Inter 1st Year Maths IA-Compound Angles Study Material PDF

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COMPOUND ANGLES Definitions and Formulae : 1.

The algebraic sum of two or more angles is called a compound a compound angle. i.e. A + B, A – B, A + B + C, A + B – C, A – B + C, B + C – A, ……. etc. are called compound angles.

2.

If A and B are any two angles then i) Sin(A + B) = sin A cos B + cos A sin B. ii) Sin(A – B) = sin A cos B – cos A sin B iii) cos(A + B) = cos A cos B – sin A sin B iv) cos(A – B) = cos A cos B + sin A sin B.

3.

If A, B, A + B, A – B are not odd multiples of

i) tan(A + B) =

ii) tan(A – B) =

4.

tan A + tan B . 1 − tan A tan B tan A − tan B . 1 + tan A tan B

If A, B, A + B and A – B are not integral multiples of

i) cot(A + B) =

ii) cot(A – B) =

5.

π/2 then

cot A cot B − 1 cot B + cot A cot A cot B + 1 . cot B − cot A

i) sin(A + B) + sin(A – B) = 2 sin A cos B ii) sin(A + B) – sin(A – B) = 2 cos A sin B iii) cos(A + B) + cos(A – B) = 2 cos A cos B iv) cos(A + B) – cos(A – B) = –2 sin A sin B v) cos(A – B) – cos(A + B) = 2sin A sin B.

6.

i) sin(A + B) sin(A – B) = sin

2

A – sin

= cos ii) cos(A + B) cos(A – B) = cos

2

iv) cot(A + B) cot(A – B) =

2

B

B – cos

– sin

= cos

iii) tan(A + B) tan(A – B) =

2

2

2

2

B

B – sin

2

tan 2 A − tan 2 B 1 − tan2 A tan 2 B cot 2Acot 2B −1 2

A

2

cot B − cot A

A

π, then

v) tan(45

o

+

θ) =

cos θ + sin θ cos θ − sin θ

= cot(45

vi) tan(45

o

–

θ) =

o

7.

o

+

1 + tan θ 1 −tan θ

cos θ − sin θ cos θ + sin θ

= cot(45

and tan(45

θ) =

–

o

θ) . tan(45

o

1 − tan θ 1 +tan θ

+

θ) =

–

θ) = 1.

i) sin(A + B + C) =

∑(cos A cos B cos C) – sin A sin B sin C

ii) cos(A + B + C) = cos A cos B cos C –

iii) tan(A + B + C) =

∑(sin A sin B sin C)

∑ (tanA)− π(tanA) 1 − ∑ (tanAtanB) VSAQ’S

Simplify the following 1. cos 100° ⋅ cos 40° + sin 100° ⋅ sin 40° Sol. L.H.S. = = cos 100° ⋅ cos 40° + sin 100° ⋅ sin 40° 1 = cos (100° – 40°) = cos 60° = = R.H.S. 2

π  π  2. tan  + θ  ⋅ tan  − θ  4  4  π π tan + tan θ tan − tan θ 4 4 Sol. ⋅ π π 1 − tan tan θ 1 + tan tan θ 4 4 + θ − θ (1 tan ) (1 tan ) = ⋅ =1 (1 − tan θ) (1 + tan θ ) 3. tan 75° + cot 75° Sol. tan 75° = 2 + 3

cot 75 ° = 2 − 3 ∴ tan 75° + cot 75° = 2 + 3 + 2 − 3 = 4

4. Express Sol.

( 3 cos 25 ° +sin 25 °) as a sine of an angle. 2

( 3 cos 25° + sin 25° ) 2

3 1 cos 25 ° + sin 25° 2 2 = sin 60° cos 25° + cos 60 ° sin 25 ° =

= sin(60° + 25° ) = sin 85° 5. tan θ in terms of tan α, if sin(θ + α) = cos (θ + α). Sol. sin(θ + α) = cos (θ + α) sin( θ + α) =1 cos(θ + α )

tan(θ + α ) = 1 tan θ + tan α =1 1 − tan θ tan α tan θ + tan α = 1 − tan θ tan α tan θ + tan θ tan α = 1− tan α tan θ[1 + tan α ] = 1− tan α ∴tan θ =

