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Class 11 RD Sharma Solutions – Chapter 7 Trigonometric Ratios of Compound Angles – Exercise 7.1 | Set 1
  • Last Updated : 04 May, 2021

Question 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 that

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

As we know, cos A = (1 – sin2A) and sin B = (1 – cos2B), where 0 <A, B < π/2

Now we find the value of cosA and sinB

cos A = √(1 – sin2A)

= √(1 – (4/5)2

= √(1 – (16/25))

= √((25 – 16)/25)

= √(9/25)

= 3/5

sin B = √(1 – cos2B)

= √(1 – (5/13)2)

= √(1 – (25/169))

= √(169 – 25)/169)

= √(144/169)

= 12/13

(i) sin (A + B)

sin (A + B) = sinA cosB + cosA sinB



= 4/5 × 5/13 + 3/5 x 12/13

= 20/65 + 36/65 

= (20 + 36)/65

= 56/65

(ii) cos (A + B)

cos (A + B) = cosA cosB – sinA sinB

= 3/5 × 5/13 –  4/5 x 12/13

= 15/65 – 48/65

= -33/65

(iii) sin (A – B)

sin (A – B) = sinA cosB – cosA sinB

= 4/5 × 5/13 – 3/5 x 12/13

= 20/65 – 36/65

= – 16/65

(iv) cos (A – B)

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

= 3/5 x 5/13 + 4/5 x 12/13

= 15/65 + 48/65

= 63/65

Question 2. (a) If sinA = 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)

Solution:

We have,

sinA = 12/13 and sinB = 4/5, where π/2 < A < and 0 < B < π/2

As we know, cosA = – √(1 – sin2A) and cosB = √(1 – sin2B)

Now we find the value of cosA and cosB

cosA = – √(1 – sin2A)

= – √(1 – (12/13)2)

= – √(1 – 144/169)

= – √((169 – 144)/169)

= – √(25/169)

= – 5/13

cosB = √(1 – sin2B)

= √(1 – (4/5)2)

= √(1 – 16/25)

= √((25 – 16)/25)

= √(9/25)

= 3/5

(i) sin (A + B)

Since, sin (A + B) = sinA cosB + cosA sinB

= 12/13 x 3/5 + (-5/13) x 4/5

= 36/65 – 20/65

= 16/65

(ii) cos (A + B)

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

= – 5/13 x 3/5 – 12/13 x 4/5

= – 15/65 – 48/65

= – 63/65

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

Solution:

We have,

sinA = 3/5, cosB = -12/13, where A and B, both lie in second quadrant.

As we know cosA = – √(1- sin2A) and sinB = √(1 – cos2B)

Now we find the value of cosA and sinB

cos A = – √(1 – sin2A)

= -√(1 – (3/5)2)

= -√(1 – 9/25)

= – √((25 – 9)/25)

= – √(16/25)

= – 4/5

sinB = √(1 – cos2B)

= √(1 – (-12/13)2)

= √(1 – 144/169)

= √((169 – 144)/169)

= √(25/169)

= 5/13

We need to find the value of sin (A + B)

Since, sin (A + B) = sinA cosB + cosA sinB

= 3/5 × (-12/13) + (-4/5) x 5/13

= -36/65 – 20/65

= -56/65

Question 3. If cosA = -24/25 and cosB = 3/5, where π < A < 3π/2 and 3π/2 < B < 2π, find the following:

(i) sin(A + B) 

(ii) cos(A + B)

Solution:

We have,

cosA = -24/25 and cosB = 3/5, where π < A < 3π/2 and 3π/2 < B < 2π

As we know that A is present in third quadrant, B is 

present in fourth quadrant, so the sine function is Negative.

