# RD Sharma Solutions for Class 11 Maths Chapter 7 Values of Trigonometric Functions at Sum or Difference of Angles

## RD Sharma Solutions Class 11 Maths Chapter 7 – Free PDF Download

RD Sharma Solutions for Class 11 Maths Chapter 7 – Values of Trigonometric Functions at Sum or Difference of Angles is provided here. In this chapter, we shall derive formulae which expresses the values of trigonometric functions at the sum or difference of two real numbers (or angles) in terms of the values of trigonometric functions at individual numbers (or angles). Experts at BYJUâ€™S have formulated the RD Sharma Class 11 Solutions for Maths in the most lucid and easy manner. RD Sharma Solutions are developed using shortcut techniques to help students grasp the concepts faster and to make learning fun. The solutions to this chapter are available in the PDF format, which can be downloaded easily from the links provided below.

Chapter 7 – Values of Trigonometric Functions at Sum or Difference of Angles contains two 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.

• Values of trigonometric functions at the sum or difference.
• Cosine of the difference and sum of two numbers.
• The sine of the difference and sum of two numbers.
• Tangent of the difference and sum of two numbers.
• Few theorems.
• Maximum and minimum values of trigonometrical expressions.
• To express the given expressions in the desired form.

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Exercise 7.1 Solutions

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EXERCISE 7.1 PAGE NO: 7.19

1. If sin A = 4/5 and cos B = 5/13, where 0 <A, B < Ï€/2, find the values of the following:
(i) sin (A + B)
(ii) cos (A + B)
(iii) sin (A â€“ B)
(iv) cos (A â€“ B)

Solution:

Given:

sin A = 4/5 and cos B = 5/13

We know that cos A = âˆš(1 – sin2 A) and sin B = âˆš(1 – cos2 B), where 0 <A, B < Ï€/2

So let us find the value of sin A and cos B

cos A = âˆš(1 – sin2 A)

= âˆš(1 â€“ (4/5)2)

= âˆš(1 â€“ (16/25))

= âˆš((25 â€“ 16)/25)

= âˆš(9/25)

= 3/5

sin B = âˆš(1 – cos2 B)

= âˆš(1 â€“ (5/13)2)

= âˆš(1 â€“ (25/169))

= âˆš(169 â€“ 25)/169)

= âˆš(144/169)

= 12/13

(i) sin (A + B)

We know that sin (A +B) = sin A cos B + cos A sin B

So,

sin (A +B) = sin A cos B + cos A sin B

= 4/5 Ã— 5/13 + 3/5 Ã— 12/13

= 20/65 + 36/65

= (20+36)/65

= 56/65

(ii) cos (A + B)

We know that cos (A +B) = cos A cos BÂ â€“Â sin A sin B

So,

cos (A + B) = cos A cos BÂ â€“Â sin A sin B

= 3/5 Ã— 5/13 â€“ 4/5 Ã— 12/13

= 15/65 â€“ 48/65

= -33/65

(iii) sin (A â€“ B)

We know that sin (A – B) = sin A cos B â€“ cos A sin B

So,

sin (A – B) = sin A cos B â€“ cos A sin B

= 4/5 Ã— 5/13 â€“ 3/5 Ã— 12/13

= 20/65 â€“ 36/65

= -16/65

(iv) cos (A â€“ B)

We know that cos (A -B) = cos A cos BÂ +Â sin A sin B

So,

cos (A -B) = cos A cos BÂ +Â sin A sin B

= 3/5 Ã— 5/13 + 4/5 Ã— 12/13

= 15/65 + 48/65

= 63/65

2. (a) If Sin A = 12/13 and sin B = 4/5, where Ï€/2<A < Ï€ and 0 <B < Ï€/2, find the following:
(i) sin (A + B) (ii) cos (A + B)

(b) If sin A = 3/5, cos B = â€“12/13, where A and B, both lie in second quadrant, find the value of sin (A +B).

Solution:

(a) Given:

Sin A = 12/13 and sin B = 4/5, where Ï€/2<A < Ï€ and 0 <B < Ï€/2

We know that cos A = – âˆš(1 – sin2 A) and cos B = âˆš(1 – sin2 B)

So let us find the value of cos A and cos B

cos A = – âˆš(1 – sin2 A)

= – âˆš(1 â€“ (12/13)2)

= – âˆš(1-144/169)

= -âˆš((169-144)/169)

= – âˆš(25/169)

= – 5/13

cos B = âˆš(1 – sin2 B)

= âˆš(1 â€“ (4/5)2)

= âˆš(1-16/25)

= âˆš((25-16)/25)

=âˆš(9/25)

= 3/5

(i) sin (A +B)

We know that sin (A + B) = sin A cos B + cos A sin B

So,

sin (A +B) = sin A cos B + cos A sin B

= 12/13 Ã— 3/5 + (-5/13) Ã— 4/5

= 36/65 â€“ 20/65

= 16/65

(ii) cos (A + B)

We know that cos (A +B) = cos A cos BÂ â€“Â sin A sin B

So,

cos (A +B) = cos A cos BÂ â€“Â sin A sin B

= -5/13 Ã— 3/5 â€“ 12/13 Ã— 4/5

= -15/65 â€“ 48/65

= – 63/65

(b) Given:

sin A = 3/5, cos B = â€“12/13, where A and B, both lie in second quadrant.

