11 Maths Chapter 5– Trigonometric Functions contains three exercises and the RD Sharma Solutions present in this page provide solutions to the questions present in each exercise. Now, let us have a look at the concepts discussed in this chapter.

• Trigonometric functions of a real number.
• Values of trigonometric functions.
• Trigonometric identities.
• Fundamental trigonometric identities.
• Signs of trigonometric functions.
• Variations in values of trigonometric functions in different quadrants.
• Values of trigonometric functions at allied angles.
• Periodic functions.
• Even and odd functions.

## Download the Pdf of RD Sharma Solutions for Class 11 Maths Chapter 5 – Trigonometric Functions

Exercise 5.1

Exercise 5.2

Exercise 5.3

### Access answers to RD Sharma Solutions for Class 11 Maths Chapter 5 – Trigonometric Functions

EXERCISE 5.1 PAGE NO: 5.18

Prove the following identities:

1. sec4 x â€“ sec2 x = tan4 x + tan2 x

Solution:

Let us consider LHS: sec4 x â€“ sec2 x

(sec2 x)2Â â€“ sec2 x

By using the formula, sec2Â Î¸ = 1 + tan2Â Î¸.

(1 + tan2 x)Â 2Â â€“ (1 + tan2 x)

1 + 2tan2 x + tan4 x â€“ 1 – tan2 x

tan4 x + tan2 x

= RHS

âˆ´ LHS = RHS

Hence proved.

2. sin6 x + cos6 x = 1 â€“ 3 sin2 x cos2 x

Solution:

Let us consider LHS: sin6 x + cos6 x

(sin2 x) 3Â + (cos2 x) 3

By using the formula, a3Â + b3Â = (a + b) (a2Â + b2Â â€“ ab)

(sin2 x + cos2 x) [(sin2 x) 2Â + (cos2 x) 2Â â€“ sin2 x cos2 x]

By using the formula, sin2 x + cos2 x = 1 and a2Â + b2Â = (a + b) 2Â â€“ 2ab

1 Ã— [(sin2 x + cos2 x) 2Â â€“ 2sin2 x cos2 x â€“ sin2 x cos2 x

12Â – 3sin2 x cos2 x

1 – 3sin2 x cos2 x

= RHS

âˆ´ LHS = RHS

Hence proved.

3. (cosec x â€“ sin x) (sec x â€“ cos x) (tan x + cot x) = 1

Solution:

Let us consider LHS: (cosec x â€“ sin x) (sec x â€“ cos x) (tan x + cot x)

By using the formulas

cosec Î¸ = 1/sin Î¸;

sec Î¸ = 1/cos Î¸;

tan Î¸ = sin Î¸ / cos Î¸;

cot Î¸ = cos Î¸ / sin Î¸

Now,

1 = RHS

âˆ´ LHS = RHS

Hence proved.

4. cosec x (sec x â€“ 1) â€“ cot x (1 â€“ cos x) = tan x â€“ sin x

Solution:

Let us consider LHS: cosec x (sec x â€“ 1) â€“ cot x (1 â€“ cos x)

By using the formulas

cosec Î¸ = 1/sin Î¸;

sec Î¸ = 1/cos Î¸;

tan Î¸ = sin Î¸ / cos Î¸;

cot Î¸ = cos Î¸ / sin Î¸

Now,

By using the formula, 1 â€“ cos2x = sin2x;

= RHS

âˆ´ LHS = RHS

Hence Proved.

5.

Solution:

Let us consider the LHS:

By using the formula,

cosec Î¸ = 1/sin Î¸;

sec Î¸ = 1/cos Î¸;

Now,

sin x

= RHS

âˆ´ LHS = RHS

Hence Proved.

6.

Solution:

Let us consider the LHS:

By using the formula,

tan Î¸ = sin Î¸ / cos Î¸;

cot Î¸ = cos Î¸ / sin Î¸

Now,

By using the formula, a3Â – b3Â = (a – b) (a2Â + b2Â + ab)

By using the formula,

cosec Î¸ = 1/sin Î¸,

sec Î¸ = 1/cos Î¸;

cosec x Ã— sec x + 1

sec x cosec x + 1

=RHS

âˆ´ LHS = RHS

Hence Proved.

