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

Answer 1:

To prove: 3sin⁻¹x = sin⁻¹ (3x − 4x³ ), x ∈ [− (1/2), (1/2)]

Let sin⁻¹x = θ, then x = sin θ.

We have,

RHS = sin^-1 (3x-4x³) = sin^-1 (3 sin θ -4sin³θ)

= sin^-1 (sin 3θ) = 3θ = 3sin^-1 x

= LHS

 

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

Answer 2:

To prove: cos ⁻¹x = cos ⁻¹ (4x³ − 3x), x ∈ [(1/2) , 1].

Let cos^-1 x = θ, then x = cos θ.

We have,

RHS = cos^-1 (4x³ -3x) = cos^-1 (4cos³θ-3cosθ) =

cos^-1 (cos 3θ) = 3θ = 3cos^-1 x

= LHS

Question 3: Prove that tan⁻¹ (2/11) + tan⁻¹ (7/24) = tan⁻¹ (1/2)

Answer 3:

 To prove: tan^-1 (2/11) + tan^-1 (7/24) = tan^-1 (1/2)

LHS = tan^-1 (2/11) + tan^-1 (7/24)

= tan⁻¹ ((48 + 77) /(264 – 14)) = tan⁻¹ (125/251) = tan⁻¹ (1/2) = RHS

Question 4: Prove that 2tan−1 1 2 + tan−1 1 7 = tan−1 31 17

Answer 4:

To prove: 2tan^-1 (1/2) + tan^-1 (1/7) = tan^-1 (31/17)

LHS = 2tan^-1 (1/2) + tan^-1 (1/7)

Question 5: Write the function , x ≠ 0, in the simplest form.

Answer 5:

Given function

Let x = tan θ

= (𝜃/2) = (1/2) tan⁻¹x

 

Question 6: Write the function , | x| > 1, in the simplest form.

Answer 6:

Given function

Let x = cosec θ

= tan^-1 (1/cot θ) = tan^-1 (tan θ = θ = cosec^-1 x

= (π/2) -sec^-1 x

 

Question 7: Write the function , x < π, in the simplest form.

Answer 7:

The given function is

Now,

 

Question 8: Write the function tan⁻¹ ((cos x−sin x) (cos x+sin x)) , 0 < x < π, in the simplest form.

Answer 8:

 The given function is tan⁻¹ ((cos x−sin x) (cos x+sin x)) Now,

Question 9: Write the function , |x| <a, in the simplest form.

Answer 9:

The given function is

Let x = a sin θ

= tan⁻¹ (tan θ) = θ = sin⁻¹ (x/a)

Question 10: Write the function in tan⁻¹ ((3a²x – x³)( a³ −3ax²)) , a > 0; , the simplest form.

Answer 10:

 The given function is tan⁻¹ ((3a²x – x³)( a³ −3ax²))

Let x = a tan θ

= tan⁻¹ (tan3θ) = 3θ = 3tan⁻¹ (x/a)

Question 11: Find the value of tan⁻¹ [2cos  (2sin⁻¹ (1/2))]

Answer 11:

The given function is tan^-1 [2cos (2sin^-1 (1/2))]

∴ tan^-1 [2cos (2sin^-1 (1/2))] = tan^-1 [2cos (2sin^-1 (sin (π/6)))]

= tan^-1 [2cos (2 × (π/6))] = tan^-1 [2cos (π/3)] = tan^-1 [2 × (1/2)]

= tan^-1 [1] = (π/4)

 

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

Answer 12:

The given function is cot(tan^-1 a + cot^-1 a).

∴ cot(tan^-1 a + cot^-1 a) = cot (π/2)             [as tan⁻¹x + cot⁻¹x = (π/2)]

= 0

Question 13: Find the value of tan (1/2) [(sin⁻¹ (2x/(1+x²))) + ((cos ⁻¹ (1−y²)/(1+y²)) ], |x| < 1, y > 0 and xy < 1.