1− tan α 1+ tan α

cos11 ° + sin11° and θ is in the third quadrant, find θ. cos11 ° − sin11° cos11 ° + sin11° Sol. tan θ = cos11 ° − sin11°  sin11°  cos11 ° 1 +  cos11 °  =  sin11°  cos11 ° 1 −  cos11 °  6. If tan θ =

1 + tan11° 1 − tan11° tan 45° + tan11° (∵ tan 45° = 1) = 1 − tan 45° tan11° tan θ = tan(45° + 11° ) =

= tan 56° = tan(180° + 56° ) = tan 236° θ = 236°

7. If 0° < A, B < 90°, such that cos A = Sol. cos A =

5 4 and sin B = , find the value of sin (A – B). 13 5

5 4 and sin B = 13 5

P

13

12

A

Q

5

R

PQ 2 = PR 2 − QR 2 = (13)2 − 52 = 169 − 25 = 144 PQ = 12 cos A =

5 12 ,sin A = 13 13

X

5

4 Y

B 3

Z

YZ 2 = XZ 2 − XY 2 = 52 − 42 = 24 − 16 = 9 YZ = 3 4 3 , cos B = 5 5 sin(A − B) = sin A cos B − cos A sin B

sin B =

=

12 3 5 4 36 20 36 − 20 16 × − × = − = = 13 5 13 5 65 65 65 65

8. What is the value of tan 20° + tan 40° + Sol. tan 20° + tan 40° + 3 tan 20° tan 40°

Consider 20° + 40° = 60° Tan(20° + 40°) = tan 60° tan 20° + tan 40° = 3 1 − tan 20° tan 40° tan 20° + tan 40° = 3(1− tan 20° tan 40° ) tan 20° + tan 40° = 3 − 3 tan 20° tan 40° tan 20° + tan 40° + 3 tan 20° tan 40° = 3

3 tan 20° tan 40°?

9. Find the value of tan 56° – tan 11° – tan 56° tan 11°. Sol. We have 56° – 11° = 45° tan(56° − 11°) = tan 45°

tan 56° − tan11° =1 1 + tan 56° tan11° tan 56 ° − tan11° = 1 + tan 56° tan11° tan 56° − tan11° − tan 56° tan11° = 1

sin(C − A) if none of sin A, sin B, sin C is zero. cos C sin A sin(C − A) sin C cos A − cos C sin A Sol. Σ =Σ sin Csin A sin Csin A sin Ccos A cos Csin A =Σ − sin Csin A sin Csin A = Σ cot A − cot C = cot A − cot C + cot B −cot A + cot C −cot B = 0 10. Evaluate Σ

11. tan 72° = tan 18° + 2 tan 54° Sol. 72° – 18° = 54° Take tan on both sides tan(72° − 18°) = tan 54° tan 72° − tan18° = tan 54° 1 + tan 72° tan18° tan 72° − tan18° = tan 54° 1 + tan(90 − 18) tan18° tan 72° − tan18° = tan 54° 1 + tan18° tan18° tan 72° − tan18° = tan 54° 1+1 tan 72° − tan18° = tan 54° 2 tan 72° − tan18° = 2 tan 54°

tan 72° = tan18° + 2 tan 54° 12. sin 750° cos 480° + cos 120° cos 60° = Sol. sin 750° = sin(2⋅360° + 30°) = sin 30° =

cos 480° = cos(360° + 120°) = cos 120° = 120° = cos(180° – 60°)