By using the formulas, sinA = √(1 – cos2A) and sinB = -√(1 – cos2B)

We find the value of sinA and sinB

sinA = – √(1 – cos2A)

= – √(1 – (-24/25)2)

= – √(1 – 576/625)

= – √((625 – 576)/625)

= – √(49/625)

= – 7/25

sinB = – √(1 – cos2B)

= – √(1 – (3/5)²)

= – √(1 – 9/25) 

= – √((25 – 9)/25)

= – √(16/25)

= – 4/5

(i) sin (A + B)

Since, sin (A + B) = sinA cosB + cosA sinB

= -7/25 x 3/5 + (-24/25) x (-4/5)

= -21/125 + 96/125

= 75/125

= 3/5

(ii) cos (A + B)

Since, cos (A + B) = cosA cosB – sinA sinB

= (-24/25) x 3/5 – (-7/25) × (-4/5)

= – 72/125 – 28/125

=- 100/125

= – 4/5

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

Solution:

We have,

tanA = 3/4 and cosB = 9/41, where π < A < 3π/2 and 0 < B < π/2

As we know that, A is present in third quadrant, B is present in first quadrant

So, tan and sin functions are positive.

Now by using the formula,

sinB = √(1 – cos2B)

We find the value of sin B.

sinB = √(1 – cos2B)

= √(1 – (9/41)2)

= √(1 – 81/1681)

= √((1681 – 81)/1681)

= √(1600/1681)

= 40/41

As we know that, tanB = sinB/cosB, so

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

= 40/9

Since, tan(A + B) = (tanA + tanB)/(1 – tanA tanB), so

= (3/4 + 40/9)/(1 – 3/4 x 40/9)

= (187/36)/(1 – 120/36)

= (187/36)/((36 – 120)/36)

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

= -187/84

Hence, tan(A + B) = -187/84

Question 5. If sinA = 1/2, cosB = 12/13, where π/2 < A < π and 3π/2 < B < 2π, find tan(A – B).

Solution:

We have,

sinA = 1/2, cosB = 12/13, where π/2 < A < π and 3π/2 < B < 2π

As we know that, A is present in second quadrant and B is present in fourth quadrant.

So, the sine function is positive, cosine and tan functions are negative in second quadrant

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

Now by using the following formulas,

cosA = -√(1 – sin2A) and sinB = -√(1 – cos2B)

We find the value of cosA and sinB

cosA = – √(1 – sin2A)

= – √(1 – (1/2)2)

= – √(1 – 1/4)

= – √((4 – 1)/4)

= – √(3/4)

= – √3/2

sinB = – √(1 – cos2B)

= – √(1 – (12/13)2)

= – √(1 – 144/169)

= – √((169 – 144)/169)

= – √(25/169)

= – 5/13

As we know, tanA = sinA/cosA and tanB = sinB / cosB

tanA = (1/2)/(-√3/2) = -1/√3 and

tanB = (-5/13)/(12/13) = -5/12

Since, tan (A – B) = (tanA – tanB) / (1 + tanA tanB), so

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

Hence, tan (A – B) = (5√3 – 12)/(5 + 12√3)

Question 6. If sinA = 1/2, cosB = √3/2, where π/2 < A < π and 0 < B < π/2, find the following:

(i) tan(A + B) 

(ii) tan(A – B) 

Solution:

We have,

SinA = 1/2 and cosB = √3/2, where π/2 < A < π and 0 < B < π/2

As we know that, A is in second quadrant, B is in first quadrant.

So, all functions are positive in first quadrant and sine function is positive, 

cosine and tan functions are negative in the second quadrant. 

So, by using the following formulas,

cosA = – √(1 – sin2A) and sinB = √(1 – cos2B)

We find the value of cosA and sinB

cosA = – √(1 – sin2A)

= – √(1 – (1/2)2)

= – √(1 – 1/4)

= – √((4 – 1)/4)

= – √(3/4)

= – √3/2

sinB = √(1 – cos2B)

= √(1 – (√3/2)2)

= √(1 – 3/4)

= √((4 – 3)/4)

= √(1/4)

= 1/2

As we know that, tanA = sinA / cosA and tanB = sinB / cosB

So, tanA = (1/2)/(-√3/2) = -1/√3 and

tanB = (1/2)/(√3/2) = 1/√3

(i) Since, tan(A + B) = (tanA + tanB)/(1 – tanA tanB), so

= (-1/√3 + 1/√3)/(1 – (-1/√3) × 1/√3)

= 0/(1 + 1/3)

= 0

Hence, tan(A + B) = 0

(ii) Since, tan(A – B) = (tanA – tanB)/(1 + tanA tanB), so

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

= ((-2/√3)/(1 – 1/3)

= ((-2/√3)/(3 – 1)/3)

