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

Answer 1:

 Given that cos ⁻¹ (cos  (13π/6))

We know that cos⁻¹ (cos x) = x if x ∈ [0, π], which is the principal value branch of cos⁻¹ x

Hence, cos ⁻¹ (cos  (13π/6)) = (π/6)

 

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

Answer 2:

Given that tan⁻¹ (tan (7π/6))

We know that tan⁻¹ (tan x) = x if, x ∈ (−(π/2), (π/2)), which is the principal value branch of tan ⁻¹ x.

∴ tan⁻¹ (tan (7π/6)) = tan⁻¹ [tan (π + (π/6))]

= tan⁻¹ (tan (π/6)) = (π/6)

Hence, tan⁻¹ (tan (7π/6)) = (π/6)

 

Question 3: Prove that 2sin−1 3 5 = tan−1 24 7.

Answer 3:

LHS= sin⁻¹ (3/5) = 2tan⁻¹                            [As sin⁻¹ (a/b) = tan⁻¹ ]

                                   [As tan⁻¹x = tan⁻¹ (2x/(1 − x²) ]

= RHS

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

Answer 4:

LHS = sin⁻¹ (8/17) + sin⁻¹ (3/5)

= tan⁻¹ (8/15) + tan⁻¹ (3/4)

                            [as tan⁻¹x + tan⁻¹y = tan⁻¹ ((x + y)/(1 − xy))]

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

Answer 5:

 LHS

= cos ⁻¹ (4/5) + cos ⁻¹ (12/13)

                                              

= tan⁻¹ (3/4) + tan⁻¹ (5/12)

                                                              [as cos⁻¹x + tan⁻¹y = tan⁻¹ ((x + y)/(1 − xy))]

                                           

= RHS

 

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

Answer 6:

LHS = cos ⁻¹ (12/13) + sin⁻¹ (3/5)

                       

= tan⁻¹ (5/12) + tan⁻¹ (3/4)

                                                              [as tan⁻¹x + tan⁻¹y = tan⁻¹ ((x + y)/(1 − xy))]

                                           

= RHS

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

Answer 7:

RHS = sin⁻¹ (5/13) + cos ⁻¹ (3/5)

= tan⁻¹ (5/12) + tan⁻¹ (4/3)

                                                              [As tan⁻¹x + tan⁻¹y = tan⁻¹ ((x + y)/(1 − xy))]

= RHS

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

Answer 8:

LHS = tan⁻¹ (1/5) + tan⁻¹ (1/7) + tan⁻¹ (1/3) + tan⁻¹ (1/8)

                                                              [As tan⁻¹x + tan⁻¹y = tan⁻¹ ((x + y)/(1 − xy))]

= tan⁻¹ (12/34) + tan⁻¹ (11/23) = tan⁻¹ (6/17) + tan⁻¹ (11/23)

= tan⁻¹ (325/325) = tan⁻¹1 = (π/4) = RHS

 

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

Answer 9:

 LHS = tan⁻¹ = (1/2) × 2tan⁻¹

= (1/2) × 2tan⁻¹

                                             [As tan⁻¹x = cos⁻¹ [(1-x²)/( 1+x²))]

= (1/2) cos ⁻¹ ((1 – x)/(1 + x))

= RHS

 

Question 10:

Answer 10:

                                          

                             [Dividing each term by cos  (y/2)]

= (π/4) – (1/2) ((π/2) − x)                                       [As  ((π/2) – x) = y]

=(x/2)

=RHS

Question 11:

Answer 11:

                         [Let x = cosy]

                         [As  1 + cos y = 2cos ² (y/2) and 1 − cos y  = 2sin² (y/2)]

                                       [Dividing each term by  cos  (y/2)]

= (π/4) – (y/2) = (π/4) – (1/2) cos ⁻¹x = RHS

 

Question 12: Prove that (9π/8) – (9/4) sin⁻¹ (1/3) = (9/4) sin⁻¹

Answer 12:

LHS = (9π/8) – (9/4) sin⁻¹ (1/3)

= (9/4) ((π/2) − sin⁻¹ (1/3)

= (9/4) (cos  ⁻¹ (1/3)                                      [As sin⁻¹x + cos ⁻¹x = (π/2)]

                                 

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

Answer 13:

Given that 2tan⁻¹ (cos x) = tan⁻¹ (2cosec x)

⇒ tan⁻¹ ((2cos x)/(1 − cos ²x)) = tan⁻¹ (2cosec x)                           [As  2tan⁻¹x = tan⁻¹ (2x/(1 − x²)]

⇒ ((2cos x)/(1 − cos ²x)) = 2cosec x

⇒ ((2cos x/sin²x)) = (2/sin x) ⇒ 2 sin x. cos x = 2sin²x

⇒ 2 sin x. cos x − 2sin²x = 0

⇒ 2 sin (cos x − sin x) = 0

⇒ 2 sin x = 0      or          cos x − sin x = 0

But sin x ≠ 0 as it does not satisfy the equation

∴ cos x − sin x = 0

⇒ cos x = sin x ⇒ tan x = 1

∴ x = (π/4)

 

Question 14: Solve for x: tan⁻¹ ((1−x)(1+x)) = (1/2) tan⁻¹x, (x > 0)

Answer 14:

Given that tan⁻¹ ((1−x)(1+x)) = (1/2) tan⁻¹x

⇒ tan⁻¹1 − tan⁻¹x = (1/2) tan⁻¹x        [Because tan⁻¹x − tan⁻¹y = tan⁻¹ ((x−y)/(1+xy))]

⇒ (π/4) = (3/2) tan⁻¹x ⇒ (π/6) = tan⁻¹x ⇒ tan (π/6) = x

∴ x =

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

(A)

(B)

(C)

(D)

Answer 15:

Given that sin(tan⁻¹x)

                            

The correct answer is D.

 

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

(A) 0, (1/2)

(B) 1, (1/2) 

(C) 0

(D) (1/2)

Answer 16:

Given that sin⁻¹ (1 − x) − 2sin⁻¹x = (π/2)

Let x = sin y

∴ sin⁻¹ (1 − sin y) − 2y = (π/2)

⇒ sin⁻¹ (1 − sin y) = ((π/2) + 2y)

⇒ 1 − sin y = sin ((π/2) + 2y)

⇒ 1 − sin y = cos 2y

⇒ 1 − sin y = 1 − 2sin²y                              [as cos2 y = 1 − 2sin²y]

⇒ 2sin²y − sin y = 0

⇒ 2x 2 − x = 0 [as  x = sin y]

⇒ (2x − 1) = 0

⇒ x = 0  or  x = (1/2)

But x = (1/2) does not satisfy the given equation.

∴ x = 0 is the solution of the given equation.

The correct answer is C.

 

Question 17: tan⁻¹ (x/y) − tan⁻¹ ((x−y)/(x+y)) is equal to

(A) (π/2)

 

(B) (π/3)

(C) (π/4)

(D) (− 3π/4)

Answer 17:tan⁻¹ (x/y) − tan⁻¹ ((x−y)/(x+y))

                             [As tan⁻¹x − tan⁻¹y = tan⁻¹ ((x – y)/(1 + xy))]

= tan⁻¹ [(x² + y²)/(x² + y²)]

= tan⁻¹1 = (π/4)

Hence, the correct answer is C.