TOPICS
Miscellaneous

Question-1 :-  Find the value of cos-1 (cos 13π/6).

Solution :-
   cos-1 (cos 13π/6)
   Here, 13π/6 is not in range of cos-1 i.e., [0, π].
   so, break of 13π/6
 = cos-1 [cos (2π + π/6)]
 = cos-1 [cos π/6]
 = π/6

Question-2 :- Find the value of tan-1 (tan 7π/6).

Solution :-
   tan-1 (tan 7π/6)
   Here, 7π/6 is not in range of cos-1 i.e., [-π/2, π/2].
   so, break of 7π/6
 = tan-1 [tan (π + π/6)]
 = tan-1 [tan π/6]
 = π/6

Question-3 :-  Prove that : 2sin-1 (3/5) = tan-1 (24/7).

Solution :-
   2sin-1 (3/5) = tan-1 (24/7)
   By taking L.H.S
   2sin-1 (3/5)   [2sin-1 x = sin-1(2x √1 - x²)]
 = sin-1 (2 x 3/5 √1 - (3/5)²)
 = sin-1 (6/5 √1 - 9/25)
 = sin-1 (6/5 √16/25)
 = sin-1 (6/5 x 4/5)
 = sin-1 (24/25)
   Let sin-1 (24/25) = θ
   sin θ = 24/25
   cos θ = √1 - sin²θ
   cos θ = √1 - (24/25)²
   cos θ = √1 - 576/625
   cos θ = √49/625
   cos θ = 7/25
   Now, tan θ = sin θ/ cos θ = 24/25 x 25/7 = 24/7
   tan θ = 24/7
       θ = tan-1 (24/7) = R.H.S

Question-4 :-  Prove that: sin-1 (8/17) + sin-1 (3/5) = tan-1 (77/36).

Solution :-
   sin-1 (8/17) + sin-1 (3/5) = tan-1 (77/36)
   L.H.S 
   sin-1 (8/17) + sin-1 (3/5)      
   By Using [sin-1 x + sin-1 y = sin-1(x√1 - y² + y√1 - x²)]
 = sin-1 (8/17 √1 - (3/5)² + 3/5 √1 - (8/17)²)
 = sin-1 (8/17 √1 - 9/25 + 3/5 √1 - 64/289)
 = sin-1 (8/17 √16/25 + 3/5 √225/289)
 = sin-1 (8/17 x 4/5 + 3/5 x 15/17)
 = sin-1 (32/85 + 9/17)
 = sin-1 [(32 + 45)/85]
 = sin-1 77/85
   Let sin-1 77/85 = θ
   sin θ = 77/85
   cos θ = √1 - sin²θ
   cos θ = √1 - (77/85)²θ
   cos θ = √1 - 5929/7225
   cos θ = √1296/7225
   cos θ = 36/85
   Now tan θ = sin θ/cos θ
   tan θ = 77/85 x 85/36 = 77/36
       θ = tan-1 77/36
 = R.H.S

Question-5 :- Prove that: cos-1 (4/5) + cos-1 (12/13) = cos-1 (33/65).

Solution :-
   cos-1 (4/5) + cos-1 (12/13) = cos-1 (33/65)
   By taking L.H.S
   cos-1 (4/5) + cos-1 (12/13)       
   By Using [cos-1 x + cos-1 y = cos-1(xy - √1 - y²1 - x²)]
 = cos-1(4/5 x 12/13 - √1 - (12/13)²1 - (4/5)²)
 = cos-1(48/65 - √1 - 144/1691 - 16/25)
 = cos-1(48/65 - √25/1699/25)
 = cos-1[48/65 - (5/13 x 3/5)]
 = cos-1[48/65 - 3/13]
 = cos-1[(48-15)/65]
 = cos-1(33/65)
 = R.H.S

Question-6 :-  Prove that : cos-1 (12/13) + sin-1 (3/5) = sin-1 (56/65).

Solution :-
   cos-1 (12/13) + sin-1 (3/5) = sin-1 (56/65)
   By taking L.H.S
   cos-1 (12/13) + sin-1 (3/5)
   Let cos-1 (12/13) = θ
   cos θ = 12/13  
   sin θ = √1 - (12/13)²
   sin θ = √1 - 144/169
   sin θ = √25/169
   sin θ = 5/13
   Now, θ = sin-1 (5/13)
   sin-1 (5/13) + sin-1 (3/5)       
   By using [sin-1 x + sin-1 y = sin-1(x√1 - y² + y√1 - x²)]
 = sin-1 (5/13 √1 - (3/5)² + 3/5 √1 - (5/13)²)
 = sin-1 (5/13 √1 - 9/25 + 3/5 √1 - 25/169)
 = sin-1 (5/13 √16/25 + 3/5 √144/169)
 = sin-1 (5/13 x 4/5 + 3/5 x 12/13)
 = sin-1 (4/13 + 36/65)
 = sin-1 [(20 + 36)/65]
 = sin-1 (56/65)
 = R.H.S

Question-7 :-  Prove that: tan-1 (63/16) = sin-1 (5/13) + cos-1 (3/5).