= ((-2/√3)/2/3)

= -√3

Hence, tan(A – B) = -√3

Question 7. Evaluate the following:

(i) sin 78° cos 18⁰- cos 78° sin 18⁰ 

(ii) cos 47° cos 13⁰ – sin 47⁰ sin 13⁰ 

(iii) sin 36° cos 9⁰+ cos 36° sin 9⁰ 

(iv) cos 80° cos 20⁰+ sin 80° sin 20⁰ 

Solution:

(i) sin 78° cos 18° – cos 78° sin 18°

Since, sinAcosB – cosAsinB = sin(A – B)

So

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

= sin 60°

= √3/2

(ii) cos 47° cos 13° – sin 47° sin 13°

Since, cosA cosB – sinA sinB = cos(A + B)

So, cos 47° cos 13° – sin 47° sin 13° = cos (47 + 13)°

= cos 60°

= 1/2

(iii) sin 36° cos 9° + cos 36° sin 9°

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

So, sin 36° cos 9° + cos 36° sin 9° = sin (36 + 9)°

= sin 45°

= 1/√2

(iv) cos 80° cos 20° + sin 80° sin 20⁰

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

So, cos 80° cos 20° + sin 80° sin 20° = cos (80 – 20)°

= cos 60°

= 1/2

Question 8. If cosA = -12/13 and cotB = 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:

We have,

cosA = -12/13 and cotB= 24/7

It is given that, A lies in second quadrant, B in the third quadrant.

So, sine function is positive in second quadrant and both sine and cosine 

functions are negative in third quadrant.

So, by using the following formulas,

sinA = √(1 – cos2A), sinB = 1/√(1 + cot2B) and cosB = -√(1 – sin2B),

We find the value of sinA and sinB

sinA = √(1 – cos2A)

= √(1 -(-12/13)2)

= √(1 – 144/169) 

= √((169 – 144)/169)

= √(25/169)

= 5/13

sinB = -1/√(1 + cot2B)

= -1/√(1 + (24/7)2)

= -1/√(1 + 576/49)

= -1/√((49 + 576)/49)

= -1/√(625/49)

= -1/(25/7)

= -7/25

cosB = -√(1 – sin2B)

= -√(1 -(-7/25)2)

= -√(1 – (49/625))

= -√((625 – 49)/625)

= -√(576/625)

= -24/25

(i) sin(A + B)

Since, sin(A + B) = sinA cosB + cosA sinB

So, 

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

= -120/325 + 84/325

= -36/325

(ii) cos(A + B)

Since, cos(A + B) = cosA cosB – sinA sinB

So, 

= -12/13 x (-24/25) – (5/13) x (-7/25)

= 288/325 + 35/325

= 323/325

(iii) tan(A + B)

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

So, 

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

=-36/323

Question 9. Prove that: cos 7π/12 + cos π/12 = sin 5π/12 – sin π/12

Solution:

As we know that, 

7π/12 = 105°, π/12 = 15°, 5π/12 = 75°

Now, L.H.S = cos 105° + cos 15°

= cos (90° + 15°) + sin (90° – 75°)

= -sin 15° + sin 75°

= sin 75° – sin 15°

= RHS

So, LHS = RHS

Hence proved.

Question 10. Prove that: (tanA + tanB) / (tanA – tanB) = sin(A + B) / sin(A – B)

Solution:

Let solve, LHS = (tanA + tanB) / (tanA – tanB)

\frac{\frac{sinA}{cosA}+\frac{sinB}{cosB}}{\frac{sinA}{cosA}-\frac{sinB}{cosB}}

\frac{\frac{sinAcosB+cosAsinB}{cosAcosB}}{\frac{sinAcosB-cosAsinB}{cosAcosB}}

As we know that, 

sin(A ± B) = sinA cosB ± cosA sinB 

So, 

= {sin(A + B)} / {sin(A – B)}

= RHS

So, LHS = RHS

Hence proved.