1 2

−1 2

= –cos 60° = – cos 120° = –

1 2

1 2

1 2 ∴ L.H.S. = = sin 750° cos 480° + cos 120° cos 60° 1 1  1 1 =  −  +−  2 2  2 2 cos 60° =

1 1 −2 −1 =− − = = = R.H.S. 4 4 4 2  4π   4π  − A  + cos  +A  = 0 13. cos A + cos   3   3   4π   4π  Sol. Consider cos  − A  + cos  + A  3 3     = cos(240 ° + A) + cos(240 ° − A)

= cos 240 ° cos A − sin 240 ° sin A + cos 240 °cos A +sin 240 °sin A = 2 cos 240 ° cos A = 2 cos(180 ° + 60 °) cos A = −2 cos 60° cos A 1 = −2 × cos A = − cos A 2

 2π   2π  3 14. cos 2 θ + cos 2  + θ  + cos 2  − θ  =  3   3  2  2π   2π  + θ  + cos 2  − θ Sol. cos 2   3   3 

= cos2 (60 ° + θ) + cos 2 (60 ° − θ) = [cos 60 ° cos θ − sin 60 °sin θ] 2 + [cos 60 ° cos θ + sin 60 °sin θ] 2 = 2(cos2 60 °cos 2 θ + sin 2 60 °sin 2 θ)[∵ (a + b)2 + (a − b)2 = 2(a2 + b2 )] 2  1 2   3   2 2   = 2   cos θ +   sin θ 2   2     

3 1  = 2  cos2 θ + sin 2 θ  4 4  2 = [cos 2 θ + 3sin2 θ ] 4 1 3 = cos 2 θ + sin 2 θ 2 2 1 3 ∴ L.H.S. = cos 2 θ + cos 2 θ + sin 2 θ 2 2 3 3 = cos 2 θ + sin 2 θ 2 2 3 =  cos 2 θ + sin 2 θ 2 3 = (∵ cos2 θ + sin 2 θ = 1) 2 = R.H.S. 1° 1° − sin 2 22 . 2 2 ° ° 1 1 Sol. Put A = sin 2 82 and B = sin 2 22 , then 2 2 1° 1° sin2 82 − sin2 22 2 2 15. Evaluate sin 2 82

= sin2 A − sin2 B = sin(A + B) sin(A − B) = sin105° sin 60° = sin(90° + 15° ) sin 60° = cos15 °sin 60° =

1+ 3 2 2

⋅

3 3+ 3 = 2 4 2

16. Prove that sin 2 52

10 10 3 +1 − sin 2 22 = 2 2 4 2

 π A  π A 17. sin 2  +  − sin 2  −  8 2  8 2  π A  π A Sol. sin 2  +  − sin2  −   8 2  8 2

[∵ sin 2 A − sin2 B = sin(A + B) sin(A − B)]  π A π A  π A π A = sin  + + −  sin  + − +  8 2 8 2  8 2 8 2  2π   2A  = sin   sin    8  2 π 1 sin A = sin sin A = sin 45° sin A = 4 2

1 1 18. cos 2 52 ° − sin 2 22 ° 2 2 1 1 Sol. cos 2 52 ° −sin 2 22 ° 2 2 [∵ cos 2 A − sin 2 B = cos(A + B) cos(A − B)] 1  1   1  1 = cos  52 ° + 22 °  cos  52 ° − 22 °  2  2   2  2 = cos 75 ° cos30 ° =

3 (cos 75°) 2

=

3  3 − 1  3− 3  = 2  2 2  4 2

  10 10 10 10  10 10 − sin 2 52 = cos 112 + 52  cos 112 − 52  2 2 2 2  2 2    (∵ cos2 A − sin2 B = cos ( A + B ) cos ( A − B ) )