We know that cos A = – âˆš(1 – sin2 A) and sin B = âˆš(1 – cos2 B)

So let us find the value of cos A and sin B

cos A = – âˆš(1 – sin2 A)

= – âˆš(1 â€“ (3/5)2)

= – âˆš(1- 9/25)

= – âˆš((25-9)/25)

= – âˆš(16/25)

= – 4/5

sin B = âˆš(1 – cos2 B)

= âˆš(1 â€“ (-12/13)2)

= âˆš(1 – 144/169)

= âˆš((169-144)/169)

= âˆš(25/169)

= 5/13

We need to find sin (A + B)

Since, sin (A + B) = sin A cos B + cos A sin B

= 3/5 Ã— (-12/13) + (-4/5) Ã— 5/13

= -36/65 â€“ 20/65

= -56/65

3. If cos A = â€“ 24/25 and cos B = 3/5, where Ï€ <A < 3Ï€/2 and 3Ï€/2 <B < 2Ï€, find the following:
(i) sin (A + B) (ii) cos (A + B)

Solution:

Given:

cos A = â€“ 24/25 and cos B = 3/5, where Ï€ <A < 3Ï€/2 and 3Ï€/2 <B < 2Ï€

We know that A is in third quadrant, B is in fourth quadrant. So sine function is negative.

By using the formulas,

sin A = – âˆš(1 – cos2 A) and sin B = -âˆš(1 – cos2 B)

So let us find the value of sin A and sin B

sin A = – âˆš(1 – cos2 A)

= – âˆš(1-(-24/25)2)

= – âˆš(1-576/625)

= – âˆš((625-576)/625)

= – âˆš(49/625)

= -7/25

sin B = -âˆš(1 – cos2 B)

= – âˆš(1-(3/5)2)

= – âˆš(1-9/25)

= – âˆš((25-9)/25)

= – âˆš(16/25)

= – 4/5

(i) sin (A + B)

We know that sin (A + B) = sin A cos B + cos A sin B

So,

sin (A + B) = sin A cos B + cos A sin B

= -7/25 Ã— 3/5 + (-24/25) Ã— (-4/5)

= -21/125 + 96/125

= 75/125

= 3/5

(ii) cos (A + B)

We know that cos (A + B) = cos A cos BÂ â€“Â sin A sin B

So,

cos (A + B) = cos A cos BÂ â€“Â sin A sin B

= (-24/25) Ã— 3/5 â€“ (-7/25) Ã— (-4/5)

= -72/125 â€“ 28/125

= -100/125

= – 4/5

4. If tan A = 3/4, cos B = 9/41, where Ï€<A < 3Ï€/2 and 0 <B < Ï€/2, find tan (A + B).

Solution:

Given:

tan A = 3/4 and cos B = 9/41, where Ï€ <A < 3Ï€/2 and 0 <B < Ï€/2

We know that, A is in third quadrant, B is in first quadrant.

So, tan function And sine function are positive.

By using the formula,

sin B = âˆš(1 – cos2 B)

Let us find the value of sin B.

sin B = âˆš(1 – cos2 B)

= âˆš(1- (9/41)2)

= âˆš(1- 81/1681)

= âˆš((1681-81)/1681)

= âˆš(1600/1681)

= 40/41

We know, tan B = sin B/cos B

= (40/41) / (9/41)

= 40/9

So, tan (A + B) = (tan A + tan B) / (1 â€“ tan A tan B)

= (3/4 + 40/9) / (1 â€“ 3/4 Ã— 40/9)

= (187/36) / (1- 120/36)

= (187/36) / ((36-120)/36)

= (187/36) / (-84/36)

= -187/84

5. If sin A = 1/2, cos B = 12/13, where Ï€/2<A < Ï€ and 3Ï€/2 <B < 2Ï€, find tan(A – B).

Solution:

Given:

sin A = 1/2, cos B = 12/13, where Ï€/2<A < Ï€ and 3Ï€/2 <B < 2Ï€

We know that, A is in second quadrant, B is in fourth quadrant.

In the second quadrant, sine function is positive, cosine and tan functions are negative.

In the fourth quadrant, sine and tan functions are negative, cosine function is positive.

By using the formulas,

cos A = – âˆš(1 – sin2 A) and sin B = -âˆš(1 – cos2 B)

So let us find the value of cos A and sin B

cos A = – âˆš(1 – sin2 A)

= – âˆš(1 â€“ (1/2)2)

= – âˆš(1- 1/4)

= – âˆš((4-1)/4)

= – âˆš(3/4)

= -âˆš3/2

sin B = -âˆš(1 – cos2 B)

= – âˆš(1-(12/13)2)

= – âˆš(1- 144/169)

= – âˆš((169-144)/169)

= – âˆš(25/169)

= – 5/13

We know, tan A = sin A / cos A and tan B = sin B / cos B

tan A = (1/2)/( -âˆš3/2) = -1/âˆš3 and

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

So, tan (A – B) = (tan A â€“ tan B) / (1 + tan A tan B)