Question 11. Prove that:

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

(ii) (cos 9⁰+ sin 9″) / (cos 9″- sin 9°) = tan 54°

(iii) (cos 8° – sin 8°) / (cos 8° + sin 8°) = tan 37⁰

Solution:

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

Let solve, LHS = (cos 11° + sin 11°)/(cos 11° – sin 11°)

Now divide the numerator and denominator by cos 11° we get,

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

= (1+ tan 11°)/(1 – 1 x tan 11°)

= (tan 45° + tan 11°)/(1 – tan 45° x tan 11°)

As we know that, 

tan(A + B) = (tanA + tanB)/(1 – tanA tanB) 

So,

(tan 45° + tan 11°)/(1 – tan 45° x tan 11°) = tan (45° + 11°)

= tan 56°

= RHS

So, LHS = RHS

Hence proved.

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

Let solve, LHS = (cos 9° + sin 9°)/(cos 9° – sin 9°)

Now divide the numerator and denominator by cos 9° we get, 

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

= (1 + tan 9°) / (1 – 1 x tan 9°)

= (tan 45° + tan 9°)/(1 – tan 45° x tan 9°)

As we know that

tan(A + B) = (tanA + tanB)/(1 – tanA tanB)

So, 

(tan 45° + tan 9°)/(1 – tan 45° x tan 9°) = tan (45° + 9°)

= tan 54°

= RHS

So, LHS = RHS

Hence proved.

(iii) (cos 8° – sin 8°)/(cos 8° + sin 8°) = tan 37⁰

Let solve, LHS = (cos 8° – sin 8°) / (cos 8° + sin 8°)

Now divide the numerator and denominator by cos 8° we get,

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

= (1 – tan 8°)/(1 + 1 x tan 8°)

= (tan 45° – tan 8°) / (1 + tan 45° x tan 8″)

As we know that

tan(A + B) = (tanA + tanB)/(1 – tanA tanB)

So,

(tan 45° – tan 8°)/(1 + tan 45° x tan 8°) = tan (45° – 8°)

= tan 37°

 = RHS

LHS = RHS

Hence proved.

Question 12. Prove that:

(i) sin (π/3 – x) cos(π/6 + x) + cos (π/3 – x) sin(π/6 + x) = 1

(ii) sin (4π/9 + 7) cos(π/9 + 7) – cos (4π/9 + 7) sin (π/9 + 7) = √3/2

(iii) sin ( 3π/8 – 5) cos (π/8 + 5) + cos (3π/8 – 5) sin(π/8 + 5) = 1

Solution:

(i) sin (π/3 – x) cos(π/6 + x) + cos (π/3 – x) sin(π/6 + x) = 1

Let solve, LHS = sin (π/3 – x) cos(π/6 + x) + cos (π/3 – x) sin(π/6 + x)

As we know that

sin(A + B) = sinA cosB + cosA sinB

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

= sin ((2π + π)/6)

= sin (π/2) 

= sin 90°

= 1

= RHS

LHS = RHS

Hence proved.

(ii) sin (4π/9 + 7) cos(π/9 + 7) – cos (4π/9 + 7) sin (π/9 + 7) = √3/2

Let solve, LHS = sin (4π/9 + 7) cos(π/9 + 7) – cos (4π/9 + 7) sin (π/9 + 7)

As we know that

sin(A – B) = sinA cosB – cosA sinB

sin (4π/9 + 7) cos(π/9 + 7) – cos (4π/9 + 7) sin (π/9 + 7) = sin (4π/9 + 7 – π/9 – 7)

= sin (3π/9)

= sin (π/3)

= sin 60° 

= √3/2 

= RHS

LHS = RHS

Hence proved.

(iii) sin ( 3π/8 – 5) cos (π/8 + 5) + cos (3π/8 – 5) sin(π/8 + 5) = 1

Let solve, LHS = sin ( 3π/8 – 5) cos (π/8 + 5) + cos (3π/8 – 5) sin(π/8 + 5) 

As we know that

sin(A + B) = sinA cosB + cosA sinB

sin ( 3π/8 – 5) cos (π/8 + 5) + cos (3π/8 – 5) sin(π/8 + 5) = sin (3π/8 – 5 + π/8 + 5)

= sin ((3π + π)/8)

= sin (4π/8)

= sin (π/2)

= sin 90°

= 1

= RHS

LHS = RHS

Hence proved.

