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

Class 11: Maths Chapter 7 solutions. Complete Class 11 Maths Chapter 7 Notes.

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

RD Sharma 11th Maths Chapter 7, Class 11 Maths Chapter 7 solutions

EXERCISE 7.1 PAGE NO: 7.19

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

Solution:

Given:

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

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

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

cos A = √(1 – sin2 A)

= √(1 – (4/5)2)

= √(1 – (16/25))

= √((25 – 16)/25)

= √(9/25)

= 3/5

sin B = √(1 – cos2 B)

= √(1 – (5/13)2)

= √(1 – (25/169))

= √(169 – 25)/169)

= √(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 × 5/13 + 3/5 × 12/13

= 20/65 + 36/65

= (20+36)/65

= 56/65

(ii) cos (A + B)

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

So,

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

= 3/5 × 5/13 – 4/5 × 12/13

= 15/65 – 48/65

= -33/65

(iii) 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 × 5/13 – 3/5 × 12/13

= 20/65 – 36/65

= -16/65

(iv) cos (A – B)

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

So,

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

= 3/5 × 5/13 + 4/5 × 12/13

= 15/65 + 48/65

= 63/65

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

(b) If sin A = 3/5, cos B = –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 π/2<A < π and 0 <B < π/2

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

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

cos A = – √(1 – sin2 A)

= – √(1 – (12/13)2)

= – √(1-144/169)

= -√((169-144)/169)

= – √(25/169)

= – 5/13

cos B = √(1 – sin2 B)

= √(1 – (4/5)2)

= √(1-16/25)

= √((25-16)/25)

=√(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 × 3/5 + (-5/13) × 4/5

= 36/65 – 20/65

= 16/65

(ii) cos (A + B)

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

So,

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

= -5/13 × 3/5 – 12/13 × 4/5

= -15/65 – 48/65

= – 63/65

(b) Given:

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

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

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

cos A = – √(1 – sin2 A)

= – √(1 – (3/5)2)

= – √(1- 9/25)

= – √((25-9)/25)

= – √(16/25)

= – 4/5

sin B = √(1 – cos2 B)

= √(1 – (-12/13)2)

= √(1 – 144/169)

= √((169-144)/169)

= √(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 × (-12/13) + (-4/5) × 5/13

= -36/65 – 20/65

= -56/65

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

Solution:

Given:

cos A = – 24/25 and cos B = 3/5, where π <A < 3π/2 and 3π/2 <B < 2π

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

By using the formulas,

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

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

sin A = – √(1 – cos2 A)

= – √(1-(-24/25)2)

= – √(1-576/625)

= – √((625-576)/625)

= – √(49/625)

= -7/25

sin B = -√(1 – cos2 B)

= – √(1-(3/5)2)

= – √(1-9/25)

= – √((25-9)/25)

= – √(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 × 3/5 + (-24/25) × (-4/5)

= -21/125 + 96/125

= 75/125

= 3/5

(ii) cos (A + B)

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

So,

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

= (-24/25) × 3/5 – (-7/25) × (-4/5)

= -72/125 – 28/125

= -100/125

= – 4/5

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

Solution:

Given:

tan A = 3/4 and cos B = 9/41, where π <A < 3π/2 and 0 <B < π/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 = √(1 – cos2 B)

Let us find the value of sin B.

sin B = √(1 – cos2 B)

= √(1- (9/41)2)

= √(1- 81/1681)

= √((1681-81)/1681)

= √(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 – tan A tan B)

= (3/4 + 40/9) / (1 – 3/4 × 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 π/2<A < π and 3π/2 <B < 2π, find tan(A – B).