7.

Solution:

Let us consider LHS:

By using the formula a3Â Â± b3Â = (a Â± b) (a2Â + b2âˆ“Â ab)

We know, sin2x + cos2x = 1.

1 – sinxÂ cosxÂ + 1 + sinx cosx

2

= RHS

âˆ´ LHS = RHS

Hence Proved.

8. (sec x sec y + tan x tan y)2Â â€“ (sec x tan y + tan x sec y)2Â = 1

Solution:

Let us consider LHS:

(sec x sec y + tan x tan y)2Â â€“ (sec x tan y + tan x sec y)2

Expanding the above equation we get,

[(sec x sec y)2Â + (tan x tan y)2Â + 2 (sec x sec y) (tan x tan y)] â€“ [(sec x tan y)2Â + (tan x sec y)2Â + 2 (sec x tan y) (tan x sec y)] [sec2 x sec2Â y + tan2 x tan2Â y + 2 (sec x sec y) (tan x tan y)] â€“ [sec2 x tan2Â y + tan2 x sec2Â y + 2 (sec2 x tan2Â y) (tan x sec y)]

sec2 x sec2Â y – sec2 x tan2Â y + tan2 x tan2Â y – tan2 x sec2Â y

sec2 x (sec2Â y – tan2Â y) + tan2 x (tan2Â y – sec2Â y)

sec2 x (sec2Â y – tan2Â y) – tan2 x (sec2Â y – tan2Â y)

We know, sec2 x â€“ tan2 x = 1.

sec2 x Ã— 1 â€“ tan2 x Ã— 1

sec2 x â€“ tan2 x

1 = RHS

âˆ´ LHS = RHS

Hence proved.

9.

Solution:

Let us Consider RHS:

= LHS

âˆ´ LHS = RHS

Hence Proved.

10.

Solution:

Let us consider LHS:

By using the formulas,

1 + tan2x = sec2x and 1 + cot2x = cosec2x

= RHS

âˆ´ LHS = RHS

Hence Proved.

11.

Solution:

Let us consider LHS:

By using the formula,

tan Î¸ = sin Î¸ / cos Î¸;

cot Î¸ = cos Î¸ / sin Î¸

Now,

By using the formula, a3Â + b3Â = (a + b) (a2Â + b2–Â ab)

We know, sin2 x + cos2 x = 1.

1 â€“ 1 + sin x cos x

Sin x cos x

= RHS

âˆ´ LHS = RHS

Hence proved.

12.

Solution:

Let us consider LHS:

By using the formula,

cosec Î¸ = 1/sin Î¸,

sec Î¸ = 1/cos Î¸;

= RHS

âˆ´ LHS = RHS

Hence proved.

13. (1 + tan Î± tan Î²) 2Â + (tan Î± â€“ tan Î²) 2Â = sec2Â Î± sec2Â Î²

Solution:

Let us consider LHS: (1 + tan Î± tan Î²) 2Â + (tan Î± â€“ tan Î²) 2

1+ tan2Â Î± tan2Â Î² + 2 tan Î± tan Î² + tan2Â Î± + tan2Â Î² â€“ 2 tan Î± tan Î²

1 + tan2Â Î± tan2Â Î² + tan2Â Î± + tan2Â Î²

tan2Â Î± (tan2Â Î² + 1) + 1 (1 + tan2Â Î²)

(1 + tan2Â Î²) (1 + tan2Â Î±)

We know, 1 + tan2Â Î¸ = sec2Â Î¸

So,

sec2Â Î± sec2Â Î²

= RHS

âˆ´ LHS = RHS

Hence proved.

EXERCISE 5.2 PAGE NO: 5.25

1. Find the values of the other five trigonometric functions in each of the following:

(i) cot x = 12/5, x in quadrant III

(ii) cos x = -1/2, x in quadrant II

(iii) tan x = 3/4, x in quadrant III

(iv) sin x = 3/5, x in quadrant I

Solution:

(i) cot x = 12/5, x in quadrant III

In third quadrant, tan x and cot x are positive. sin x, cos x, sec x, cosec x are negative.