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

Solution:

Let solve, LHS = (tan 69°+tan 66°)/(1-tan 69° tan 66°)

As we know that

tan(A + B) = (tanA + tanB) / (1 – tanA tanB)

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

= tan (69 +66)°

= tan 135⁰

= – tan 45º

= -1

= RHS

LHS = RHS

Hence proved.

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

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

Solution:

(i) We have,

tanA = 5/6 and tanB = 1/11

As we know that

tan(A + B) = (tanA + tanB)/(1 – tanA tanB)

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

= (55 + 6)/(66 – 5)

= 61/61

= 1

= tan 45° or tan π/4

So, tan(A + B) = tan π/4

(A + B) = π/4

Hence proved.

(ii) We have,

tanA = m/(m – 1) and tanB = 1/(2m – 1)

As we know that

tan(A – B) = (tanA – tanB) / (1 + tanA tanB)

\frac{\frac{m}{m-1}-\frac{1}{2m-1}}{1+\frac{m}{m-1}\frac{1}{2m-1}}

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

= (2m2 – 2m + 1)/(2m2 – 2m + 1)

= 1

= tan 45° or tan π/4

So, tan(A – B) = tan π/4

(A – B) = π/4

Hence proved.

Question 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 solve, LHS = cos2 π/4 sin2 π/12

As we know that

cos2 A – 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 x √3/2

= √3/4 = RHS

LHS = RHS 

Hence proved.

(ii) sin2 (n + 1)A – sin2nA = sin (2n + 1)A sinA

Let solve, LHS = sin2 (n+1)A – sin2nA

As we know that

sin2A – sin2B = sin(A + B) sin(A – B)

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

So,

sin2(n + 1)A – sin2nA = sin((n + 1)A + nA) sin((n + 1)A – nA)

= sin(nA + A + nA) sin (nA + A – nA)

= sin(2nA + A) sinA

= sin(2n + 1)A sinA 

= RHS

LHS = RHS

Hence proved.

Question 16. Prove that:

(i) {sin(A + B) + sin(A – B)}/{cos(A + B) + cos(A – B)} = tanA

Solution:

Prove: {sin(A + B) + sin(A – B)}/{cos(A + B) + cos(A – B)} = tanA

Proof:

Let solve, LHS = {sin(A + B) + sin(A – B)}/{cos(A + B) + cos(A – B)}

As we know that, sin (A ± B) = sinA cosB ± cosA sinB and cos(A ± B) = cosA cosB ± sinA sinB

So,

\frac{sin(A+B)+sin(A-B)}{cos(A+B)+cos(A-B)}

\frac{sinAcosB+cosAsinB+sinAcosB-cosAsinB}{cosAcosB-sinAsinB+cosAcosB+sinAsinB}      

= (2cosA cosB)(2cosA cosB)

= tan A

= RHS

LHS = RHS

Hence proved.

(ii) {sin(A – B)/cosA.cosB} + {sin(B – C)/cosB.cosC} + {sin(C – A)/cosC.cosA} = 0

Solution:

Prove: {sin(A – B)/cosA.cosB} + {sin(B – C)/cosB.cosC} + {sin(C – A)/cosC.cosA} = 0

Proof:

Let solve, LHS = {sin(A – B)/cosA.cosB} + {sin(B – C)/cosB.cosC} + {sin(C – A)/cosC.cosA}

As we know that

sin(A – B) = sinA cosB – cosA sinB

So, 

= {sin(A – B)/cosA.cosB} + {sin(B – C)/cosB.cosC} + {sin(C – A)/cosC.cosA}

= (sinA cosB – cosA sinB)/(cosA cosB) + (sinB cosC – cosB sinC)/(cosB cosC) + 

   (sinC cosA – cosC sinA)/(cosC cosA)

= (sinA cosB)/(cosA sinB) – (cosA sinB)/(cosB cosC) + (sinB cosC)/(cosB cosC) – 

  (cosB sinC)/(cosB cosC) + (sinC cosA)/(cosC cosA)

= tanA – tanB + tanB – tanC + tanC – tanA

= 0

= RHS

LHS = RHS

Hence proved.