Solution:

Given:

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

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

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

cos A = – √(1 – sin2 A)

= – √(1 – (1/2)2)

= – √(1- 1/4)

= – √((4-1)/4)

= – √(3/4)

= -√3/2

sin B = -√(1 – cos2 B)

= – √(1-(12/13)2)

= – √(1- 144/169)

= – √((169-144)/169)

= – √(25/169)

= – 5/13

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

tan A = (1/2)/( -√3/2) = -1/√3 and

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

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

= ((-1/√3) – (-5/12)) / (1 + (-1/√3) × (-5/12))

= ((-12+5√3)/12√3) / (1 + 5/12√3)

= ((-12+5√3)/12√3) / ((12√3 + 5)/12√3)

= (5√3 – 12) / (5 + 12√3)

6. If sin A = 1/2, cos B = √3/2, where π/2<A < π and 0 <B < π/2, find the following:
(i) tan (A + B) (ii) tan (A – B)

Solution:

Given:

Sin A = 1/2 and cos B = √3/2, where π/2 <A < π and 0 <B < π/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 = – √(1 – sin2 A) and sin B = √(1 – cos2 B)

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

cos A = – √(1 – sin2 A)

= – √(1 – (1/2)2)

= – √(1- 1/4)

= – √((4-1)/4)

= – √(3/4)

= -√3/2

sin B = √(1 – cos2 B)

= √(1-(√3/2)2)

= √(1- 3/4)

= √((4-3)/4)

= √(1/4)

= 1/2

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

tan A = (1/2)/( -√3/2) = -1/√3 and

tan B = (1/2)/(√3/2) = 1/√3

(i) tan (A + B) = (tan A + tan B) / (1 – tan A tan B)

= (-1/√3 + 1/√3) / (1 – (-1/√3) × 1/√3)

= 0 / (1 + 1/3)

= 0

(ii) tan (A – B) = (tan A – tan B) / (1 + tan A tan B)

= ((-1/√3) – (1/√3)) / (1 + (-1/√3) × (1/√3))

= ((-2/√3) / (1 – 1/3)

= ((-2/√3) / (3-1)/3)

= ((-2/√3) / 2/3)

= – √3

7. Evaluate the following:
(i) sin 780 cos 180 – cos 780 sin 180 

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

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

Solution:

(i) sin 780 cos 180 – cos 780 sin 180

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

sin 780 cos 180 – cos 780 sin 180 = sin(78 – 18) °

= sin 60°

= √3/2

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

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

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

= cos 60°

= 1/2

(iii) sin 360 cos 90 + cos 360 sin 90

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

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

= sin 45°

= 1/√2

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

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

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

= cos 60°

= ½

8. If cos A = –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 sine function is positive.

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

By using the formulas,

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

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

sin A = √(1 – cos2 A)

= √(1 – (-12/13)2)

= √(1 – 144/169)

= √((169-144)/169)

= √(25/169)

= 5/13

sin B = – 1/√(1 + cot2 B)

= – 1/√(1 + (24/7)2)

= – 1/√(1 + 576/49)

= -1/√((49+576)/49)

= -1/√(625/49)

= -1/(25/7)

= -7/25

cos B = -√(1 – sin2 B)

= -√(1-(-7/25)2)

= -√(1-(49/625))

= -√((625-49)/625)

= -√(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 × (-24/25) + (-12/13) × (-7/25)

= -120/325 + 84/325

= -36/325

(ii) cos (A + B)

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

So,

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

= -12/13 × (-24/25) – (5/13) × (-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 7π/12 + cos π/12 = sin 5π/12 – sin π/12

Solution:

We know that, 7π/12 = 105°, π/12 = 15°; 5π/12 = 75°

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

cos (90° + 15°) + sin (90° – 75°)

-sin 15° + sin 75°

sin 75° – sin 15°

= RHS

∴ LHS = RHS

Hence proved.

10. Prove that: (tan A + tan B) / (tan A – tan B) = sin (A + B) / sin (A – B)  

Solution:

Let us consider LHS: (tan A + tan B) / (tan A – tan B)

= RHS

∴ LHS = RHS

Hence proved.

11. Prove that:
(i) (cos 11o + sin 11o) / (cos 11o – sin 11o) = tan 56o  

(ii) (cos 9o + sin 9o) / (cos 9o – sin 9o) = tan 54o

(iii) (cos 8o – sin 8o) / (cos 8o + sin 8o) = tan 37o

Solution:

(i) (cos 11o + sin 11o) / (cos 11o – sin 11o) = tan 56o

Let us consider LHS:

(cos 11o + sin 11o) / (cos 11o – sin 11o)

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

(cos 11o + sin 11o) / (cos 11o – sin 11o) = (1 + tan 11o) / (1 – tan 11o)

= (1 + tan 11o) / (1- 1×tan 11o)

= (tan 45o + tan 11o) / (1 – tan 45× tan 11o)

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

So,

(tan 45o + tan 11o) / (1 – tan 45× tan 11o) = tan (45o + 11o)

= tan 56o

= RHS

∴ LHS = RHS

Hence proved.