By using the formulas,

tan x = 1/cot x

= 1/(12/5)

= 5/12

cosec x = –âˆš(1 + cot2 x)

= –âˆš(1 + (12/5)2)

= –âˆš(25+144)/25

= –âˆš(169/25)

= -13/5

sin x = 1/cosec x

= 1/(-13/5)

= -5/13

cos x = – âˆš(1 – sin2 x)

= – âˆš(1 â€“ (-5/13)2)

= – âˆš(169-25)/169

= – âˆš(144/169)

= -12/13

sec x = 1/cos x

= 1/(-12/13)

= -13/12

âˆ´ sin x = -5/13, cos x = -12/13, tan x = 5/12, cosec x = -13/5, sec x = -13/12

(ii) cos x = -1/2, x in quadrant II

In second quadrant, sin x and cosec x are positive. tan x, cot x, cos x, sec x are negative.

By using the formulas,

sin x = âˆš(1 â€“ cos2 x)

= âˆš(1 â€“ (-1/2)2)

= âˆš(4-1)/4

= âˆš(3/4)

= âˆš3/2

tan x = sin x/cos x

= (âˆš3/2)/(-1/2)

= -âˆš3

cot x = 1/tan x

= 1/-âˆš3

= -1/âˆš3

cosec x = 1/sin x

= 1/(âˆš3/2)

= 2/âˆš3

sec x = 1/cos x

= 1/(-1/2)

= -2

âˆ´ sin x = âˆš3/2, tan x = -âˆš3, cosec x = 2/âˆš3, cot x = -1/âˆš3 sec x = -2

(iii) tan x = 3/4, x in quadrant III

In third quadrant, tan x and cot x are positive. sin x, cos x, sec x, cosec x are negative.

By using the formulas,

sin x = âˆš(1 â€“ cos2 x)

= – âˆš(1-(-4/5)2)

= – âˆš(25-16)/25

= – âˆš(9/25)

= – 3/5

cos x = 1/sec x

= 1/(-5/4)

= -4/5

cot x = 1/tan x

= 1/(3/4)

= 4/3

cosec x = 1/sin x

= 1/(-3/5)

= -5/3

sec x = -âˆš(1 + tan2 x)

= – âˆš(1+(3/4)2)

= – âˆš(16+9)/16

= – âˆš (25/16)

= -5/4

âˆ´ sin x = -3/5, cos x = -4/5, cosec x = -5/3, sec x = -5/4, cot x = 4/3

(iv) sin x = 3/5, x in quadrant I

In first quadrant, all trigonometric ratios are positive.

So, by using the formulas,

tan x = sin x/cos x

= (3/5)/(4/5)

= 3/4

cosec x = 1/sin x

= 1/(3/5)

= 5/3

cos x = âˆš(1-sin2 x)

= âˆš(1 â€“ (-3/5)2)

= âˆš(25-9)/25

= âˆš(16/25)

= 4/5

sec x = 1/cos x

= 1/(4/5)

= 5/4

cot x = 1/tan x

= 1/(3/4)

= 4/3

âˆ´ cos x = 4/5, tan x = 3/4, cosec x = 5/3, sec x = 5/4, cot x = 4/3

2. If sin x = 12/13Â and lies in the second quadrant, find the value of sec x + tan x.

Solution:

Given:

Sin x = 12/13 and x lies in the second quadrant.

We know, in second quadrant, sin x and cosec x are positive and all other ratios are negative.

By using the formulas,

Cos x = âˆš(1-sin2 x)

= – âˆš(1-(12/13)2)

= – âˆš(1- (144/169))

= – âˆš(169-144)/169

= -âˆš(25/169)

= – 5/13

We know,

tan x = sin x/cos x

sec x = 1/cos x

Now,

tan x = (12/13)/(-5/13)

= -12/5

sec x = 1/(-5/13)

= -13/5

Sec x + tan x = -13/5 + (-12/5)

= (-13-12)/5

= -25/5

= -5

âˆ´ Sec x + tan x = -5

3. If sin x = 3/5,Â tanÂ y = 1/2Â and Ï€/2 < x< Ï€< y< 3Ï€/2Â find the value of 8 tan x -âˆš5 sec y.

Solution:

Given:

sin x = 3/5, tan y = 1/2 andÂ Ï€/2 < x< Ï€< y< 3Ï€/2

We know that, x is in second quadrant and y is in third quadrant.