= ((-1/âˆš3) â€“ (-5/12)) / (1 + (-1/âˆš3) Ã— (-5/12))

= ((-12+5âˆš3)/12âˆš3) / (1 + 5/12âˆš3)

= ((-12+5âˆš3)/12âˆš3) / ((12âˆš3 + 5)/12âˆš3)

= (5âˆš3 â€“ 12) / (5 + 12âˆš3)

6. If sin A = 1/2, cos B = âˆš3/2, where Ï€/2<A < Ï€ and 0 <B < Ï€/2, find the following:
(i) tan (A + B) (ii) tan (A – B)

Solution:

Given:

Sin A = 1/2 and cos B = âˆš3/2, where Ï€/2 <A < Ï€ and 0 <B < Ï€/2

We know that, A is in second quadrant, B is in first quadrant.

In the second quadrant, sine function is positive. cosine and tan functions are negative.

In first quadrant, all functions are positive.

By using the formulas,

cos A = – âˆš(1 – sin2 A) and sin B = âˆš(1 – cos2 B)

So let us find the value of cos A and sin B

cos A = – âˆš(1 – sin2 A)

= – âˆš(1 â€“ (1/2)2)

= – âˆš(1- 1/4)

= – âˆš((4-1)/4)

= – âˆš(3/4)

= -âˆš3/2

sin B = âˆš(1 – cos2 B)

= âˆš(1-(âˆš3/2)2)

= âˆš(1- 3/4)

= âˆš((4-3)/4)

= âˆš(1/4)

= 1/2

We know, tan A = sin A / cos A and tan B = sin B / cos B

tan A = (1/2)/( -âˆš3/2) = -1/âˆš3 and

tan B = (1/2)/(âˆš3/2) = 1/âˆš3

(i) tan (A + B) = (tan A + tan B) / (1 â€“ tan A tan B)

= (-1/âˆš3 + 1/âˆš3) / (1 â€“ (-1/âˆš3) Ã— 1/âˆš3)

= 0 / (1 + 1/3)

= 0

(ii) tan (A – B) = (tan A â€“ tan B) / (1 + tan A tan B)

= ((-1/âˆš3) â€“ (1/âˆš3)) / (1 + (-1/âˆš3) Ã— (1/âˆš3))

= ((-2/âˆš3) / (1 – 1/3)

= ((-2/âˆš3) / (3-1)/3)

= ((-2/âˆš3) / 2/3)

= – âˆš3

7. Evaluate the following:
(i) sin 780Â cos 180Â â€“ cos 780Â sin 180Â

(ii) cos 470Â cos 130Â – sin 470Â sin 130
(iii) sin 360Â cos 90Â + cos 360Â sin 90Â

(iv) cos 800Â cos 200Â + sin 800Â sin 200

Solution:

(i)Â sin 780Â cos 180Â â€“ cos 780Â sin 180

We know that sin (A – B) = sin A cos B â€“ cos A sin B

sin 780Â cos 180Â â€“ cos 780Â sin 180Â = sin(78 â€“ 18) Â°

= sin 60Â°

= âˆš3/2

(ii)Â cos 470Â cos 130Â – sin 470Â sin 130

We know that cos A cos B â€“ sin A sin B = cos (A + B)

cos 470Â cos 130Â – sin 470Â sin 130Â = cos (47 + 13) Â°

= cos 60Â°

= 1/2

(iii) sin 360Â cos 90Â + cos 360Â sin 90

We know that sin (A +B) = sin A cos B + cos A sin B

sin 360Â cos 90Â + cos 360Â sin 90Â = sin (36 + 9) Â°

= sin 45Â°

= 1/âˆš2

(iv) cos 800Â cos 200Â + sin 800Â sin 200

We know that cos A cos B + sin A sin B = cos (A – B)

cos 800Â cos 200Â + sin 800Â sin 200Â = cos (80 – 20) Â°

= cos 60Â°

= Â½

8. If cos A = â€“12/13 and cot B = 24/7, where A lies in the second quadrant and B in the third quadrant, find the values of the following:
(i) sin (A + B) (ii) cos (A + B) (iii) tan (A + B)

Solution:

Given:

cos A = -12/13 and cot B = 24/7

We know that, A lies in second quadrant, B in the third quadrant.

In the second quadrant sineÂ function is positive.

In the third quadrant, both sine and cosine functions are negative.

By using the formulas,

sin A = âˆš(1 – cos2 A), sin B = – 1/âˆš(1 + cot2 B) and cos B = -âˆš(1 – sin2 B),

So let us find the value of sin A and sin B

sin A = âˆš(1 â€“ cos2 A)

= âˆš(1 â€“ (-12/13)2)

= âˆš(1 â€“ 144/169)

= âˆš((169-144)/169)

= âˆš(25/169)

= 5/13

sin B = – 1/âˆš(1 + cot2 B)

= – 1/âˆš(1 + (24/7)2)

= – 1/âˆš(1 + 576/49)

= -1/âˆš((49+576)/49)

= -1/âˆš(625/49)

= -1/(25/7)

= -7/25

cos B = -âˆš(1 – sin2 B)

= -âˆš(1-(-7/25)2)

= -âˆš(1-(49/625))

