INVERSE TRIGONOMETRIC FUNCTIONS

Exercise 1

Q1: Find the principal values of the following:

1. sin⁻¹(−1/2)

We know sin(−θ) = −sin(θ)
and sin(π/6) = 1/2

∴ sin⁻¹(−1/2) = −π/6

Answer: −π/6

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2. cos⁻¹(√3/2)

We know cos(π/6) = √3/2

∴ cos⁻¹(√3/2) = π/6

Answer: π/6

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3. cosec⁻¹(2)

We know cosec(π/6) = 2

∴ cosec⁻¹(2) = π/6

Answer: π/6

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4. tan⁻¹(−√3)

We know tan(−θ) = −tan(θ)
and tan(π/3) = √3

∴ tan⁻¹(−√3) = −π/3

Answer: −π/3

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5. cos⁻¹(−1/2)

We know cos(2π/3) = −1/2

∴ cos⁻¹(−1/2) = 2π/3

Answer: 2π/3

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6. tan⁻¹(−1)

We know tan(π/4) = 1 and tan(−π/4) = −1

∴ tan⁻¹(−1) = −π/4

Answer: −π/4

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7. sec⁻¹(2/√3)

We know sec(π/6) = 2/√3

∴ sec⁻¹(2/√3) = π/6

Answer: π/6

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8. cot⁻¹(√3)

We know cot(π/6) = √3

∴ cot⁻¹(√3) = π/6

Answer: π/6

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9. cos⁻¹(−1/√2)

We know cos(3π/4) = −1/√2

∴ cos⁻¹(−1/√2) = 3π/4

Answer: 3π/4

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10. cosec⁻¹(−√2)

We know cosec(−π/4) = −√2

∴ cosec⁻¹(−√2) = −π/4

Answer: −π/4

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Q2: Find the values of the following:

11. tan⁻¹(1) + cos⁻¹(−1/2) + sin⁻¹(−1/2)

We know
tan⁻¹(1) = π/4
cos⁻¹(−1/2) = 2π/3
sin⁻¹(−1/2) = −π/6

∴ Required value = π/4 + 2π/3 − π/6
= (3π + 8π − 2π) / 12
= 9π/12
= 3π/4

Answer: 3π/4

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12. cos⁻¹(1/2) + 2 sin⁻¹(1/2)

We know
cos⁻¹(1/2) = π/3
sin⁻¹(1/2) = π/6

∴ Required value = π/3 + 2 × (π/6)
= π/3 + π/3
= 2π/3

Answer: 2π/3

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13. If sin⁻¹(x) = y, then

By definition of principal value of sin⁻¹(x),
the range is [−π/2, π/2]

Answer: (B) −π/2 ≤ y ≤ π/2

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14. tan⁻¹(√3) − sec⁻¹(−2) is equal to

We know
tan⁻¹(√3) = π/3

Now,
sec⁻¹(−2) means the angle whose secant is −2.
Since sec(θ) = −2 ⇒ cos(θ) = −1/2

So, θ = 2π/3 (because cos(2π/3) = −1/2 and range of sec⁻¹x is [0, π] – {π/2}).

∴ sec⁻¹(−2) = 2π/3

Now,
tan⁻¹(√3) − sec⁻¹(−2) = π/3 − 2π/3 = −π/3

Answer: −π/3

Option (B)

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EXERCISE 2.2

Q1: Prove the following:

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

Proof:
Let θ = sin⁻¹ x. Then x = sin θ and θ ∈ [−π/2, π/2]. For x ∈ [−1/2, 1/2] we have θ ∈ [−π/6, π/6].
Use the triple-angle identity for sine:
sin 3θ = 3 sin θ − 4 sin³ θ.
Substitute sin θ = x to get
sin 3θ = 3x − 4x³.
Since 3θ ∈ [−π/2, π/2], the principal value satisfies
sin⁻¹(3x − 4x³) = sin⁻¹(sin 3θ) = 3θ = 3 sin⁻¹ x.
Hence 3 sin⁻¹ x = sin⁻¹(3x − 4x³) for x ∈ [−1/2, 1/2].