In second quadrant, cos x and tan x are negative.

In third quadrant, sec y is negative.

By using the formula,

cos x = – âˆš(1-sin2 x)

tan x = sin x/cos x

Now,

cos x = – âˆš(1-sin2 x)

= – âˆš(1 â€“ (3/5)2)

= – âˆš(1 â€“ 9/25)

= – âˆš((25-9)/25)

= – âˆš(16/25)

= – 4/5

tan x = sin x/cos x

= (3/5)/(-4/5)

= 3/5 Ã— -5/4

= -3/4

We know that sec y = – âˆš(1+tan2 y)

= – âˆš(1 + (1/2)2)

= – âˆš(1 + 1/4)

= – âˆš((4+1)/4)

= – âˆš(5/4)

= – âˆš5/2

Now, 8 tan x – âˆš5 sec y = 8(-3/4) – âˆš5(-âˆš5/2)

= -6 + 5/2

= (-12+5)/2

= -7/2

âˆ´ 8 tan x – âˆš5 sec y = -7/2

4. If sin x + cos x = 0 and x lies in the fourth quadrant, find sin x and cos x.

Solution:

Given:

Sin x + cos x = 0 and x lies in fourth quadrant.

Sin xÂ = -cos x

Sin x/cos x = -1

So, tan x = -1 (since, tan x = sin x/cos x)

We know that, in fourth quadrant, cos x and sec x are positive and all other ratios are negative.

By using the formulas,

Sec x = âˆš(1 + tan2 x)

Cos x = 1/sec x

Sin x = – âˆš(1- cos2 x)

Now,

Sec x = âˆš(1 + tan2 x)

= âˆš(1 + (-1)2)

= âˆš2

Cos x = 1/sec x

= 1/âˆš2

Sin x = – âˆš(1 â€“ cos2 x)

= – âˆš(1 â€“ (1/âˆš2)2)

= – âˆš(1 â€“ (1/2))

= – âˆš((2-1)/2)

= – âˆš(1/2)

= -1/âˆš2

âˆ´ sin x = -1/âˆš2 and cos x = 1/âˆš2

5. If cos x = -3/5 and Ï€<x<3Ï€/2 find the values of other five trigonometric functions and hence evaluateÂ

Solution:

Given:

cos x= -3/5 and Ï€ <x < 3Ï€/2

We know that in the third quadrant, tan x and cot x are positive and all other rations are negative.

By using the formulas,

Sin x = – âˆš(1-cos2 x)

Tan x = sin x/cos x

Cot x = 1/tan x

Sec x = 1/cos x

Cosec x = 1/sin x

Now,

Sin x = – âˆš(1-cos2 x)

= – âˆš(1-(-3/5)2)

= – âˆš(1-9/25)

= – âˆš((25-9)/25)

= – âˆš(16/25)

= – 4/5

Tan x = sin x/cos x

= (-4/5)/(-3/5)

= -4/5 Ã— -5/3

= 4/3

Cot x = 1/tan x

= 1/(4/3)

= 3/4

Sec x = 1/cos x

= 1/(-3/5)

= -5/3

Cosec x = 1/sin x

= 1/(-4/5)

= -5/4

âˆ´
= [(-5/4) + (3/4)] / [(-5/3) â€“ (4/3)]

= [(-5+3)/4] / [(-5-4)/3]

= [-2/4] / [-9/3]

= [-1/2] / [-3]

= 1/6

EXERCISE 5.3 PAGE NO: 5.39

1. Find the values of the following trigonometric ratios:

(i) sin 5Ï€/3

(ii) sin 17Ï€

(iii) tan 11Ï€/6

(iv) cos (-25Ï€/4)

(v) tan 7Ï€/4

(vi) sin 17Ï€/6

(vii) cos 19Ï€/6

(viii) sin (-11Ï€/6)