= -âˆš((625-49)/625)

= -âˆš(576/625)

= -24/25

So, now let us find

(i) sin (A + B)

We know that sin (A + B) = sin A cos B + cos A sin B

So,

sin (A + B) = sin A cos B + cos A sin B

= 5/13 Ã— (-24/25) + (-12/13) Ã— (-7/25)

= -120/325 + 84/325

= -36/325

(ii) cos (A + B)

We know that cos (A + B) = cos A cos BÂ â€“Â sin A sin B

So,

cos (A + B) = cos A cos BÂ â€“Â sin A sin B

= -12/13 Ã— (-24/25) â€“ (5/13) Ã— (-7/25)

= 288/325 + 35/325

= 323/325

(iii) tan (A + B)

We know that tan (A + B) = sin (A+B) / cos (A+B)

= (-36/325) / (323/325)

= -36/323

9. Prove that: cos 7Ï€/12 + cos Ï€/12 = sin 5Ï€/12 â€“ sin Ï€/12

Solution:

We know that, 7Ï€/12 = 105Â°, Ï€/12 = 15Â°; 5Ï€/12 = 75Â°

Let us consider LHS: cos 105Â° + cos 15Â°

cos (90Â° + 15Â°) + sin (90Â° – 75Â°)

-sin 15Â° + sin 75Â°

sin 75Â° – sin 15Â°

= RHS

âˆ´ LHS = RHS

Hence proved.

10. Prove that: (tan A + tan B) / (tan A â€“ tan B) = sin (A + B) / sin (A – B) Â

Solution:

Let us consider LHS: (tan A + tan B) / (tan A â€“ tan B)

= RHS

âˆ´ LHS = RHS

Hence proved.

11. Prove that:
(i) (cos 11o + sin 11o) / (cos 11o â€“ sin 11o) = tan 56o Â

(ii) (cos 9o + sin 9o) / (cos 9o â€“ sin 9o) = tan 54o

(iii) (cos 8o â€“ sin 8o) / (cos 8o + sin 8o) = tan 37o

Solution:

(i) (cos 11o + sin 11o) / (cos 11o â€“ sin 11o) = tan 56o

Let us consider LHS:

(cos 11o + sin 11o) / (cos 11o â€“ sin 11o)

Now let us divide the numerator and denominator by cos 11o we get,

(cos 11o + sin 11o) / (cos 11o â€“ sin 11o) = (1 + tan 11o) / (1 â€“ tan 11o)

= (1 + tan 11o) / (1- 1Ã—tan 11o)

= (tan 45o + tan 11o) / (1 â€“ tan 45o Ã— tan 11o)

We know that tan (A+B) = (tan A + tan B) / (1 â€“ tan A tan B)

So,

(tan 45o + tan 11o) / (1 â€“ tan 45o Ã— tan 11o) = tan (45o + 11o)

= tan 56o

= RHS

âˆ´ LHS = RHS

Hence proved.

(ii) (cos 9o + sin 9o) / (cos 9o â€“ sin 9o) = tan 54o

Let us consider LHS:

(cos 9o + sin 9o) / (cos 9o â€“ sin 9o)

Now let us divide the numerator and denominator by cos 9o we get,

(cos 9o + sin 9o) / (cos 9o â€“ sin 9o) = (1 + tan 9o) / (1 â€“ tan 9o)

= (1 + tan 9o) / (1 â€“ 1 Ã— tan 9o)

= (tan 45o + tan 9o) / (1 â€“ tan 45o Ã— tan 9o)

We know that tan (A+B) = (tan A + tan B) / (1 â€“ tan A tan B)

So,

(tan 45o + tan 9o) / (1 â€“ tan 45o Ã— tan 9o) = tan (45o + 9o)

= tan 54o

= RHS

âˆ´ LHS = RHS

Hence proved.

(iii) (cos 8o â€“ sin 8o) / (cos 8o + sin 8o) = tan 37o

Let us consider LHS:

(cos 8o â€“ sin 8o) / (cos 8o + sin 8o)

Now let us divide the numerator and denominator by cos 8o we get,

(cos 8o â€“ sin 8o) / (cos 8o + sin 8o) = (1 â€“ tan 8o) / (1 + tan 8o)

= (1 â€“ tan 8o) / (1 + 1Ã—tan 8o)

= (tan 45o â€“ tan 8o) / (1 + tan 45o Ã—tan 8o)

We know that tan (A+B) = (tan A + tan B) / (1 â€“ tan A tan B)

So,

(tan 45o â€“ tan 8o) / (1 + tan 45o Ã—tan 8o) = tan (45o â€“ 8o)

= tan 37o

= RHS

âˆ´ LHS = RHS

Hence proved.

12. Prove that:

(i)

(ii)

(iii)

Solution:

(i)

= sin 90o

= 1

= RHS

âˆ´ LHS = RHS

Hence proved.

(ii)

= sin 60o

= âˆš3/2

= RHS

âˆ´ LHS = RHS

Hence proved.

(iii)

= sin 90o

= 1

= RHS

âˆ´ LHS = RHS

Hence proved.