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

Proof:
Let φ = cos⁻¹ x. Then x = cos φ and φ ∈ [0, π]. For x ∈ [1/2, 1] we have φ ∈ [0, π/3].
Use the triple-angle identity for cosine:
cos 3φ = 4 cos³ φ − 3 cos φ.
Substitute cos φ = x to get
cos 3φ = 4x³ − 3x.
Since 3φ ∈ [0, π], the principal value satisfies
cos⁻¹(4x³ − 3x) = cos⁻¹(cos 3φ) = 3φ = 3 cos⁻¹ x.
Hence 3 cos⁻¹ x = cos⁻¹(4x³ − 3x) for x ∈ [1/2, 1].

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Q2: Write the following functions in the simplest form

3. tan⁻¹((√(1 + x²) − 1) / x), x ≠ 0

Let x = tanθ.
Then √(1 + x²) = secθ.

So,
(√(1 + x²) − 1) / x
= (secθ − 1) / tanθ
= (1 − cosθ) / sinθ
= tan(θ / 2).

Hence,
tan⁻¹((√(1 + x²) − 1) / x) = θ / 2 = (1/2) tan⁻¹x.

Answer: (1/2) tan⁻¹x

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4. tan⁻¹(√((1 − cos x) / (1 + cos x))), 0 < x < π

We know that
tan²(x/2) = (1 − cos x) / (1 + cos x).

So,
√((1 − cos x) / (1 + cos x)) = tan(x / 2).

Hence,
tan⁻¹(√((1 − cos x) / (1 + cos x))) = tan⁻¹(tan(x / 2)) = x / 2.

Answer: x / 2

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5. tan⁻¹((cos x − sin x) / (cos x + sin x)), −π/4 < x < 3π/4

Divide numerator and denominator by cos x:

= (1 − tan x) / (1 + tan x)
= tan(π/4 − x).

Hence,
tan⁻¹((cos x − sin x) / (cos x + sin x)) = π/4 − x.

Answer: π/4 − x

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6. tan⁻¹(x / √(a² − x²)), |x| < a

Let x = a sinθ.
Then √(a² − x²) = a cosθ.

So,
x / √(a² − x²) = tanθ.

Hence,
tan⁻¹(x / √(a² − x²)) = θ = sin⁻¹(x / a).

Answer: sin⁻¹(x / a)

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7. tan⁻¹((3a²x − x³) / (a³ − 3ax²)), a > 0, −a/√3 < x < a/√3

Let x = a tanθ.

Then
(3a²x − x³) / (a³ − 3ax²)
= (3 tanθ − tan³θ) / (1 − 3 tan²θ)
= tan(3θ).

Hence,
tan⁻¹((3a²x − x³) / (a³ − 3ax²)) = 3θ = 3 tan⁻¹(x / a).

Answer: 3 tan⁻¹(x / a)

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Find the values of each of the following:

8. tan⁻¹[ 2 cos ( 2 sin⁻¹ (1/2) ) ]

Solution:
sin⁻¹(1/2) = π/6
2 × sin⁻¹(1/2) = 2 × (π/6) = π/3
cos(π/3) = 1/2
2 × cos(π/3) = 1

∴ tan⁻¹(1) = π/4

Answer: π/4

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Find the value of

9. tan ½ [ sin⁻¹ ( 2x / (1 + x²) ) + cos⁻¹ ( (1 − y²) / (1 + y²) ) ],
where |x| < 1, y > 0 and xy < 1.

Solution:

1. Let x = tan α.
   Then α = tan⁻¹x and since |x| < 1, α lies between −π/4 and π/4.

Now, sin(2α) = (2 tan α) / (1 + tan²α) = 2x / (1 + x²).
Therefore, sin⁻¹(2x / (1 + x²)) = 2α.

2. Let y = tan β.
   Then β = tan⁻¹y and y > 0 ⇒ β lies between 0 and π/2.

Now, cos(2β) = (1 − tan²β) / (1 + tan²β) = (1 − y²) / (1 + y²).
Therefore, cos⁻¹((1 − y²) / (1 + y²)) = 2β.

3. Substituting these results,

½ [ sin⁻¹(2x / (1 + x²)) + cos⁻¹((1 − y²) / (1 + y²)) ]
= ½ [ 2α + 2β ]
= α + β.