(ii) (cos 9o + sin 9o) / (cos 9o – sin 9o) = tan 54o

Let us consider LHS:

(cos 9o + sin 9o) / (cos 9o – sin 9o)

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

(cos 9o + sin 9o) / (cos 9o – sin 9o) = (1 + tan 9o) / (1 – tan 9o)

= (1 + tan 9o) / (1 – 1 × tan 9o)

= (tan 45o + tan 9o) / (1 – tan 45o × tan 9o)

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

So,

(tan 45o + tan 9o) / (1 – tan 45o × tan 9o) = tan (45o + 9o)

= tan 54o

= RHS

∴ LHS = RHS

Hence proved.

(iii) (cos 8o – sin 8o) / (cos 8o + sin 8o) = tan 37o

Let us consider LHS:

(cos 8o – sin 8o) / (cos 8o + sin 8o)

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

(cos 8o – sin 8o) / (cos 8o + sin 8o) = (1 – tan 8o) / (1 + tan 8o)

= (1 – tan 8o) / (1 + 1×tan 8o)

= (tan 45o – tan 8o) / (1 + tan 45o ×tan 8o)

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

So,

(tan 45o – tan 8o) / (1 + tan 45o ×tan 8o) = tan (45o – 8o)

= tan 37o

= RHS

∴ LHS = RHS

Hence proved.

12. Prove that:

(i)

(ii)

(iii)

Solution:

(i)

= sin 90o

= 1

= RHS

∴ LHS = RHS

Hence proved.

(ii)

= sin 60o

= √3/2

= RHS

∴ LHS = RHS

Hence proved.

(iii)

= sin 90o

= 1

= RHS

∴ LHS = RHS

Hence proved.

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

Solution:

Let us consider LHS:

(tan 69o + tan 66o) / (1 – tan 69o tan 66o)

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

Here, A = 69o and B = 66o

So,

(tan 69o + tan 66o) / (1 – tan 69o tan 66o) = tan (69 + 66)o

= tan 135o

= – tan 45o

= – 1

= RHS

∴ LHS = RHS

Hence proved.

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

(ii) If tan A = m/(m–1) and tan B = 1/(2m – 1), then prove that A – B = π/4

Solution:

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

Given:

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

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

= [(5/6) + (1/11)] / [1 – (5/6) × (1/11)]

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

= 61/61

= 1

= tan 45or tan π/4

So, tan (A + B) = tan π/4

∴ (A + B) = π/4

Hence proved.

(ii) If tan A = m/(m–1) and tan B = 1/(2m – 1), then prove that A – B = π/4

Given:

tan A = m/(m–1) and tan B = 1/(2m – 1)

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

= (2m2 – m – m + 1) / (2m2 – m – 2m + 1 + m)

= (2m2 – 2m + 1) / (2m2 – 2m + 1)

= 1

= tan 45o or tan π/4

So, tan (A – B) = tan π/4

∴ (A – B) = π/4

Hence proved.

15. prove that:
(i) cos2 π/4 – sin2 π/12 = √3/4

(ii) sin(n + 1) A – sin2nA = sin (2n + 1) A sin A

Solution:

(i) cos2 π/4 – sin2 π/12 = √3/4

Let us consider LHS:

cos2 π/4 – sin2 π/12

We know that, cos2A – sinB = cos (A + B) cos (A – B)

So,

cos2 π/4 – sin2 π/12 = cos (π/4 + π/12) cos (π/4 – π/12)

= cos 4π/12 cos 2π/12

= cos π/3 cos π/6

= 1/2 × √3/2

= √3/4

= RHS

∴ LHS = RHS

Hence proved.