13. Prove that: (tan 69o + tan 66o) / (1 â€“ tan 69o tan 66o) = -1

Solution:

Let us consider LHS:

(tan 69o + tan 66o) / (1 â€“ tan 69o tan 66o)

We know that, tan (A + B) = (tan A + tan B) / (1 â€“ tan A tan B)

Here, A = 69o and B = 66o

So,

(tan 69o + tan 66o) / (1 â€“ tan 69o tan 66o) = tan (69 + 66)o

= tan 135o

= – tan 45o

= – 1

= RHS

âˆ´ LHS = RHS

Hence proved.

14. (i) If tan A = 5/6 and tan B = 1/11, prove that A + B = Ï€/4

(ii) If tan A = m/(mâ€“1) and tan B = 1/(2m â€“ 1), then prove that A â€“ B = Ï€/4

Solution:

(i) If tan A = 5/6 and tan B = 1/11, prove that A + B = Ï€/4

Given:

tan A = 5/6 and tan B = 1/11

We know that, tan (A + B) = (tan A + tan B) / (1 â€“ tan A tan B)

= [(5/6) + (1/11)] / [1 â€“ (5/6) Ã— (1/11)]

= (55+6) / (66-5)

= 61/61

= 1

= tan 45o or tan Ï€/4

So, tan (A + B) = tan Ï€/4

âˆ´ (A + B) = Ï€/4

Hence proved.

(ii) If tan A = m/(mâ€“1) and tan B = 1/(2m â€“ 1), then prove that A â€“ B = Ï€/4

Given:

tan A = m/(mâ€“1) and tan B = 1/(2m â€“ 1)

We know that, tan (A – B) = (tan A – tan B) / (1 + tan A tan B)

= (2m2 â€“ m â€“ m + 1) / (2m2 â€“ m â€“ 2m + 1 + m)

= (2m2 â€“ 2m + 1) / (2m2 â€“ 2m + 1)

= 1

= tan 45o or tan Ï€/4

So, tan (A – B) = tan Ï€/4

âˆ´ (A – B) = Ï€/4

Hence proved.

15. prove that:
(i) cos2Â Ï€/4 – sin2 Ï€/12 = âˆš3/4

(ii) sin2 (n + 1) A â€“ sin2nA = sin (2n + 1) A sin A

Solution:

(i) cos2Â Ï€/4 – sin2 Ï€/12 = âˆš3/4

Let us consider LHS:

cos2Â Ï€/4 – sin2 Ï€/12

We know that, cos2A â€“ sin2 B = cos (A + B) cos (A â€“ B)

So,

cos2Â Ï€/4 – sin2 Ï€/12 = cos (Ï€/4 + Ï€/12) cos (Ï€/4 â€“ Ï€/12)

= cos 4Ï€/12 cos 2Ï€/12

= cos Ï€/3 cos Ï€/6

= 1/2 Ã— âˆš3/2

= âˆš3/4

= RHS

âˆ´ LHS = RHS

Hence proved.

(ii) sin2 (n + 1) A â€“ sin2nA = sin (2n + 1) A sin A

Let us consider LHS:

sin2 (n + 1) A â€“ sin2nA

We know that, sin2A â€“ sin2 B = sin (A + B) sin (A â€“ B)

Here, A = (n + 1) A and B = nA

So,

sin2 (n + 1) A â€“ sin2n A = sin ((n + 1) A + nA) sin ((n + 1) A â€“ nA)

= sin (nA +A + nA) sin (nA +A â€“ nA)

= sin (2nA +A) sin (A)

= sin (2n + 1) A sin A

= RHS

âˆ´ LHS = RHS

Hence proved.

16. Prove that:

(i)

(ii)

(iii)

(iv) sin2 B = sin2 A + sin2 (A-B) â€“ 2sin A cos B sin (A – B)

(v) cos2 A + cos2 B â€“ 2 cos A cos B cos (A +B) = sin2 (A + B)

(vi)

Solution:

(i)

= tan A

= RHS

âˆ´ LHS = RHS

Hence proved.

(ii)

Let us consider LHS:

= tan A â€“ tan B + tan B â€“ tan C + tan C â€“ tan A

= 0

= RHS

âˆ´ LHS = RHS

Hence proved.

(iii)

= cotB â€“ cotA + cot C â€“ cotB + cotA â€“ cot C

= 0

= RHS

âˆ´ LHS = RHS

Hence proved.

(iv) sin2 B = sin2 A + sin2 (A-B) â€“ 2sin A cos B sin (A – B)

Let us consider RHS:

sin2A + sin2 (A -B) â€“ 2 sin A cos B sin (A – B)

sin2A + sin (A -B) [sin (A â€“B) â€“ 2 sin A cos B]

We know that, sin (A â€“B) = sin A cos B â€“ cos A sin B

So,

sin2A + sin (A -B) [sin A cos B â€“ cos A sin BÂ â€“ 2 sin A cos B]

sin2A + sin (A -B) [-sin A cos B â€“ cos A sin B]

sin2A â€“ sin (A -B) [sin A cos B + cos A sin B]

We know that, sin (A +B) = sin A cos B + cos A sin B

So,

sin2A â€“ sin (A â€“ B) sin (A + B)

sin2 A â€“ sin2 A + sin2 B

sin2 B

= LHS

âˆ´ LHS = RHS

Hence proved.