19. cos2 112

= cos165 0. cos 60 0 =

=−

1  3 + 1   2  2 2 

1 1 cos (180 0 − 15 0 ) = − cos15 0 2 2

20. Prove that tan 3A tan 2A tan A = tan 3A − tan 2A −tan A Solution: We know that tan 3A = tan ( 2A + A ) tan 2 A + tan A tan 3A = 1 − tan 2 A tan A tan 3A − tan 2A tan A tan 3 = tan 2A + tan A tan 3A − tan 2 − tan A = tan A tan 2A tan 3A

21. Find the expansion of sin(A + B + C). Sol. sin(A + B + C) = sin[(A + B) − C]

= sin(A + B) cos C − cos(A − B) sin C = (sin A cos B + cos A sin B) cos C −[cos(A − B) cos B − sin A sin B]sin c = sin A cos B cos C + cos A sin B cos C − cos A cos Bsin C +sin A sin Bsin C

22. Find the expansion of cos (A – B – C). Sol. cos (A – B – C) = cos[(A – B) – C] = cos(A − B) cos C + sin(A − B)sin C

= (cos A cos B + sin A sin B) cos C + (sin A cos B − cos A sin B]sin C = cos A cos B cos C +sin A sin Bcos C +sin A cos Bsin C −cos A sin Bsin C 23. For what values of x in the first quadrant Sol.

2 tan x > 0 ⇒ tan 2x > 0 1 − tan 2 x π ⇒ 0 < 2x < (since x in the first quadrant) 2 π ⇒ 0 < x< 4

Therefore,

π 2 tan x is positive for 0 < x < 2 4 1 − tan x

2 tan x is positive? 1 − tan2 x

SAQ’S 24. Prove that sin 4

π 8

+ sin

4

3π 3 4 5π 4 7π + sin + sin = 8 8 8 2

Solution:

sin4

π 8

+ sin4

3π 5π 7π + sin4 + sin4 8 8 8

4

4

  2 π π   2 π π  +  sin 2 − 8  +  sin  2 + 8      π π π + cos4 + cos 4 + sin 4 8 8 8  π π π π  2  sin 4 + cos4  = 2  sin 2 + cos 2 8 8 8 8  

 π sin 8  π sin4 8

= 2 − 4sin

2

π 8

cos 2

2

π      +  sin π − 8     

4

2 π   2π cos 2   − 2sin 8 8  

π 8 2

π π π 1 3  = 2 −  2sin cos  = 2 − sin 2 = 2 − = 8 8 4 2 2  5π 7π 3 + cos 4 = 8 8 8 8 2 1 26. Prove that (i) sin A sin ( 60 − A) sin 600 + A = sin 3A 4 1 (ii) cos A cos (60 + A )cos 60 0 − A = cos3 A 4 0 0 (iii) tan A tan (60 + A ) tan ( 60 − A ) = tan 3A 25. Prove that cos 4

π

+ cos 4 3

π

+ cos 4

(

)

(

)

3 16 π π π π 2 3 4 1 cos cos (v) cos cos = 9 9 9 9 16

(iv) sin 200 sin 400 sin 600 sin 800 =

27. Prove that tan θ + 2 tan 2θ + 4 tan 4θ + 8cot 8θ = cot θ

7 π −3 π , where < α < π and 0 < β < , then find the values of and sin β = 2 5 25 2 tan(α + β) and sin(α + β). −3 π , where < α < π Sol. cos α = 2 5 α in II quadrant π 7 sin β = , where 0 < β < 25 2 β in I Quadrant 28. If cos α =

A

5

4

α

B

3

C

AB2 = AC2 − BC2 = 52 − 32 = 25 − 9 = 16 ∴ AB = 4 3 4 4 cos α = − ,sin α = , tan α = − 5 5 3 X

25

7

β

Y

24

Z

YZ2 = XZ2 − XY2 = 25 2 − 7 2 = 625 − 49 = 576 YZ = 24 7 24 7 , cos β = , tan β = 25 25 24 tan α + tan β ∴tan( α + β) = 1 − tan α tan β −4 7 − 32 + 7 + 4 = 3 24 = 4 7 7 1+ × 1+ 3 24 18 −25 3 − 25 18 = 24 = × =− 18 + 7 24 25 4 18 sin(α + β ) = sinα cosβ + cos α sin β sin β =

