# RD Sharma Solutions for Class 11 Chapter 7 - Values of Trigonometric Functions at Sum or Difference of Angles Exercise 7.1

In Exercise 7.1 of Chapter 7, we shall discuss problems based on values of trigonometric functions at sum or difference and 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 interesting for students to solve. The links to the solutions of this exercise can be accessed in the RD Sharma Class 11 Maths pdf which is available in the below-mentioned links.

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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 + 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 Â

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.