Question-1 :-  Prove that 3sin^-1 x = sin^-1 (3x - 4x³), x ∈ [-1/2, 1/2].

Solution :-

   Let x = sin θ and θ = sin-1 x.
   R.H.S
   sin-1 (3x - 4x³)
 = sin-1 (3sin θ - 4sin³ θ)
 = sin-1 (sin 3θ)
 = 3θ = 3sin-1 x
    

Question-2 :-  Prove that 3cos^-1 x = cos^-1 (4x³ - 3x), x ∈ [1/2, 1].

Solution :-

   Let x = cos θ and θ = cos-1 x.
   R.H.S
   cos-1 (4x³ - 3x)
 = cos-1 (4cos³ θ - 3cos θ)
 = cos-1 (cos 3θ)
 = 3θ = 3cos-1 x

Question-3 :-  Prove that .

Solution :-

Question-4 :-  Prove that .

Solution :-

Question-5 :-  Write the function in simplest form : .

Solution :-

Question-6 :-  Write the function in simplest form : .

Solution :-

Question-7 :-  Write the function in simplest form : .

Solution :-

Question-8 :-  Write the function in simplest form : .

Solution :-

Question-9 :-  Write the function in simplest form : .

Solution :-

Question-10 :-  Write the function in simplest form : .

Solution :-

Question-11 :-  Find the value of.

Solution :-

Question-12 :-  Find the value of cot (tan^-1a + cot^-1a).

Solution :-

   cot(tan-1a + cot-1a)
 = cot(π/2)   ∴ [tan-1a + cot-1a = π/2]
 = 0
    

Question-13 :-  Find the value of.

Solution :-

Question-14 :-  Find the value of x, if sin (sin^-1 1/5 + cos^-1 x) = 1.

Solution :-

    sin (sin-1 1/5 + cos-1 x) = 1
    sin (sin-1 1/5 + cos-1 x) = sin π/2
    sin-1 1/5 + cos-1 x = sin-1 (sin π/2)
    sin-1 1/5 + cos-1 x  = π/2    ∴ [sin-1x + cos-1x = π/2]
    sin-1 1/5 = π/2 - cos-1 x
    sin-1 1/5 = sin-1 x
    1/5 = x or x = 1/5

Question-15 :-  Find the value of.

Solution :-

Question-16 :-  Find the value of sin^-1(sin 2π/3).

Solution :-

   sin-1(sin 2π/3)
   Range of sin-1 is [-π/2, π/2], so 2π/3 is not in Range. Then we break 2π/3.
 = sin-1[sin(π - π/3)]
 = sin-1[sin π/3]
 = π/3
    

Question-17 :-  Find the value of tan^-1(tan 3π/4).

Solution :-

   tan-1(tan 3π/4)
   Range of tan-1 is [-π/2, π/2], so 3π/4 is not in Range. Then we break 3π/4.
 = tan-1[tan(π - π/4)]
 = tan-1[-tan π/4]
 = tan-1[tan (-π/4)]
 = -π/4

Question-18 :-  Find the value of tan(sin^-1 3/5 + cot^-1 3/2).

Solution :-

   tan(sin-1 3/5 + cot-1 3/2)
   
   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
   Then, tan θ₁ = sin θ₁/cos θ₁ = 3/5 x 5/4 = 3/4 
   tan θ₁ = 3/4
       θ₁ = tan-1 3/4
   
   Secondly,
   Let cot-1 3/2 = θ₂
   cot θ₂ = 3/2
   tan θ₂ = 1/cot θ₂
   tan θ₂ = 2/3
       θ₂ = tan-1 2/3

   Now, Accordin to question
   tan(θ₁ + θ₂)
 = tan(tan-1 3/4 + tan-1 2/3)
 = tan[tan-1(3/4 + 2/3) ÷ (1 - 3/4 x 2/3)]
 = tan[tan-1(9 + 8)/12 ÷ (1 - 1/2)]
 = tan[tan-1 (17/12 ÷ 1/2)]
 = tan[tan-1 (17/12 x 2/1)]
 = tan[tan-1 (17/6)]
 = 17/6

Question-19 :-  cos^-1(cos 7π/6) is equal to
(A) 7π/6   (B) 5π/6  (C) π/3  (D) π/6

Solution :-

   cos-1(cos 7π/6)
   Range of cos-1 is [0, π], so 7π/6 is not in Range. Then we break 7π/6.
 = cos-1[cos(π + π/6)]
 = cos-1[-cos π/6]
 = cos-1[cos (-π/6)]  {Range = [0, π]}
 = cos-1[cos (π - π/6)]
 = cos-1[cos (5π/6)]
 = 5π/6     
   Option is (B)
    

Question-20 :-  sin [π/3 + sin^-1(-1/2)] is equal to
(A) 1/2   (B) 1/3  (C) 1/4  (D) 1

Solution :-

   sin [π/3 + sin-1(-1/2)]
 = sin [π/3 + sin-1(-sin π/6)]
 = sin [π/3 + sin-1(sin (-π/6))]
 = sin [π/3 - π/6)]
 = sin [(2π - π)/6]
 = sin [π/6]
 = 1/2

Question-21 :-  tan^-1(√3) - cot^-1(-√3) is equal to
(A) π   (B) -π/2  (C) 0  (D) 2√3

Solution :-

   tan-1(√3) - cot-1(-√3)
   Let tan-1(√3) = θ₁
   tan θ₁ = √3      {Range = [-π/2, π/2]}
   tan θ₁ = tan π/3
       θ₁ = π/3
    
   Let cot-1(-√3) = θ₂ 
   cot θ₂ = -√3
   cot θ₂ = -cot π/6   
   cot θ₂ = cot (-π/6)  {Range = [0, π]}
   cot θ₂ = cot (π - π/6)
   cot θ₂ = cot (5π/6)
       θ₂ = 5π/6

   Now, θ₁ - θ₂
 = π/3 - 5π/6
 = (2π - 5π)/6
 = -3π/6
 = -π/2
   Option (B) is correct.