4 24 − 3 7 = × + × 5 25 5 25 96 21 75 3 = − = = 125 125 125 5

29. If 0 < A < B <

24 4 π and cos(A − B) = , then find the value of and sin(A + B) = 4 25 5

tan 2A. Sol. sin(A + B) =

24 25

25

24

A+B 7

24 7 4 cos(A − B) = 5 tan(A + B) =

5

3

A–B 4

3 4 Now 2A = (A + B) + (A – B) tan 2A = tan [ (A + B) + (A − B)] tan(A − B) =

=

tan(A + B) + tan(A − B) 1 − tan(A + B) tan(A − B)

24 3 + 96 + 21 − 117 = 7 4 = = 24 3 28 − 72 44 1− × 7 4 30. If A + B , A are acute angles such that sin ( A + B) =

24 3 and tan A = find the value of 25 4

cos B Solution: sin ( A + B ) =

24 25

∴cos ( A + B ) =

7 25

3 3 4 ∴ sin A = & cos A = 4 5 5 cos B = cos( A + B − A) = cos( A + B ) cos A + sin ( A + B ) sin A 7 4 24 3 100 4 × + × = = 25 5 25 5 125 5 tan A =

31. If tan α – tan β = m and cot α – cot β = n, then prove that cot( α − β) =

1 1 − . m n

Sol. We have tan α – tan β = m 1 1 − =m cot α cot β

cot β − cot α =m cot α cot β cot α cot β 1 = cot β − cot α m cot α − cot β= n

∴

...(1)

− (cot β − cot α) = n cot β − cot α = −n 1 1 =− cot β − cot α n L.H.S. = cot( α − β) =

...(2) cot α cot β cot β −cot α

=

cot α cot β 1 + cot β − cot α cot β − cot α

=

1 1 − (∵ from(1) & (2)) = R.H.S. m n

32. If tan(α – β) =

7 4 and tan α = , where α and β are in the firs quadrant prove that 24 3

α + β = π/2. 7 4 Sol. tan(α – β) = and tan α = 24 3 7

25 2β 34

tan(α − β ) =

tan α − tan β 1+ tanα tanβ

4 − tan β 7 ⇒ 3 = 4 1+ tanβ 24 3 4 − 3 tan β 7 3 ⇒ = 3+ 4 tanβ 24 3 4 − 3 tan β 7 ⇒ = 3 + 4 tan β 24

⇒ 24[4 − 3 tan β ] = 7(3+ 4 tanβ ) ⇒ 96 − 72 tanβ = 21+ 28 tanβ ⇒ 96 − 21 = 28 tanβ + 72 tanβ ⇒ 100 tan β = 75 75 3 = 100 4 tan α + tan β ∴tan( α + β) = 1 − tan α tan β 4 3 + = 3 4 =α 4 3 1− ⋅ 3 4 π tan(α + β ) = tan 2 π ∴α + β = 2 ∴tan β =

sin(α + β ) a + b = , then prove that a tan β = b tan α . sin(α − β) a − b sin(α + β ) a + b Sol. Given that = sin( α − β) a − b 33. If

By using componendo and dividendo, we get sin(α + β ) + sin(α − β ) a + b + a − b 2a a = = = sin(α + β ) − sin(α −β ) a+ b− a+ b 2b b

⇒

sin(α + β ) + sin(α − β ) a = sin(α + β ) − sin(α −β ) b

⇒

sin α cosβ + cos α sin β+ sin α cos β − cos α sin β a = sin α cos β + cos α sin β − sin α cos β + cos α sin β b

2sin α cos β a = 2 cos α sin β b a ⇒ tan α cot β = b ⇒ b tan α = a tan β hence, a tan β = b tanα