(ii) sin(n + 1) A – sin2nA = sin (2n + 1) A sin A

Let us consider LHS:

sin(n + 1) A – sin2nA

We know that, sin2A – sinB = sin (A + B) sin (A – B)

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

So,

sin(n + 1) A – sin2n A = sin ((n + 1) A + nA) sin ((n + 1) A – nA)

= sin (nA +A + nA) sin (nA +A – nA)

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

= sin (2n + 1) A sin A

= RHS

∴ LHS = RHS

Hence proved.

16. Prove that:

(i)

(ii)

(iii)

(iv) sinB = sinA + sin(A-B) – 2sin A cos B sin (A – B)

(v) cosA + cosB – 2 cos A cos B cos (A +B) = sin(A + B)

(vi)

Solution:

(i)

tan A

= RHS

∴ LHS = RHS

Hence proved.

(ii)

Let us consider LHS:

tan A – tan B + tan B – tan C + tan C – tan A

= 0

= RHS

∴ LHS = RHS

Hence proved.

(iii)

= cotB – cotA + cot C – cotB + cotA – cot C

= 0

= RHS

∴ LHS = RHS

Hence proved.

(iv) sinB = sinA + sin(A-B) – 2sin A cos B sin (A – B)

Let us consider RHS:

sin2A + sin(A -B) – 2 sin A cos B sin (A – B)

sin2A + sin (A -B) [sin (A –B) – 2 sin A cos B]

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

So,

sin2A + sin (A -B) [sin A cos B – cos A sin B – 2 sin A cos B]

sin2A + sin (A -B) [-sin A cos B – cos A sin B]

sin2A – 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 – sin (A – B) sin (A + B)

sinA – sinA + sinB

sinB

= LHS

∴ LHS = RHS

Hence proved.

(v) cosA + cosB – 2 cos A cos B cos (A + B) = sin(A + B)

Let us consider LHS:

cos2A + cos2B – 2 cos A cos B cos (A +B)

cos2A + 1 – sin2B – 2 cos A cos B cos (A +B)

1 + cos2A – sin2B – 2 cos A cos B cos (A +B)

We know that, cos2A – sin2B = cos (A +B) cos (A –B)

So,

1 + cos (A +B) cos (A –B) – 2 cos A cos B cos (A +B)

1 + cos (A +B) [cos (A –B) – 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 B – 2 cos A cos B]

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

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

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

So,

1 – cos(A + B)

sin(A + B)

= RHS

∴ LHS = RHS

Hence proved.

(vi)

∴ LHS = RHS

Hence proved.

17. Prove that:
(i) tan 8x – tan 6x – tan 2x = tan 8x tan 6x tan 2x

(ii) tan π/12 + tan π/6 + tan π/12 tan π/6 = 1

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

(iv) tan 13x – tan 9x – tan 4x = tan 13x tan 9x tan 4x

Solution:

(i) tan 8x – tan 6x – tan 2x = tan 8x tan 6x tan 2x

Let us consider LHS:

tan 8x – tan 6x – tan 2x

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

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

So,

tan 8x = (tan 6x + tan 2x) / (1 – tan 6x tan 2x)

By cross-multiplying we get,

tan 8x (1 – tan 6x tan 2x) = tan 6x + tan 2x

tan 8x – tan 8x tan 6x tan2x = tan 6x + tan 2x

Upon rearranging we get,

tan 8x – tan 6x – tan 2x = tan 8x tan 6x tan 2x

= RHS

∴ LHS = RHS

Hence proved.

(ii) tan π/12 + tan π/6 + tan π/12 tan π/6 = 1

We know,

π/12 = 15° and π/6 = 30°

So, we have 15° + 30° = 45°

Tan (15° + 30°) = tan 45°

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

So,

(tan 15o + tan 30o) / (1 – tan 15o tan 30o) = 1

tan 15° + tan 30° = 1 – tan 15° tan 30°

Upon rearranging we get,

tan15° + tan30° + tan15° tan30° = 1

Hence proved.

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

We know 36° + 9° = 45°

So we have,

tan (36° + 9°) = tan 45°

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

So,

(tan 36o + tan 9o) / (1 – tan 36o tan 9o) = 1

tan 36° + tan 9° = 1 – tan 36° tan 9°

Upon rearranging we get,

tan 36° + tan 9° + tan 36° tan 9° = 1

Hence proved.