4. Hence,
   tan(½ [ sin⁻¹(2x / (1 + x²)) + cos⁻¹((1 − y²) / (1 + y²)) ])
   = tan(α + β)
   = (tan α + tan β) / (1 − tan α tan β)
   = (x + y) / (1 − xy).

Final Answer: (x + y) / (1 − xy)

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Find the values of each of the expressions

10. sin⁻¹( sin(2π/3) )

sin(2π/3) = √3/2.
Principal value of sin⁻¹ for √3/2 is π/3 (because sin range is [−π/2, π/2]).

Answer: π/3

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11. tan⁻¹( tan(3π/4) )

tan(3π/4) = −1.
Principal value of tan⁻¹ for −1 is −π/4 (range of tan⁻¹ is (−π/2, π/2)).

Answer: −π/4

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12. tan( sin⁻¹(3/5) + cot⁻¹(3/2) )

Let A = sin⁻¹(3/5). Then sin A = 3/5 ⇒ cos A = 4/5 ⇒ tan A = 3/4.
Let B = cot⁻¹(3/2). Then cot B = 3/2 ⇒ tan B = 2/3.

Use tan(A + B) = (tan A + tan B) / (1 − tan A · tan B).

Compute numerator: 3/4 + 2/3 = (9 + 8)/12 = 17/12.
Compute denominator: 1 − (3/4)(2/3) = 1 − 1/2 = 1/2.

So tan(A + B) = (17/12) ÷ (1/2) = (17/12) × 2 = 17/6.

Answer: 17/6

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13. cos⁻¹( cos(7π/6) )

cos(7π/6) = −√3/2.
Principal value of cos⁻¹ for −√3/2 (range [0, π]) is 5π/6.

Answer: 5π/6

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14. sin( π/3 − sin⁻¹(−1/2) )

sin⁻¹(−1/2) = −π/6.
So π/3 − (−π/6) = π/3 + π/6 = π/2.
sin(π/2) = 1.

Answer: 1

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15. tan⁻¹(√3) − cot⁻¹(−√3)

tan⁻¹(√3) = π/3.
Find cot⁻¹(−√3). Principal range of cot⁻¹ is (0, π). Angle in (0, π) whose cot is −√3 is 5π/6 (because cot 5π/6 = −√3). So cot⁻¹(−√3) = 5π/6.

Therefore difference = π/3 − 5π/6 = −π/2.

Answer: −π/2

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Miscellaneous Exercise on Chapter 2

Find the value of the following:

1. cos⁻¹( cos(13π/6) )

13π/6 = 2π + π/6,

so cos(13π/6) = cos(π/6) = √3/2.

Principal value of cos⁻¹ for √3/2 (range 0 to π) is π/6.

Answer: π/6

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2. tan⁻¹( tan(7π/6) )

7π/6 = π + π/6, and tan(π + θ) = tan θ,

so tan(7π/6) = tan(π/6) = 1/√3.

Principal value of tan⁻¹ for 1/√3 (range −π/2 to π/2) is π/6.

Answer: π/6

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Prove that

3. 2 sin⁻¹(3/5) = tan⁻¹(24/7)

Let A = sin⁻¹(3/5). Then sin A = 3/5, cos A = 4/5, so tan A = 3/4.
Use the double-angle formula for tangent:
tan 2A = 2 tan A / (1 − tan² A) = 2·(3/4) / (1 − 9/16) = (3/2) / (7/16) = 24/7.
Thus 2A = tan⁻¹(24/7). Therefore 2 sin⁻¹(3/5) = tan⁻¹(24/7).

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

Let A = sin⁻¹(8/17). Then sin A = 8/17, cos A = 15/17, tan A = 8/15.
Let B = sin⁻¹(3/5). Then sin B = 3/5, cos B = 4/5, tan B = 3/4.
Now use tan(A + B) = (tan A + tan B) / (1 − tan A·tan B).

Numerator = 8/15 + 3/4 = (32 + 45)/60 = 77/60.
Denominator = 1 − (8/15)(3/4) = 1 − 24/60 = 36/60 = 3/5.

Hence tan(A + B) = (77/60) ÷ (36/60) = 77/36.