⇒

3π , then show that 4 (1 – tan A) (1 + tan B) = 2. 3π Sol. A − B = 4 A − B = 135° 34. If A − B =

tan(A − B) = tan135° = tan(90° + 45)° = − cot 45° = − 1 ∴

tan A − tan B = −1 1 + tan A tan B tan A − tan B = − (1+ tan A tan B) tan A − tan B = − 1− tan A tan B tan A − tan B+ tan A tan B= − 1 tan B − tan A − tan A tan B = 1 ...(1) L.H.S. = (1 − tan A)(1 + tan B) = 1+ (tan B− tan A − tan A tan B) = 1+ 1 (∵ from(1)) = 2 = R.H.S.

π and if none of A, B, C is an odd multiple of π/2, then prove that 2 cot A + cot B + cot C = cotA cotB cotC. π Sol. A + B + C = 2 π A+ B= − C 2 π  cot(A + B) = cot  − C  2  − cot A cot B 1 = tan C cot B +cot A cot A cot B − 1 1 = cot B +cot A cot C cot C[cot A cot B −1] = cot B + cot A 35. If A + B + C =

cot A cot B cot C −cot C cot A +cot B cot A cot B cot C = cot A + cot B +cot C ∴cot A +cot B +cot C =cot A cot Bcot C

π and if none of A, B, C is an odd multiple of π/2, then prove that, 2 tan A tan B + tan B tan C + tan C tan A = 1. π Sol. A + B + C = 2 π A+ B= − C 2 π  tan(A + B) = tan  − C  2   36. If A + B + C =

tan A + tan B = cot C 1 − tan A tan B tan A + tan B 1 ⇒ = − 1 tan A tan B tan C ⇒ tan C[tan A + tan B] = 1− tan A tan B ⇒

⇒ tan C tan A + tan C tan B = 1− tan A tan B ⇒ tan A tan B + tan B tan C+ tan C tan A= 1 cos(B + C) = 2. cos B cos C cos(B +C) Sol. L.H.S. = Σ cos Bcos C cos B cos C −sin Bsin C =Σ cos B cos C cos B cos C sin Bsin C =Σ − cos Bcos C cos Bcos C = Σ(1 − tan B tan C) 37. Σ

= 1 − tan B tan C + 1− tan C tan A +

1− tan A tan B

= 3 − (tan A tan B + tan B tan C + tan C tan A) = 3 − 1 (∵ from(b)) = 2 = R.H.S.

38. Prove that sin2 α + cos2 (α + β) + 2 sin α sin β cos(α + β) is independent of α. Sol. Given expression, sin2α + cos2 (α+β) + 2 sinα sinβ cos(α+β)

= sin 2 α + 1 − sin 2 (α + β ) + 2sin α sinβ cos(α + β ) = 1+ [sin2 α − sin2 (α + β )]+ 2sinα sinβ cos(α + β ) = 1+ sin(α + α + β ) sin(α − α −β ) + 2sin α sinβ cos(α + β )

= 1+ sin(2α + β ) sin(−β ) + 2sinα sinβ cos(α + β ) = 1− sin(2α + β ) sinβ + [2sinα cos(α + β )]sinβ = 1− sin(2α + β ) sin α + [sin(α + α + β )+ sin(α − α − β )]sinβ = 1− sin(2α + β ) sinα + [sin(2α + β )− sinβ ]sinβ = 1− sin(2α + β ) sin α + sin(2α + β ) sinβ − sin2 β = 1− sin2 β = cos 2 β Thus the given expression is independent of α.