(iv) tan 13x – tan 9x – tan 4x = tan 13x tan 9x tan 4x

Let us consider LHS:

tan 13x – tan 9x – tan 4x

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

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

So,

tan 13x = (tan 9x + tan 4x) / (1 – tan 9x tan 4x)

By cross-multiplying we get,

tan 13x (1 – tan 9x tan 4x) = tan 9x + tan 4x

tan 13x – tan 13x tan 9x tan 4x = tan 9x + tan 4x

Upon rearranging we get,

tan 13x – tan 9x – tan 4x = tan 13x tan 9x tan 4x

= RHS

∴ 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 – 5 cos x
(ii) 12 cos x + 5 sin x + 4
(iii) 5 cos x + 3 sin (π/6 – x) + 4
(iv) sin x – cos x + 1

Solution:

We know that the maximum value of A cos α + B sin α + C is C + √(A2 +B2),

And the minimum value is C – √(a2 +B2).

(i) 12 sin x – 5 cos x

Given: f(x) = 12 sin x – 5 cos x

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

((-5)2 + 122) ≤ 12 sin x – 5 cos x ≤ ((-5)2 + 122)

(25+144) ≤ 12 sin x – 5 cos x ≤ (25+144)

169 ≤ 12 sin x – 5 cos x ≤ 169

-13 ≤ 12 sin x – 5 cos x ≤ 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 – (122 + 52) ≤ 12 cos x + 5 sin x + 4 ≤ 4 + (122 + 52)

4 – (144+25) ≤ 12 cos x + 5 sin x + 4 ≤ 4 + (144+25)

4 –169 ≤ 12 cos x + 5 sin x + 4 ≤ 4 + 169

-9 ≤ 12 cos x + 5 sin x + 4 ≤ 17

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

(iii) 5 cos x + 3 sin (π/6 – x) + 4 

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

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

f(x) = 5 cos x + 3 sin (π/6 – x) + 4 

= 5 cos x + 3 (sin π/6 cos x – cos π/6 sin x) + 4

= 5 cos x + 3/2 cos x – 33/2 sin x + 4

= 13/2 cos x – 33/2 sin x + 4

So, here A = 13/2, B = – 33/2, C = 4

4 – [(13/2)2 + (-33/2)2] ≤ 13/2 cos x – 33/2 sin x + 4 ≤ 4 + [(13/2)2 + (-33/2)2]

4 – [(169/4) + (27/4)] ≤ 13/2 cos x – 33/2 sin x + 4 ≤ 4 + [(169/4) + (27/4)]

4 – 7 ≤ 13/2 cos x – 33/2 sin x + 4 ≤ 4 + 7

-3 ≤ 13/2 cos x – 33/2 sin x + 4 ≤ 11

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

(iv) sin x – cos x + 1

Given: f(x) = sin x – cos x + 1

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

1 – [(-1)2 + 12] ≤ sin x – cos x + 1 ≤ 1 + [(-1)2 + 12]

1 – (1+1) ≤ sin x – cos x + 1 ≤ 1 + (1+1)

1 – 2 ≤ sin x – cos x + 1 ≤ 1 + 2

Hence, the maximum and minimum values of f(x) are 1 – 2 and 1 + 2 respectively.

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

(i) √3 sin x – cos x

(ii) cos x – sin x

(iii) 24 cos x + 7 sin x

Solution:

(i) √3 sin x – cos x

Let f(x) = √3 sin x – cos x

Dividing and multiplying by √((√3)2 + 12) i.e. by 2

f(x) = 2(√3/2 sin x – 1/2 cos x)

Sine expression:

f(x) = 2(cos π/6 sin x – sin π/6 cos x) (since, √3/2 = cos π/6 and 1/2 = sin π/6)

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

f(x) = 2 sin (x – π/6)

Again,

f(x) = 2(√3/2 sin x – 1/2 cos x)

Cosine expression:

f(x) = 2(sin π/3 sin x – cos π/3 cos x)

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

f(x) = -2 cos(π/3 + x)