So A + B = tan⁻¹(77/36). Thus sin⁻¹(8/17) + sin⁻¹(3/5) = tan⁻¹(77/36).

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5. cos⁻¹(4/5) + cos⁻¹(12/13) = cos⁻¹(33/65)

Let A = cos⁻¹(4/5). Then cos A = 4/5, sin A = 3/5.
Let B = cos⁻¹(12/13). Then cos B = 12/13, sin B = 5/13.
Use cos(A + B) = cos A cos B − sin A sin B:

cos(A + B) = (4/5)(12/13) − (3/5)(5/13) = (48 − 15)/65 = 33/65.

Hence A + B = cos⁻¹(33/65). Therefore cos⁻¹(4/5) + cos⁻¹(12/13) = cos⁻¹(33/65). □

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6. cos⁻¹(12/13) + sin⁻¹(3/5) = sin⁻¹(56/65)

Let A = cos⁻¹(12/13). Then cos A = 12/13, sin A = 5/13.
Let B = sin⁻¹(3/5). Then sin B = 3/5, cos B = 4/5.
Use sin(A + B) = sin A cos B + cos A sin B:

sin(A + B) = (5/13)(4/5) + (12/13)(3/5) = 4/13 + 36/65 = (20 + 36)/65 = 56/65.

Hence A + B = sin⁻¹(56/65). Therefore cos⁻¹(12/13) + sin⁻¹(3/5) = sin⁻¹(56/65). □

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7. tan⁻¹(63/16) = sin⁻¹(5/13) + cos⁻¹(3/5)

Let A = sin⁻¹(5/13). Then sin A = 5/13, cos A = 12/13, tan A = 5/12.
Let B = cos⁻¹(3/5). Then cos B = 3/5, sin B = 4/5, tan B = 4/3.
Use tan(A + B) = (tan A + tan B) / (1 − tan A·tan B).

Numerator = 5/12 + 4/3 = 5/12 + 16/12 = 21/12 = 7/4.
Denominator = 1 − (5/12)(4/3) = 1 − 20/36 = 1 − 5/9 = 4/9.

Hence tan(A + B) = (7/4) ÷ (4/9) = (7/4)·(9/4) = 63/16.

So A + B = tan⁻¹(63/16). Therefore tan⁻¹(63/16) = sin⁻¹(5/13) + cos⁻¹(3/5).

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8. Prove that
tan⁻¹(√x) = ½ cos⁻¹((1 − x) / (1 + x)), for x ∈ [0, 1].

Proof:
Let θ = tan⁻¹(√x). Then tan θ = √x and θ ∈ [0, π/4]. Hence tan²θ = x.
Use the identity for cosine of a double angle in terms of tangent:
cos 2θ = (1 − tan²θ) / (1 + tan²θ) = (1 − x) / (1 + x).
Therefore 2θ = cos⁻¹((1 − x) / (1 + x)), so θ = ½ cos⁻¹((1 − x) / (1 + x)).
Thus tan⁻¹(√x) = ½ cos⁻¹((1 − x) / (1 + x)). □

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9. Prove that
cot⁻¹( [√(1 + sin x) + √(1 − sin x)] / [√(1 + sin x) − √(1 − sin x)] ) = x/2, for x ∈ (0, π/4).

Proof:
Write sin x in half-angle form: sin x = 2 sin(x/2) cos(x/2). Then
1 + sin x = [sin(x/2) + cos(x/2)]² and 1 − sin x = [cos(x/2) − sin(x/2)]².
Since x ∈ (0, π/4), both sin(x/2) and cos(x/2) are positive, so
√(1 + sin x) = sin(x/2) + cos(x/2),
√(1 − sin x) = cos(x/2) − sin(x/2).
Now compute the fraction:
[√(1 + sin x) + √(1 − sin x)] / [√(1 + sin x) − √(1 − sin x)]
= [ (s + c) + (c − s) ] / [ (s + c) − (c − s) ]
= (2c) / (2s) = cot(x/2).
Hence the left side is cot⁻¹(cot(x/2)).

Because x/2 ∈ (0, π/8) lies in the principal range (0, π) for cot⁻¹, we get cot⁻¹(cot(x/2)) = x/2.