π 2π 3π 7π ⋅ cot ⋅ cot ...cot =1 . 16 16 16 16 2π 3π 7π π ⋅cot ...cot Sol. cot ⋅cot 16 16 16 16 π 7π  2π 6π   3π 5π  4π  =  cot ⋅ cot ⋅cot ⋅cot  cot  cot  ⋅cot 16  16 16   16 16  16  16 39. Prove that cot

2π 3π  π π  π π    π 2π    π 3π  =  cot ⋅ cot  −    cot ⋅ cot  −   cot ⋅cot  −   ⋅cot 4  2 16    16  2 16    16  2 16    16 π π  2π 2π   =  cot ⋅ tan   cot ⋅ tan  16   16 16   16 = 1× 1× 1× 1 = 1

3π 3π    cot ⋅ tan  ⋅1 16   16

40. Prove that tan 70° – tan 20° = 2 tan 50°. Sol. tan 50° = tan(70° – 20°) tan 70° − tan 20° = 1 + tan 70° tan 20° ⇒ tan 70 ° − tan 20°

= tan 50°(1 + tan 70° ⋅ tan 20°) = tan 50°(1 + tan 70° ⋅ tan(90° − 70° )] = tan 50°[1 + tan 70° ⋅ cot 70°] = tan 50°[1 + 1] = 2 tan 50° ∴ tan 70° − tan 20° = 2 tan 50°

41. If A + B = 45°, then prove that i) (1 + tan A) (1 + tan B) = 2 ii) (cot A – 1)(cot B – 1) = 2. Sol. i) A + B = 45° ⇒ tan(A + B) = tan 45° = 1

tan A + tan B =1 1 − tan A tan B ⇒ tan A + tan B = 1− tan A tan B ⇒

⇒ tan A + tan B + tan A tan B = 1...(1) Now, (1 + tan A)(1+ tan B) = 1+ tan A+ tan B+ tan A tan B= 2 (from(1)) ii) A + B = 45° ⇒ cot(A + B) = cot 45° = 1 cot A cot B− 1 ⇒ =1 cot B + cot A ⇒ cot A cot B −1 = cot A + cot B ⇒ cot A cot B − cot A − cot B =1...(2) Now, (cot A −1)(cot B −1) = cot A cot B − cot A −cot B +1 = 2 (from(2)) 42. If A, B, C are the angles of a triangle and if none of them is equal to π/2, then prove that i) tan A + tan B + tan C = tan A tan B tan C ii) cotA cotB + cotB cotC + cot C cot A =1 Sol. i) Given A + B + C = π ⇒ A +B = π−C

⇒ tan(A + B) = tan(π − C) tan A + tan B = − tan C 1 − tan A tan B ⇒ tan A + tan B = − tan C(1 − tan A tan B)

⇒

⇒ tan A + tan B = − tan C + tan A tan B tan C ⇒ tan A + tan B + tan C = tan A tan B tan C 1 etc., in (i) above, we get ii) Replacing tan A by cot A 1 1 1 1 + + = cot A cot B cot C cot A cot B cot C ⇒ cot A cot B + cot Bcot C +cot Ccot A =1

LAQ’S 43. Let ABC be a triangle such that cot A + cot B + cot C = 3 . Then prove that ABC is an equilateral triangle. Sol. Given that A + B + C = 180° We get Σ cot A cot B = 1 Now, Σ(cot A – cot B)2 = Σcot2 A + cot2 B – 2 cot A cot B 2 2 2 = 2 cot A + 2 cot B + 2 cot C − 2 cot A cot B −2 cot B cot C −2 cot Cco t A

(on expanding) 2 = 2{(cot A + cot B + cot C) − 2(cot A cot B) − 2 cot B cot C −2 cot Ccot A}

− 2(cot A cot B + cot B cot C + cot Ccot A) = 2(cot A + cot B + cot C) 2 − 6(cot A cot B +cot Bcot C +cot Ccot A) = 2 ⋅3 − 6 = 0 ⇒ cot A = cot B = cot C

⇒ cot A = cot B = cot C =

3 1 = 3 3

(since cot A + cot B+ cot C = 3)

⇒ A + B + C = 60° (Since each angle lies in the interval [0,180°]...