(v) cos2 A + cos2 B â€“ 2 cos A cos B cos (A + B) = sin2 (A + B)

Let us consider LHS:

cos2A + cos2B â€“ 2 cos A cos B cos (A +B)

cos2A + 1 â€“ sin2B – 2 cos A cos B cos (A +B)

1 +Â cos2A â€“ sin2B – 2 cos A cos B cos (A +B)

We know that, cos2A â€“ sin2B = cos (A +B) cos (A â€“B)

So,

1 +Â cos (A +B) cos (A â€“B) – 2 cos A cos B cos (A +B)

1 + cos (A +B) [cos (A â€“B) â€“ 2 cos A cos B]

We know that, cos (A – B) = cos A cos B + sin A sin B.

So,

1 + cos (A +B) [cos A cos B + sin A sin BÂ â€“ 2 cos A cos B]

1 + cos (A +B) [-cos A cos B + sin A sin B]

1 â€“ cos (A +B) [cos A cos B â€“ sin A sin B]

We know that, cos (A +B) = cos A cos B â€“ sin A sin B.

So,

1 â€“ cos2 (A + B)

sin2 (A + B)

= RHS

âˆ´ LHS = RHS

Hence proved.

(vi)

âˆ´ LHS = RHS

Hence proved.

17. Prove that:
(i) tan 8x â€“ tan 6x â€“ tan 2x = tan 8x tan 6x tan 2x

(ii) tan Ï€/12 + tan Ï€/6 + tan Ï€/12 tan Ï€/6 = 1

(iii) tan 36oÂ + tan 9oÂ + tan 36oÂ tan 9o =Â 1

(iv) tan 13x â€“ tan 9x â€“ tan 4x = tan 13x tan 9x tan 4x

Solution:

(i) tan 8x â€“ tan 6x â€“ tan 2x = tan 8x tan 6x tan 2x

Let us consider LHS:

tan 8x â€“ tan 6x â€“ tan 2x

tan 8x = tan(6x + 2x)

We know that, tan (A + B) = (tan A + tan B) / (1 â€“ tan A tan B)

So,

tan 8x = (tan 6x + tan 2x) / (1 â€“ tan 6x tan 2x)

By cross-multiplying we get,

tan 8x (1 â€“ tan 6x tan 2x) = tan 6x + tan 2x

tan 8x â€“ tan 8x tan 6x tan2x = tan 6x + tan 2x

Upon rearranging we get,

tan 8x â€“ tan 6x â€“ tan 2x = tan 8x tan 6x tan 2x

= RHS

âˆ´ LHS = RHS

Hence proved.

(ii) tan Ï€/12 + tan Ï€/6 + tan Ï€/12 tan Ï€/6 = 1

We know,

Ï€/12 = 15Â° and Ï€/6 = 30Â°

So, we have 15Â° + 30Â° = 45Â°

Tan (15Â° + 30Â°) = tan 45Â°

We know that, tan (A + B) = (tan A + tan B) / (1 â€“ tan A tan B)

So,

(tan 15o + tan 30o) / (1 â€“ tan 15o tan 30o) = 1

tan 15Â° + tan 30Â° = 1 â€“ tan 15Â° tan 30Â°

Upon rearranging we get,

tan15Â° + tan30Â° + tan15Â° tan30Â° = 1

Hence proved.

(iii) tan 36oÂ + tan 9oÂ + tan 36oÂ tan 9o =Â 1

We know 36Â° + 9Â° = 45Â°

So we have,

tan (36Â° + 9Â°) = tan 45Â°

We know that, tan (A + B) = (tan A + tan B) / (1 â€“ tan A tan B)

So,

(tan 36o + tan 9o) / (1 â€“ tan 36o tan 9o) = 1

tan 36Â° + tan 9Â° = 1 â€“ tan 36Â° tan 9Â°

Upon rearranging we get,

tan 36Â° + tan 9Â° + tan 36Â° tan 9Â° = 1

Hence proved.

(iv) tan 13x â€“ tan 9x â€“ tan 4x = tan 13x tan 9x tan 4x

Let us consider LHS:

tan 13x â€“ tan 9x â€“ tan 4x

tan 13x = tan (9x + 4x)

We know that, tan (A + B) = (tan A + tan B) / (1 â€“ tan A tan B)

So,

tan 13x = (tan 9x + tan 4x) / (1 â€“ tan 9x tan 4x)

By cross-multiplying we get,

tan 13x (1 â€“ tan 9x tan 4x) = tan 9x + tan 4x

tan 13x â€“ tan 13x tan 9x tan 4x = tan 9x + tan 4x

Upon rearranging we get,

tan 13x â€“ tan 9x â€“ tan 4x = tan 13x tan 9x tan 4x

= RHS

âˆ´ LHS = RHS

Hence proved.

EXERCISE 7.2 PAGE NO: 7.26

1. Find the maximum and minimum values of each of the following trigonometrical expressions:

(i) 12 sin x â€“ 5 cos x
(ii) 12 cos x + 5 sin x + 4
(iii) 5 cos x + 3 sin (Ï€/6 – x) + 4
(iv) sin x â€“ cos x + 1

Solution:

We know that the maximum value of A cos Î± + B sin Î± + C is C + âˆš(A2Â +B2),

And the minimum value is C – âˆš(a2Â +B2).