Solution :-
   tan-1 (63/16) = sin-1 (5/13) + cos-1 (3/5)
   R.H.S 
   sin-1 (5/13) + cos-1 (3/5) 
   Firstly,    
   Let sin-1 5/13 = θ₁
   sin θ₁ = 5/13
   cos θ₁ = √1 - sin²θ₁
   cos θ₁ = √1 - (5/13)²
   cos θ₁ = √1 - 25/169
   cos θ₁ = √144/169
   cos θ₁ = 12/13
   Now tan θ₁ = sin θ₁/cos θ₁
   tan θ₁ = 5/13 x 13/12 = 5/12
       θ₁ = tan-1 5/12
         
   Secondly, 
   Let cos-1 (3/5) = θ₂
   sin θ₂ = 3/5
   sin θ₂ = √1 - cos²θ₂
   sin θ₂ = √1 - (3/5)²
   sin θ₂ = √1 - 9/25
   sin θ₂ = √16/25
   sin θ₂ = 4/5
   Now tan θ₂ = sin θ₂/cos θ₂
   tan θ₂ = 4/5 x 5/3 = 4/3
       θ₂ = tan-1 4/3
   
   Now, θ₁ + θ₂, We have, tan (θ₁ + θ₂) = [(tan θ₁ + tan θ₂) ÷ (1 - tan θ₁ tan θ₂)]
 = tan-1 (5/12) + tan-1 (4/3)
 = tan-1[(5/12 + 4/3) ÷ (1 - 5/12 x 4/3)]
 = tan-1[(5 + 16)/12 ÷ (1 - 5/9)]
 = tan-1[21/12 ÷ 4/9)]
 = tan-1[7/4 x 9/4]
 = tan-1(63/16)
 = L.H.S

Question-8 :- Prove that: Prove that: tan-1 (1/5) + tan-1 (1/7) + tan-1 (1/3) + tan-1 (1/8) = π/4.

Solution :-
   answer

Question-9 :-  Prove that : tan-1 √x = 1/2 . cos-1 (1 - x)/(1 + x), x ∈ [0,1].

Solution :-
   tan-1 √x = 1/2 . cos-1 (1 - x)/(1 + x)
   2tan-1 √x = cos-1 (1 - x)/(1 + x)
   By Taking L.H.S
   2tan-1 √x
   We using Property, 2tan-1 x = cos-1 (1 - x²)/ (1 + x²)
 = cos-1 [1 - (√x)²]/ [1 + (√x)²]
 = cos-1 (1 - x)/(1 + x)
 = R.H.S

Question-10 :-  Prove that: answer

Solution :-
   answer

Question-11 :- Prove that:question

Solution :-
   answer
9

Question-12 :-  Prove that : 9π/8 - 9/4 sin-1 1/3 = 9/4 sin-1(2√2)/3.

Solution :-
  9π/8 - 9/4 sin-1 1/3 = 9/4 sin-1(2√2)/3
  Taking out common value 9/4 of both sides then,
  9/4 x (π/2 - sin-1 1/3) = 9/4 sin-1(2√2)/3
  π/2 - sin-1 1/3 = sin-1(2√2)/3
  cos-1 1/3 = sin-1 (2√2)/3
  By taking of L.H.S
  let cos-1 1/3 = θ
  cos θ = 1/3
  sin θ = √1 - cos²θ
  sin θ = √1 - (1/3)²
  sin θ = √1 - 1/9
  sin θ = √8/9
  sin θ = (2√2)/3
      θ = sin-1(2√2)/3 = R.H.S

Question-13 :- Solve: 2tan-1(cos x) = tan-1(2 cosec x)

Solution :-
  2tan-1(cos x) = tan-1(2 cosec x)
  tan-1[(2 cos x) ÷ (1 - cos²x)] = tan-1(2 cosec x)
  (2 cos x) ÷ (1 - cos²x) = 2 cosec x
  2 cos x ÷ sin²x = 2 ÷ sin x
  cos x = sin x
  tan x = 1
  tan x = tan π/4
      x = π/4

Question-14 :- Solve: tan-1[(1 - x) ÷ (1 + x)] = 1/2 tan-1x, (x > 0).

Solution :-
  tan-1[(1 - x) ÷ (1 + x)] = 1/2 tan-1x
  tan-1 1 - tan-1 x = 1/2 tan-1x
  tan-1 (tan π/4) = 1/2 tan-1x + tan-1 x
  π/4 = 3/2 tan-1x
  tan-1x = π/6
  x = tan (π/6)
  x = 1/√3

Question-15 :-  sin(tan-1x), |x|<1 is equal to
question

Solution :-
  sin(tan-1x)
  Let tan-1x = θ
  x = tan θ
  sin θ = x/(√1 + x²)
      θ = sin-1[x/(√1 + x²)]
  tan-1x = sin-1[x/(√1 + x²)]
  sin(tan-1x) = sin(sin-1[x/(√1 + x²)])
  sin(tan-1x) = x/(√1 + x²)
  Option is D.

Question-16 :-  sin-1(1 - x) - 2 sin-1x = π/2, then x is equal to
question

Solution :-
  answer

Question-17 :-  question
answer

Solution :-
answer
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