TOPICS
Miscellaneous

Example-1 :-  Find the value of sin-1 (sin 3π/5).

Solution :-
   sin-1 (sin 3π/5)
   Here, 3π/5 is not in range of sin-1 i.e., [-π/2, π/2]
 = sin-1 [sin (π - 3π/5)]
 = sin-1 [sin 2π/5]
 = 2π/5

Example-2 :-  Show that sin-1 (3/5) - sin-1 (8/17) = cos-1 (84/85).

Solution :-
   sin-1 (3/5) - sin-1 (8/17) = cos-1 (84/85)
   Firstly,
   Let sin-1 (3/5) = θ₁
   sin θ₁ = 3/5
   cos θ₁ = √1 - sin²θ₁
   cos θ₁ = √1 - (3/5)²
   cos θ₁ = √1 - 9/25
   cos θ₁ = √16/25
   cos θ₁ = 4/5
          
   Secondly, 
   Let sin-1 (8/17) = θ₂
   sin θ₂ = 8/17
   cos θ₂ = √1 - sin²θ₂
   cos θ₂ = √1 - (8/17)²
   cos θ₂ = √1 - 64/289
   cos θ₂ = √225/289
   cos θ₂ = 15/17

   Now, cos(θ₁ - θ₂) = cos θ₁ cos θ₂ + sin θ₁ sin θ₂
 = 4/5 x 15/17 + 3/5 x 8/17
 = 12/17 + 24/85
 = (60 + 24)/85
 = 84/85

   cos(θ₁ - θ₂) = 84/85
       θ₁ - θ₂ = cos-1 (84/85)

Example-3 :-  Show that sin-1 (12/13) + cos-1 (4/5) + tan-1 (63/16) = π.

Solution :-
   sin-1 (12/13) + cos-1 (4/5) + tan-1 (63/16) = π
   By using of L.H.S
   Firstly,
   Let sin-1 (12/13) = θ₁
   sin θ₁ = 12/13
   cos θ₁ = √1 - sin²θ₁
   cos θ₁ = √1 - (12/13)²
   cos θ₁ = √1 - 144/169
   cos θ₁ = √25/169
   cos θ₁ = 5/13
   Now, tan θ₁ = sin θ₁/ cos θ₁ = 12/13 x 13/5 = 12/5 
   tan θ₁ = 12/5 

   Secondly,
   Let cos-1 (4/5) = θ₂
   cos θ₂ = 4/5 
   sin θ₂ = √1 - cos²θ₂
   sin θ₂ = √1 - (4/5)²
   sin θ₂ = √1 - 16/25
   sin θ₂ = √9/25
   sin θ₂ = 3/5
   Now, tan θ₂ = sin θ₂/ cos θ₂ = 3/5 x 5/4 = 3/4
   tan θ₂ = 3/4

   We have, tan (θ₁ + θ₂) = [(tan θ₁ + tan θ₂) ÷ (1 - tan θ₁ tan θ₂)]
 = [(12/5 + 3/4) ÷ (1 - 12/5 x 3/4)]
 = [(48 + 15)/20 ÷ (1 - 36/20)]
 = 63/20 ÷ (-16/20)
 = 63/20 x (-20/16)
 = -63/16
   tan (θ₁ + θ₂) = -63/16
   
   Thirdly,
   tan-1 (63/16) = θ₃
   tan θ₃ = 63/16
   
   tan (θ₁ + θ₂) = -tan θ₃
   i.e., tan (θ₁ + θ₂) = tan (θ₃) or tan (θ₁ + θ₂) = tan (π – θ₃)
   Therefore, θ₁ + θ₂ = -θ₃ or θ₁ + θ₂ = π – θ₃
   Since θ₁, θ₂ and θ₃ are positive, θ₁ + θ₂ ≠ -θ₃ 
   Hence, θ₁ + θ₂ + θ₃ = π = R.H.S

Example-4 :-  Simplify miscellaneous if a/b tan x > -1.

Solution :-
    class 12

Example-5 :-  Solve tan-1 2x + tan-1 3x = π/4.

Solution :-
   tan-1 2x + tan-1 3x = π/4
   tan-1 [(2x + 3x) ÷ (1 - 2x.3x)] = π/4
   5x ÷ (1 - 6x²) = tan π/4
   5x ÷ (1 - 6x²) = 1
   5x = 1 - 6x² 
   6x² + 5x - 1 = 0
   6x² + 6x - x - 1 = 0
   6x(x + 1) - 1(x + 1) = 0
   (x + 1)(6x - 1) = 0
   x = -1, x = 1/6 
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