(i) 12 sin x â€“ 5 cos x

Given: f(x) = 12 sin x â€“ 5 cos x

Here, A = -5, B = 12 and C = 0

âˆš((-5)2 + 122) â‰¤ 12 sin x â€“ 5 cos x â‰¤ âˆš((-5)2 + 122)

âˆš(25+144) â‰¤ 12 sin x â€“ 5 cos x â‰¤ âˆš(25+144)

âˆš169 â‰¤ 12 sin x â€“ 5 cos x â‰¤ âˆš169

-13Â â‰¤Â 12 sinÂ x – 5 cosÂ x â‰¤ 13

Hence, the maximum and minimum values of f(x) are 13 and -13 respectively.

(ii) 12 cos x + 5 sin x + 4

Given: f(x) = 12 cos x + 5 sin x + 4

Here, A = 12, B = 5 and C = 4

4 – âˆš(122 + 52) â‰¤ 12 cos x + 5 sin x + 4 â‰¤ 4 + âˆš(122 + 52)

4 – âˆš(144+25) â‰¤ 12 cos x + 5 sin x + 4 â‰¤ 4 + âˆš(144+25)

4 –âˆš169 â‰¤ 12 cos x + 5 sin x + 4 â‰¤ 4 + âˆš169

-9Â â‰¤Â 12 cos x + 5 sin x + 4 â‰¤ 17

Hence, the maximum and minimum values of f(x) are -9 and 17 respectively.

(iii) 5 cos x + 3 sin (Ï€/6 – x) + 4Â

Given: f(x) = 5 cos x + 3 sin (Ï€/6 – x) + 4Â

We know that, sin (A – B) = sin A cos B â€“ cos A sin B

f(x) = 5 cos x + 3 sin (Ï€/6 – x) + 4Â

= 5 cos x + 3 (sin Ï€/6 cos x â€“ cos Ï€/6 sin x) + 4

= 5 cos x + 3/2 cos x – 3âˆš3/2 sin x + 4

= 13/2 cos x – 3âˆš3/2 sin x + 4

So, here A = 13/2, B = – 3âˆš3/2, C = 4

4 – âˆš[(13/2)2 + (-3âˆš3/2)2] â‰¤ 13/2 cos x – 3âˆš3/2 sin x + 4 â‰¤ 4 + âˆš[(13/2)2 + (-3âˆš3/2)2]

4 – âˆš[(169/4) + (27/4)] â‰¤ 13/2 cos x – 3âˆš3/2 sin x + 4 â‰¤ 4 + âˆš[(169/4) + (27/4)]

4 â€“ 7 â‰¤ 13/2 cos x – 3âˆš3/2 sin x + 4 â‰¤ 4 + 7

-3 â‰¤ 13/2 cos x – 3âˆš3/2 sin x + 4 â‰¤ 11

Hence, the maximum and minimum values of f(x) are -3 and 11 respectively.

(iv) sin x â€“ cos x + 1

Given: f(x) = sin x â€“ cos x + 1

So, here A = -1, B = 1 And c = 1

1 – âˆš[(-1)2 + 12] â‰¤ sin x â€“ cos x + 1 â‰¤ 1 + âˆš[(-1)2 + 12]

1 – âˆš(1+1) â‰¤ sin x â€“ cos x + 1 â‰¤ 1 + âˆš(1+1)

1 – âˆš2 â‰¤ sin x â€“ cos x + 1 â‰¤ 1 + âˆš2

Hence, the maximum and minimum values of f(x) are 1 – âˆš2 and 1 + âˆš2 respectively.

2. Reduce each of the following expressions to the Sine and Cosine of a single expression:

(i) âˆš3 sin x â€“ cos x

(ii) cos x â€“ sin x

(iii) 24 cos x + 7 sin x

Solution:

(i) âˆš3 sin x â€“ cos x

Let f(x) = âˆš3 sin x â€“ cos x

Dividing and multiplying by âˆš((âˆš3)2 + 12) i.e. by 2

f(x) = 2(âˆš3/2 sin x â€“ 1/2 cos x)

Sine expression:

f(x) = 2(cos Ï€/6 sin x â€“ sin Ï€/6 cos x) (since, âˆš3/2 = cos Ï€/6 and 1/2 = sin Ï€/6)

We know that, sin A cos B â€“ cos A sin B = sin (A â€“ B)

f(x) = 2 sin (x – Ï€/6)

Again,

f(x) = 2(âˆš3/2 sin x â€“ 1/2 cos x)

Cosine expression:

f(x) = 2(sin Ï€/3 sin x â€“ cos Ï€/3 cos x)

We know that, cos A cos B â€“ sin A sin B = cos (A + B)

f(x) = -2 cos(Ï€/3 + x)

(ii) cos x â€“ sin x

Let f(x) = cos x â€“ sin x

Dividing and multiplying by âˆš(12 + 12) i.e. by âˆš2,

f(x) = âˆš2(1/âˆš2 cos x â€“ 1/âˆš2 sin x)