(ii) cos x – sin x

Let f(x) = cos x – sin x

Dividing and multiplying by √(12 + 12) i.e. by √2,

f(x) = √2(1/√2 cos x – 1/√2 sin x)

Sine expression:

f(x) = √2(sin π/4 cos x – cos π/4 sin x) (since, 1/√2 = sin π/4 and 1/√2 = cos π/4)

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

f(x) = √2 sin (π/4 – x)

Again,

f(x) = √2(1/√2 cos x – 1/√2 sin x)

Cosine expression:

f(x) = 2(cos π/4 cos x – sin π/4 sin x)

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

f(x) = √2 cos (π/4 + x)

(iii) 24 cos x + 7 sin x

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

Dividing and multiplying by √((√24)2 + 72) = √625 i.e. by 25,

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

Sine expression:

f(x) = 25(sin α cos x + cos α sin x) where, sin α = 24/25 and cos α = 7/25

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

f(x) = 25 sin (α + x)

Cosine expression:

f(x) = 25(cos α cos x + sin α sin x) where, cos α = 24/25 and sin α = 7/25

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

f(x) = 25 cos (α – x)

3. Show that Sin 100o – Sin 10o is positive.

Solution:

Let f(x) = sin 100° – sin 10°

Dividing And multiplying by √(12 + 12) i.e. by √2,

f(x) = √2(1/√2 sin 100o – 1/√2 sin 10o)

f(x) = √2(cos π/4 sin (90+10)o – sin π/4 sin 10o) (since, 1/√2 = cos π/4 and 1/√2 = sin π/4)

f(x) = √2(cos π/4 cos 10o – sin π/4 sin 10o)

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

f(x) = √2 cos (π/4 + 10o)

∴ f(x) = √2 cos 55°

4. Prove that (2√3 + 3) sin x + 2√3 cos x lies between – (2√3 + √15) and (2√3 + √15).

Solution:

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

Here, A = 2√3, B = 2√3 + 3 and C = 0

– √[(2√3)2 + (2√3 + 3)2] ≤ (2√3 + 3) sin x + 2√3 cos x ≤ √[(2√3)2 + (2√3 + 3)2]

– √[12+12+9+12√3] ≤ (2√3 + 3) sin x + 2√3 cos x ≤ √[12+12+9+12√3]

– √[33+12√3] ≤ (2√3 + 3) sin x + 2√3 cos x ≤ √[33+12√3]

– √[15+12+6+12√3] ≤ (2√3 + 3) sin x + 2√3 cos x ≤ √[15+12+6+12√3]

We know that (12√3 + 6 < 12√5) because the value of √5 – √3 is more than 0.5

So if we replace, (12√3 + 6 with 12√5) the above inequality still holds.

So by rearranging the above expression √(15+12+12√5)we get, 2√3 + √15

– 2√3 + √15 ≤ (2√3 + 3) sin x + 2√3 cos x ≤ 2√3 + √15

Hence proved.

RD Sharma Solutions for Class 11 Maths Chapter 7: Download PDF

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

Download PDF: RD Sharma Solutions for Class 11 Maths Chapter 7–Values of Trigonometric Functions at Sum or Difference of Anglesma Solutions for Class 11 Maths Chapter 1–Sets PDF

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About RD Sharma

RD Sharma isn’t the kind of author you’d bump into at lit fests. But his bestselling books have helped many CBSE students lose their dread of maths. Sunday Times profiles the tutor turned internet star
He dreams of algorithms that would give most people nightmares. And, spends every waking hour thinking of ways to explain concepts like ‘series solution of linear differential equations’. Meet Dr Ravi Dutt Sharma — mathematics teacher and author of 25 reference books — whose name evokes as much awe as the subject he teaches. And though students have used his thick tomes for the last 31 years to ace the dreaded maths exam, it’s only recently that a spoof video turned the tutor into a YouTube star.

R D Sharma had a good laugh but said he shared little with his on-screen persona except for the love for maths. “I like to spend all my time thinking and writing about maths problems. I find it relaxing,” he says. When he is not writing books explaining mathematical concepts for classes 6 to 12 and engineering students, Sharma is busy dispensing his duty as vice-principal and head of department of science and humanities at Delhi government’s Guru Nanak Dev Institute of Technology.