(ix) cosec (-20Ï€/3)

(x) tan (-13Ï€/4)

(xi) cos 19Ï€/4

(xii) sin 41Ï€/4

(xiii) cos 39Ï€/4

(xiv) sin 151Ï€/6

Solution:

(i) sin 5Ï€/3

5Ï€/3 = (5Ï€/3 Ã— 180)o

= 300o

= (90Ã—3 + 30)o

Since, 300o lies in IV quadrant in which sine function is negative.

sin 5Ï€/3 = sin (300)o

= sin (90Ã—3 + 30)o

= – cos 30o

= – âˆš3/2

(ii) sin 17Ï€

Sin 17Ï€ = sin 3060o

= sin (90Ã—34 + 0)o

Since, 3060o lies in the negative direction of x-axis i.e., on boundary line of II and III quadrants.

Sin 17Ï€ = sin (90Ã—34 + 0)o

= – sin 0o

= 0

(iii) tan 11Ï€/6

tan 11Ï€/6 = (11/6 Ã— 180)o

= 330o

Since, 330o lies in the IV quadrant in which tangent function is negative.

tan 11Ï€/6 = tan (300)o

= tan (90Ã—3 + 60)o

= – cot 60o

= – 1/âˆš3

(iv) cos (-25Ï€/4)

cos (-25Ï€/4) = cos (-1125)o

= cos (1125)o

Since, 1125o lies in the I quadrant in which cosine function is positive.

cos (1125)o = cos (90Ã—12 + 45)o

= cos 45o

= 1/âˆš2

(v) tan 7Ï€/4

tan 7Ï€/4 = tan 315o

= tan (90Ã—3 + 45)o

Since, 315o lies in the IV quadrant in which tangent function is negative.

tan 315o = tan (90Ã—3 + 45)o

= – cot 45o

= -1

(vi) sin 17Ï€/6

sin 17Ï€/6 = sin 510o

= sin (90Ã—5 + 60)o

Since, 510o lies in the II quadrant in which sine function is positive.

sin 510o = sin (90Ã—5 + 60)o

= cos 60o

= 1/2

(vii) cos 19Ï€/6

cos 19Ï€/6 = cos 570o

= cos (90Ã—6 + 30)o

Since, 570o lies in III quadrant in which cosine function is negative.

cos 570o = cos (90Ã—6 + 30)o

= – cos 30o

= – âˆš3/2

(viii) sin (-11Ï€/6)

sin (-11Ï€/6) = sin (-330o)

= – sin (90Ã—3 + 60)o

Since, 330o lies in the IV quadrant in which the sine function is negative.

sin (-330o) = – sin (90Ã—3 + 60)o

= – (-cos 60o)

= – (-1/2)

= 1/2

(ix) cosec (-20Ï€/3)

cosec (-20Ï€/3) = cosec (-1200)o

= – cosec (1200)o

= – cosec (90Ã—13 + 30)o

Since, 1200o lies in the II quadrant in which cosec function is positive.

cosec (-1200)o = – cosec (90Ã—13 + 30)o

= – sec 30o

= -2/âˆš3

(x) tan (-13Ï€/4)

tan (-13Ï€/4) = tan (-585)o

= – tan (90Ã—6 + 45)o

Since, 585o lies in the III quadrant in which the tangent function is positive.

tan (-585)o = – tan (90Ã—6 + 45)o

= – tan 45o

= -1

(xi) cos 19Ï€/4

cos 19Ï€/4 = cos 855o

= cos (90Ã—9 + 45)o

Since, 855o lies in the II quadrant in which the cosine function is negative.

cos 855o = cos (90Ã—9 + 45)o

= – sin 45o

= – 1/âˆš2

(xii) sin 41Ï€/4

sin 41Ï€/4 = sin 1845o

= sin (90Ã—20 + 45)o

Since, 1845o lies in the I quadrant in which the sine function is positive.

sin 1845o = sin (90Ã—20 + 45)o

= sin 45o

= 1/âˆš2

(xiii) cos 39Ï€/4

cos 39Ï€/4 = cos 1755o

= cos (90Ã—19 + 45)o

Since, 1755o lies in the IV quadrant in which the cosine function is positive.