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10. Prove that
tan⁻¹( [√(1 + x) − √(1 − x)] / [√(1 + x) + √(1 − x)] ) = π/4 − ½ cos⁻¹ x, for −1/√2 ≤ x ≤ 1.
(Hint: put x = cos 2θ.)

Proof:
Put x = cos 2θ. Then cos 2θ = x, so 2θ = cos⁻¹ x and θ = ½ cos⁻¹ x. Using half-angle formulas,
√(1 + x) = √(1 + cos 2θ) = √(2 cos²θ) = √2 cos θ,
√(1 − x) = √(1 − cos 2θ) = √(2 sin²θ) = √2 sin θ.
Substitute into the given ratio:
[√(1 + x) − √(1 − x)] / [√(1 + x) + √(1 − x)]
= [√2(cos θ − sin θ)] / [√2(cos θ + sin θ)]
= (cos θ − sin θ) / (cos θ + sin θ)
= tan(π/4 − θ).
Therefore the left side equals tan⁻¹(tan(π/4 − θ)) = π/4 − θ (θ lies in a range making this the principal value). But θ = ½ cos⁻¹ x, so the expression equals π/4 − ½ cos⁻¹ x.

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Solve the following equations:

11. 2 tan⁻¹(cos x) = tan⁻¹(2 cosec x)

Let tan⁻¹(cos x) = θ
⇒ tan θ = cos x

Now,
Left side = 2 tan⁻¹(cos x) = tan⁻¹(2 tan θ / (1 – tan² θ))

Substitute tan θ = cos x:

tan⁻¹[(2 cos x) / (1 – cos² x)]
= tan⁻¹[(2 cos x) / (sin² x)]
= tan⁻¹(2 cot x cosec x)

But we are given that
2 tan⁻¹(cos x) = tan⁻¹(2 cosec x)

So comparing both sides:
tan⁻¹(2 cot x cosec x) = tan⁻¹(2 cosec x)

⇒ cot x = 1
⇒ tan x = 1
⇒ x = nπ + π/4, n ∈ Z

Final Answer:
x = nπ + π/4, n ∈ Z

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12. tan⁻¹((1 − x)/(1 + x)) = ½ tan⁻¹ x, (x > 0)

Let θ = tan⁻¹ x. Then x = tan θ and θ ∈ (0, π/2).

Use the identity
tan(π/4 − θ) = (1 − tan θ) / (1 + tan θ).

Hence the left side equals
tan⁻¹((1 − x)/(1 + x)) = tan⁻¹(tan(π/4 − θ)) = π/4 − θ,
since π/4 − θ lies in (−π/4, π/4).

The equation becomes
π/4 − θ = (1/2) θ.

Solve for θ:
π/4 = (3/2) θ ⇒ θ = (2/3) · (π/4) = π/6.

Thus x = tan θ = tan(π/6) = 1/√3.

Answer: x = 1 / √3 (which is > 0, so valid).

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13: sin (tan⁻¹x), |x| < 1 is equal to

(A) x / √(1 − x²)
(B) 1 / √(1 − x²)
(C) 1 / √(1 + x²)
(D) x / √(1 + x²)

Solution:
Let θ = tan⁻¹x
⇒ tan θ = x

We know,
sin θ = (tan θ) / √(1 + tan²θ)

Substitute tan θ = x,
sin θ = x / √(1 + x²)

Hence,
sin (tan⁻¹x) = x / √(1 + x²)

Correct Answer: (D)

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

(A) 0, ½
(B) 1, ½
(C) 0
(D) ½

Solution:
Let sin⁻¹x = θ
⇒ x = sin θ

Then the equation becomes,
sin⁻¹(1 − sin θ) = π/2 + 2θ

Now, take sin on both sides:
1 − sin θ = sin(π/2 + 2θ)
⇒ 1 − sin θ = cos(2θ)
⇒ 1 − sin θ = 1 − 2sin²θ

Simplify:
sin θ = 2sin²θ
⇒ 2sin²θ − sin θ = 0
⇒ sin θ (2sin θ − 1) = 0

Hence,
sin θ = 0 or sin θ = ½

⇒ x = 0 or x = ½

Correct Answer: (A) 0, ½

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