In Exercise 7.1 of Chapter 7, we shall discuss problems based on values of trigonometric functions at sum or difference and a few theorems. Students wishing to clear their doubts pertaining to this exercise can utilise the RD Sharma Class 11 Solutions. The solutions are formulated in a comprehensive manner to make it easy for students to solve. The links to the solutions of this exercise can be accessed in the RD Sharma Class 11 Maths PDFs, which are available in the below-mentioned links.
RD Sharma Solutions for Class 11 Maths Exercise 7.1 Chapter 7 – Values of Trigonometric Functions at Sum or Difference of Angles
Also, access another exercise of RD Sharma Solutions for Class 11 Maths Chapter 7 – Values of Trigonometric Functions at Sum or Difference of Angles
Access answers to RD Sharma Solutions for Class 11 Maths Exercise 7.1 Chapter 7 – Values of Trigonometric Functions at Sum or Difference of Angles
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 the 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 the 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 the third quadrant, and B is in the 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 the third quadrant, and B is in the first quadrant.
So, the 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 the second quadrant, and B is in the fourth quadrant.
In the second quadrant, the sine function is positive, and the 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 the second quadrant, and B is in the first quadrant.
In the second quadrant, the sine function is positive. The cosine and tan functions are negative.
In the 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 the second quadrant and B in the third quadrant.
In the second quadrant, the 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, and 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 45o × tan 11o)
We know that tan (A+B) = (tan A + tan B) / (1 – tan A tan B)
So,
(tan 45o + tan 11o) / (1 – tan 45o × 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, and 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, and 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 45o or 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) sin2 (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 – sin2 B = 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) sin2 (n + 1) A – sin2nA = sin (2n + 1) A sin A
Let us consider LHS.
sin2 (n + 1) A – sin2nA
We know that, sin2A – sin2 B = sin (A + B) sin (A – B)
Here, A = (n + 1) A and B = nA
So,
sin2 (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) sin2 B = sin2 A + sin2 (A-B) – 2sin A cos B sin (A – B)
(v) cos2 A + cos2 B – 2 cos A cos B cos (A +B) = sin2 (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) sin2 B = sin2 A + sin2 (A-B) – 2sin A cos B sin (A – B)
Let us consider RHS.
sin2A + sin2 (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)
sin2 A – sin2 A + sin2 B
sin2 B
= LHS
∴ LHS = RHS
Hence proved.
(v) cos2 A + cos2 B – 2 cos A cos B cos (A + B) = sin2 (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 – cos2 (A + B)
sin2 (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.
Comments