cos 1755o = cos (90Ã—19 + 45)o

= sin 45o

= 1/âˆš2

(xiv) sin 151Ï€/6

sin 151Ï€/6 = sin 4530o

= sin (90Ã—50 + 30)o

Since, 4530o lies in the III quadrant in which the sine function is negative.

sin 4530o = sin (90Ã—50 + 30)o

= – sin 30o

= -1/2

2. prove that:
(i) tan 225oÂ cot 405oÂ + tan 765oÂ cot 675oÂ = 0

(ii) sin 8Ï€/3 cos 23Ï€/6 + cos 13Ï€/3 sin 35Ï€/6 = 1/2

(iii) cos 24o + cos 55o + cos 125o + cos 204o + cos 300o = 1/2

(iv) tan (-125o) cot (-405o) â€“ tan (-765o) cot (675o) = 0

(v) cos 570o sin 510o + sin (-330o) cos (-390o) = 0

(vi) tan 11Ï€/3 â€“ 2 sin 4Ï€/6 â€“ 3/4 cosec2 Ï€/4 + 4 cos2 17Ï€/6 = (3 – 4âˆš3)/2

(vii) 3 sin Ï€/6 sec Ï€/3 â€“ 4 sin 5Ï€/6 cot Ï€/4 = 1

Solution:

(i) tan 225oÂ cot 405oÂ + tan 765oÂ cot 675oÂ = 0

Let us consider LHS:

tan 225Â° cot 405Â° + tan 765Â° cot 675Â°

tan (90Â° Ã— 2 + 45Â°) cot (90Â° Ã— 4 + 45Â°) + tan (90Â° Ã— 8 + 45Â°) cot (90Â° Ã— 7 + 45Â°)

We know that when n is odd, cotÂ â†’Â tan.

tan 45Â° cot 45Â° + tan 45Â° [-tan 45Â°]

tan 45Â° cot 45Â° – tan 45Â° tan 45Â°

1 Ã— 1 â€“ 1 Ã— 1

1 â€“ 1

0 = RHS

âˆ´ LHS = RHS

Hence proved.

(ii) sin 8Ï€/3 cos 23Ï€/6 + cos 13Ï€/3 sin 35Ï€/6 = 1/2

Let us consider LHS:

sin 8Ï€/3 cos 23Ï€/6 + cos 13Ï€/3 sin 35Ï€/6

sin 480Â° cos 690Â° + cos 780Â° sin 1050Â°

sin (90Â° Ã— 5 + 30Â°) cos (90Â° Ã— 7 + 60Â°) + cos (90Â° Ã— 8 + 60Â°) sin (90Â° Ã— 11 + 60Â°)

We know that when n is odd, sinÂ â†’Â cos and cosÂ â†’Â sin.

cos 30Â° sin 60Â° + cos 60Â° [-cos 60Â°]

âˆš3/2 Ã— âˆš3/2 â€“ 1/2 Ã— 1/2

3/4 – 1/4

2/4

1/2

= RHS

âˆ´ LHS = RHS

Hence proved.

(iii) cos 24o + cos 55o + cos 125o + cos 204o + cos 300o = 1/2

Let us consider LHS:

cos 24o + cos 55o + cos 125o + cos 204o + cos 300o

cos 24Â° + cos (90Â° Ã— 1 â€“ 35Â°) + cos (90Â° Ã— 1 + 35Â°) + cos (90Â° Ã— 2 + 24Â°) + cos (90Â° Ã— 3 + 30Â°)

We know that when n is odd, cosÂ â†’Â sin.

cos 24Â° + sin 35Â° – sin 35Â° – cos 24Â° + sin 30Â°

0 + 0 + 1/2

1/2

= RHS

âˆ´ LHS = RHS

Hence proved.

(iv) tan (-125o) cot (-405o) â€“ tan (-765o) cot (675o) = 0

Let us consider LHS:

tan (-125o) cot (-405o) â€“ tan (-765o) cot (675o)

We know that tan (-x) = -tan (x) and cot (-x) = -cot (x).