Answer 13:

 The given function is tan (1/2) [(sin^-1 (2x /(1+x 2))) + ((cos^-1 (1-y²)/(1+y²)) ]

∴ tan (1/2) [(sin^-1 (2x/(1+x²))) + ((cos^-1 (1-y²)/(1+y²)))]

= tan (1/2) [2tan^-1 x + 2tan^-1y]                   [As  2tan⁻¹x = sin⁻¹ (2x/(1+x²)) = cos ⁻¹ (1−x²)/(1+x²)) ]

= tan (1/2) [2(tan⁻¹x + tan⁻¹y)] = tan[tan⁻¹x + tan⁻¹y]

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

 

Question 14: If sin(sin⁻¹ (1/5) +cos⁻¹x) = 1, then find the value of x.

Answer 14:

Since, sin (sin^-1 (1/5) + cos^-1 x) = 1

∴ (sin^-1 (1/5) + cos^-1 x) = sin^-11

⇒ (sin^-1 (1/5) + cos^-1 x) = (π/2)

⇒ sin⁻¹ (1/5) = sin⁻¹x                  [as sin−1x + sin −1x = π 2 ]

⇒ x = (1/5)

 

Question 15: If tan⁻¹ ((x−1)/(x−2)) + tan⁻¹ ((x+1)/(x+2)) = (π/4) , then find the value of x.

Answer 15:

Given that tan⁻¹ ((x−1)/(x−2)) + tan⁻¹ ((x+1)/(x+2)) = (π/4)

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

⇒ (x² + 2x − x − 2 + x² + x − 2x – 2) / ((x² – 4 − (x² − 1)) = 1 ⇒ ((2x² – 4)/ −3) = 1

⇒ 2x² − 4 = −3 ⇒ x² = (1/2) ⇒ x = ± .

Question 16: Find the values of sin⁻¹ (sin (2π/3)).

Answer 16:

Given that sin^-1 (sin (2π/3)).

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

∴ sin^-1 (sin (2π/3)) = sin^-1 (sin {π-(π/3)}) = sin^-1 (sin (π/3)) = (π/3) ∈ [-(π/2), (π/2)]

Hence, sin^-1 (sin (2π/3)) = (π/3)

 

Question 17: Find the values of tan⁻¹ (tan (3π/4)).

Answer 17:

Given that tan^-1 (tan (3π/4))

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 (3π/4)) = tan⁻¹ (tan {π – (π/4)}) = tan⁻¹ (− tan (π/4))

= tan⁻¹ (tan {− (π/4)}) = − (π/4) ∈ (− (π/2) , (π/2))

Hence, tan⁻¹ (tan (3π/4)) = − (π/4)

 

Question 18: Find the values of tan (sin⁻¹ (3/5) + cot⁻¹ (3/2)).

Answer 18:

Given that tan (sin⁻¹ (3/5) + cot⁻¹ (3/2))

∴ tan (sin⁻¹ (3/5) + cot⁻¹ (3/2)) = tan (tan⁻¹ + tan⁻¹ (2/3))

[as sin⁻¹ (a/b) = tan⁻¹ and cot⁻¹ (a/b) = tan⁻¹ (b/a)]

= tan (tan⁻¹ (3/4) + tan⁻¹ (2/3))

= 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

Answer 19:

 Given that cos^-1 (cos (7π/6))

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

∴ cos^-1 (cos (7π/6)) = cos^-1 [cos (2π- (5π/6))] = cos^-1 (cos (5π/6)) = (5π/6) ∈ [0, π]

Hence, cos^-1 (cos (7π/6)) = (5π/6)

The correct answer 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

Answer 20:

Given that sin ((π/3) -sin^-1 (- (1/2)))

We know that the range of the principal value branch of sin^–1 is [- (π/2), (π/2)].

∴ 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 (3π/6) = sin (π/2) = 1

Hence, sin ((π/3) − sin⁻¹ (−(1/2))) = 1

The correct answer is D.

Question 21: tan⁻¹ − cot⁻¹ (−) is equal to

(A) π

(B) – (π/2)

(C) 0

(D)

Answer 21:

Given that tan^-1-cot^-1 (-)

We know that the range of the principal value branch of tan^–1 is (- (π/2), (π/2)) and cot^–1 is (0, π).

= (π/3) – (5π/6) = ((2π − 5π) / 6) = − (3π/6) = − (π/2)

Hence, tan⁻¹ − cot⁻¹ (−) = − (π/2)

The correct answer is B.