Sine expression:

f(x) = âˆš2(sin Ï€/4 cos x â€“ cos Ï€/4 sin x) (since, 1/âˆš2 = sin Ï€/4 and 1/âˆš2 = cos Ï€/4)

We know that sin A cos B â€“ cos A sin B = sin (A â€“ B)

f(x) = âˆš2 sin (Ï€/4 – x)

Again,

f(x) = âˆš2(1/âˆš2 cos x â€“ 1/âˆš2 sin x)

Cosine expression:

f(x) = 2(cos Ï€/4 cos x â€“ sin Ï€/4 sin x)

We know that cos A cos B â€“ sin A sin B = cos (A + B)

f(x) = âˆš2 cos (Ï€/4 + x)

(iii) 24 cos x + 7 sin x

Let f(x) = 24 cos x + 7 sin x

Dividing and multiplying by âˆš((âˆš24)2Â + 72) = âˆš625 i.e. by 25,

f(x) = 25(24/25 cos x + 7/25 sin x)

Sine expression:

f(x) = 25(sin Î± cos x + cos Î± sin x)Â where, sin Î± = 24/25 and cos Î± = 7/25

We know that sin A cos B + cos A sin B = sin (A + B)

f(x) = 25 sin (Î± + x)

Cosine expression:

f(x) = 25(cos Î± cos x + sin Î± sin x)Â where, cos Î± = 24/25 and sin Î± = 7/25

We know that cos A cos B + sin A sin B = cos (A – B)

f(x) = 25 cos (Î± – x)

3. Show that Sin 100oÂ â€“ Sin 10oÂ is positive.

Solution:

Let f(x) = sin 100Â° â€“ sin 10Â°

Dividing And multiplying by âˆš(12 + 12) i.e. by âˆš2,

f(x) = âˆš2(1/âˆš2 sin 100o â€“ 1/âˆš2 sin 10o)

f(x) = âˆš2(cos Ï€/4 sin (90+10)o â€“ sin Ï€/4 sin 10o) (since, 1/âˆš2 = cos Ï€/4 and 1/âˆš2 = sin Ï€/4)

f(x) = âˆš2(cos Ï€/4 cos 10o â€“ sin Ï€/4 sin 10o)

We know that cos A cos B â€“ sin A sin B = cos (A + B)

f(x) = âˆš2 cos (Ï€/4 + 10o)

âˆ´Â f(x) = âˆš2 cos 55Â°

4. Prove that (2âˆš3 + 3) sin x + 2âˆš3 cos xÂ lies between â€“ (2âˆš3 + âˆš15) and (2âˆš3 + âˆš15).

Solution:

Let f(x) = (2âˆš3 + 3) sin x + 2âˆš3 cos x

Here, A = 2âˆš3, B = 2âˆš3 + 3 and C = 0

– âˆš[(2âˆš3)2 + (2âˆš3 + 3)2] â‰¤ (2âˆš3 + 3) sin x + 2âˆš3 cos x â‰¤ âˆš[(2âˆš3)2 + (2âˆš3 + 3)2]

– âˆš[12+12+9+12âˆš3] â‰¤ (2âˆš3 + 3) sin x + 2âˆš3 cos x â‰¤ âˆš[12+12+9+12âˆš3]

– âˆš[33+12âˆš3] â‰¤ (2âˆš3 + 3) sin x + 2âˆš3 cos x â‰¤ âˆš[33+12âˆš3]

– âˆš[15+12+6+12âˆš3] â‰¤ (2âˆš3 + 3) sin x + 2âˆš3 cos x â‰¤ âˆš[15+12+6+12âˆš3]

We know that (12âˆš3 + 6 < 12âˆš5) because the value of âˆš5 – âˆš3 is more than 0.5

So if we replace, (12âˆš3 + 6 with 12âˆš5) the above inequality still holds.

So by rearranging the above expression âˆš(15+12+12âˆš5)we get, 2âˆš3 + âˆš15

– 2âˆš3 + âˆš15 â‰¤ (2âˆš3 + 3) sin x + 2âˆš3 cos x â‰¤ 2âˆš3 + âˆš15

Hence proved.

## Frequently Asked Questions on RD Sharma Solutions for Class 11 Maths Chapter 7

### Where to download RD Sharma Solutions for Class 11 Maths Chapter 7?

RD Sharma Solutions for Class 11 Maths Chapter 7 can be downloaded at BYJUâ€™S website. It can be available in free PDF. To download go to BYJUâ€™S RD Sharma Solutions website/select the Class 11/opt the subject as Maths/click on to the desired chapter that is Chapter 7 Values of Trigonometric Functions at Sum or Difference of Angles.

### What are the applications of learning RD Sharma Solutions for Class 11 Maths Chapter 7?

Applications of learning RD Sharma Solutions for Class 11 Maths Chapter 7 are finding the height and distance of different objects without measuring. By learning these concepts students will be able to answer all the questions based on trigonometry as well as it may help in writing class tests and board exams. Students can download this PDF from the provided links. The expert faculty at BYJUâ€™S provides students with PDFs 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. The step-wise solutions prepared by subject experts help students to solve problems comfortably.

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