[-tan (225Â°)] [-cot (405Â°)] â€“ [-tan (765Â°)] cot (675Â°)

tan (225Â°) cot (405Â°) + tan (765Â°) cot (675Â°)

tan (90Â° Ã— 2 + 45Â°) cot (90Â° Ã— 4 + 45Â°) + tan (90Â° Ã— 8 + 45Â°) cot (90Â° Ã— 7 + 45Â°)

tan 45Â° cot 45Â° + tan 45Â° [-tan 45Â°]

1 Ã— 1 + 1 Ã— (-1)

1 â€“ 1

0

= RHS

âˆ´ LHS = RHS

Hence proved.

(v) cos 570o sin 510o + sin (-330o) cos (-390o) = 0

Let us consider LHS:

cos 570o sin 510o + sin (-330o) cos (-390o)

We know that sin (-x) = -sin (x) and cos (-x) = +cos (x).

cos 570oÂ sin 510oÂ + [-sin (330o)] cos (390o)

cos 570oÂ sin 510oÂ – sin (330o) cos (390o)

cos (90Â° Ã— 6 + 30Â°) sin (90Â° Ã— 5 + 60Â°) â€“ sin (90Â° Ã— 3 + 60Â°) cos (90Â° Ã— 4 + 30Â°)

We know that cos is negative at 90Â° + Î¸ i.e. in Q2Â and when n is odd, sinÂ â†’Â cos and cosÂ â†’Â sin.

-cos 30Â° cos 60Â° – [-cos 60Â°] cos 30Â°

-cos 30Â° cos 60Â° + cos 60Â° cos 30Â°

0

= RHS

âˆ´ LHS = RHS

Hence proved.

(vi) tan 11Ï€/3 â€“ 2 sin 4Ï€/6 â€“ 3/4 cosec2 Ï€/4 + 4 cos2 17Ï€/6 = (3 – 4âˆš3)/2

Let us consider LHS:

tan 11Ï€/3 â€“ 2 sin 4Ï€/6 â€“ 3/4 cosec2 Ï€/4 + 4 cos2 17Ï€/6

tan (11 Ã— 180o)/3 â€“ 2 sin (4 Ã— 180o)/6 â€“ 3/4 cosec2 180o/4 + 4 cos2 (17 Ã— 180o)/6

tan 660o â€“ 2 sin 120o – 3/4 (cosec 45o)2 + 4 (cos 510o)2

tan (90Â° Ã— 7 + 30Â°) â€“ 2 sin (90Â° Ã— 1 + 30Â°) â€“ 3/4 [cosec 45Â°]2Â + 4 [cos (90Â° Ã— 5 + 60Â°)]2

We know that tan and cos is negative at 90Â° + Î¸ i.e. in Q2Â and when n is odd, tanÂ â†’Â cot, sinÂ â†’Â cos and cosÂ â†’Â sin.

[-cot 30Â°] â€“ 2 cos 30Â° – 3/4 [cosec 45Â°]2Â + [-sin 60Â°]2

– cot 30Â° – 2 cos 30Â° – 3/4 [cosec 45Â°]2Â + [sin 60Â°]2

-âˆš3 – 2âˆš3/2 â€“ 3/4 (âˆš2)2 + 4 (âˆš3/2)2

-âˆš3 – âˆš3 â€“ 6/4 + 12/4

(3 – 4âˆš3)/2

= RHS

âˆ´ LHS = RHS

Hence proved.

(vii) 3 sin Ï€/6 sec Ï€/3 â€“ 4 sin 5Ï€/6 cot Ï€/4 = 1

Let us consider LHS:

3 sin Ï€/6 sec Ï€/3 â€“ 4 sin 5Ï€/6 cot Ï€/4

3 sin 180o/6 sec 180o/3 â€“ 4 sin 5(180o)/6 cot 180o/4

3 sin 30Â° sec 60Â° – 4 sin 150Â° cot 45Â°

3 sin 30Â° sec 60Â° – 4 sin (90Â° Ã— 1 + 60Â°) cot 45Â°

We know that when n is odd, sinÂ â†’Â cos.