(iii) {sin(A – B)/sinA.sinB} + {sin(B – C)/sinB.sinC} + {sin(C – A)/sinC.sinA} = 0

Solution:

Prove: {sin(A – B)/sinA.sinB} + {sin(B – C)/sinB.sinC} + {sin(C – A)/sinC.sinA} = 0

Proof:

Let solve, LHS = {sin(A – B)/sinA.sinB} + {sin(B – C)/sinB.sinC} + {sin(C – A)/sinC.sinA}

As we know that

sin(A – B) = sinA cosB – cosA sinB

= {sin(A – B)/sinA.sinB} + {sin(B – C)/sinB.sinC} + {sin(C – A)/sinC.sinA}

= (sinA cosB – cosA sinB)/(sinA sinB) + (sinB cosC – cosB sinC)/(sinB sinC) + 

    (sinC cosA- cosC sinA)/(sinC sinA)

= (sinA.cosB)/(sinA.sinB) – (cosA.sinB)/(sinA.sinB) + (sinB.cosC)/(sinB.sinC) – 

   (cosB.sinC)/(sinB.sinC) +(sinC.cosA)/(sinC.sinA) – (cosC.sinA)/(sinC.sinA)

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

= 0

= RHS

LHS = RHS

Hence proved.

(iv) sin2B = sin2A + sin2(A – B) – 2sinA cosB sin(A – B)

Solution:

Prove: sin2B = sin2A + sin2(A – B) – 2sinA cosB sin(A – B)

Proof:

Let’s solve RHS = sin2A + sin2(A – B) – 2 sinA cosB sin(A – B)

= sin2A+ sin(A -B)[sin(A – B) – 2 sinA cosB]

As we know that

sin(A – B) = sinA cosB –  cosA sinB

So,

= sin2A + sin(A -B)[sinA cosB – cosA sinB – 2 sinA cosB]

= sin2A + sin(A-B)[-sinA cosB – cosA sinB] 

= sin2A – sin(A -B)[sinA cosB + cosA sinB]

As we know that

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

So,

= sin2A – sin(A – B) sin(A + B).

= sin2A – sin2A + sin2B

= sin2B

= LHS

LHS = RHS

Hence proved.

(v) cos2A + cos2B – 2 cosA cosB cos(A + B) = sin2(A + B)

Solution:

Prove: cos2A + cos2B – 2 cosA cosB cos(A + B) = sin2(A + B)

Proof:

Let solve LHS = cos2A + cos2B – 2 cosA cosB cos(A + B)

= cos2A + 1 – sin2B – 2 cosA cosB cos(A + B)

= 1 + cos2A – sin2B – 2 cosA cosB cos(A + B)

As we know that cos2A – sin2B = cos(A + B) cos(A – B)

So,

= 1 + cos(A + B) cos(A – B) – 2 cosA cosB cos(A + B)

= 1 + cos(A + B)[cos(A – B) – 2 cosA cosB]

Also, 

cos(A – B) = cosA cosB + sinA sinB

So,

= 1 + cos(A +B)[cosA cosB + sinA sinB – 2 cosA cosB]

= 1 + cos(A +B)[-cosA cosB + sinA sinB]

= 1 cos(A +B)[cosA cosB – sinA sinB]

Also,

cos(A + B) = cosA cosB – sinA sinB

So,

1 – cos2(A + B) = sin2(A + B) = RHS

LHS = RHS

Hence proved.

(vi) tan(A + B)/cot(A – B) = (tan2A – tan2B)/(1 – tan2A tan2B)

Solution:

Prove: tan(A + B)/cot(A – B) = (tan2A – tan2B)/(1 – tan2A tan2B)

Proof:

Let solve LHS = tan(A + B)/cot(A – B)

As we knwo that

tan (A ± B) = (tanA ± tanB) / (1 ± tanA tanB)

So, 

 \frac{tan(A+B)}{1/tan(A-B)}=\frac{\frac{tanA+tanB}{1-tanAtanB}}{1/\frac{tanA-tanB}{1+tanAtanB}}

\frac{tanA+tanB}{1-tanAtanB}\frac{tanA-tanB}{1+tanAtanB}

We know that, (x+y) (x – y) = x2– y2

So,

\frac{tanA+tanB}{1-tanAtanB}\frac{tanA-tanB}{1+tanAtanB}=\frac{tan^2A-tan^2B}{1-tan^2Atan^2B}    

= RHS

LHS = RHS 

Hence proved.

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