3 sin 30Â° sec 60Â° – 4 cos 60Â° cot 45Â°

3 (1/2) (2) â€“ 4 (1/2) (1)

3 â€“ 2

1

= RHS

âˆ´ LHS = RHS

Hence proved.

3. Prove that:

(i)

(ii)

(iii)

(iv)

(v)

Solution:

(i)

1 = RHS

âˆ´ LHS = RHS

Hence proved.

(ii)

1 + 1

2 = RHS

âˆ´ LHS = RHS

Hence proved.

(iii)

1 = RHS

âˆ´ LHS = RHS

Hence proved.

(iv)

{1 + cot x â€“ (-cosec x)} {1 + cot x + (-cosec x)}

{1 + cot x + cosec x} {1 + cot x â€“ cosec x}

{(1 + cot x) + (cosec x)} {(1 + cot x) â€“ (cosec x)}

By using the formula, (a + b) (a â€“ b) = a2Â â€“ b2

(1 + cot x)2Â â€“ (cosec x)2

1 + cot2 x + 2 cot x â€“ cosec2 x

We know that 1 + cot2 x = cosec2 x

cosec2 x + 2 cot x â€“ cosec2 x

2 cot x = RHS

âˆ´ LHS = RHS

Hence proved.

(v)

1 = RHS

âˆ´ LHS = RHS

Hence proved.

4. Prove that: sin2 Ï€/18 + sin2 Ï€/9 + sin2 7Ï€/18 + sin2 4Ï€/9 = 2

Solution:

Let us consider LHS:

sin2 Ï€/18 + sin2 Ï€/9 + sin2 7Ï€/18 + sin2 4Ï€/9

sin2 Ï€/18 + sin2 2Ï€/18 + sin2 7Ï€/18 + sin2 8Ï€/18

sin2 Ï€/18 + sin2 2Ï€/18 + sin2 (Ï€/2 – 2Ï€/18) + sin2 (Ï€/2 â€“ Ï€/18)

We know that when n is odd, sinÂ â†’Â cos.

sin2 Ï€/18 + sin2 2Ï€/18 + cos2 2Ï€/18 + cos2 2Ï€/18

when rearranged,

sin2 Ï€/18 + cos2 2Ï€/18 + sin2 Ï€/18 + cos2 2Ï€/18

We know that sin2Â + cos2x = 1.

So,

1 + 1

2 = RHS

âˆ´ LHS = RHS

Hence proved.

## Frequently Asked Questions on RD Sharma Solutions for Class 11 Maths Chapter 5

### How RD Sharma Solutions for Class 11 Maths Chapter 5 helpful for board exams?

RD Sharma Solutions for Class 11 Maths Chapter 5 provides answers with detailed descriptions as per term limit particularised by the Board for self-evaluation. Solving these solutions will provide excellent practice for the students so they can finish the paper on time. So, itâ€™s clear that the RD Sharma Solutions for Class 11 Maths Chapter 5 are essential to score high in examinations. Students can get acquainted with writing exams and will be able to face exams more confidently.

### Mention the topics that are covered in RD Sharma Solutions for Class 11 Maths Chapter 5?

The topics that are covered in RD Sharma Solutions for Class 11 Maths Chapter 5 are
Trigonometric functions of a real number.
Values of trigonometric functions.
Trigonometric identities.
Fundamental trigonometric identities.
Signs of trigonometric functions.
Variations in values of trigonometric functions in different quadrants.
Values of trigonometric functions at allied angles.
Periodic functions.
Even and odd functions.
By learning these RD Sharma solutions thoroughly students will be able to solve complex problems easily.

### Is it necessary to practice all the exercises present in Chapter 5 of RD Sharma Solutions for Class 11 Maths?

Yes, it is compulsory to practice all the exercises present in Chapter 5 of RD Sharma Solutions for Class 11 Maths. Because all exercises contain numerous questions to solve which may come in their finals. This makes them more confident towards the syllabus they have. The expert faculty at BYJUâ€™S provides students with PDF of solutions to solve the problems according to the CBSE syllabus. The solutions for exercise wise problems are prepared after vast research is conducted on each topic.