Revision notes, worked problems, a question bank, assertion and reason questions and answers to the NCERT exercises. Tap a group to open it; opening one closes the others.
From Chapter 1, f⁻¹ exists only if f is one-one and onto. Over its natural domain, sin x repeats its values, so it is not one-one, and its inverse does not exist.
If we restrict sine to [−π/2, π/2] it becomes one-one and onto [−1, 1], so it has an inverse. Other intervals such as [π/2, 3π/2] would work equally well and give other branches. The branch we use when nothing else is said is the principal value branch.
| Function | Domain | Range (principal branch) |
|---|---|---|
| sin⁻¹ | [−1, 1] | [−π/2, π/2] |
| cos⁻¹ | [−1, 1] | [0, π] |
| tan⁻¹ | R | (−π/2, π/2) |
| cot⁻¹ | R | (0, π) |
| sec⁻¹ | R − (−1, 1) | [0, π] − {π/2} |
| cosec⁻¹ | R − (−1, 1) | [−π/2, π/2] − {0} |
The domain of an inverse function is the range of the original, and its range is the restricted domain of the original. That is how to recall the table.
The graph of the inverse is the reflection of the graph of the original in the line y = x, because (a, b) on one becomes (b, a) on the other. Each graph below shows the principal value branch only.
Example: principal value of cot⁻¹(−1/√3). Let it be y, so cot y = −1/√3 = −cot(π/3) = cot(π − π/3) = cot(2π/3), and 2π/3 ∈ (0, π). So the value is 2π/3.
| x | sin⁻¹ x | cos⁻¹ x | tan⁻¹ x |
|---|---|---|---|
| 0 | 0 | π/2 | 0 |
| 1/2 | π/6 | π/3 | — |
| 1/√2 | π/4 | π/4 | — |
| √3/2 | π/3 | π/6 | — |
| 1 | π/2 | 0 | π/4 |
| 1/√3 | — | — | π/6 |
| √3 | — | — | π/3 |
| −1/2 | −π/6 | 2π/3 | — |
| −1 | −π/2 | π | −π/4 |
| −√3 | — | — | −π/3 |
A dash means the value is not a standard angle or is outside the domain of that function.
| Function | Rule |
|---|---|
| sin⁻¹(−x) | −sin⁻¹ x |
| tan⁻¹(−x) | −tan⁻¹ x |
| cosec⁻¹(−x) | −cosec⁻¹ x |
| cos⁻¹(−x) | π − cos⁻¹ x |
| cot⁻¹(−x) | π − cot⁻¹ x |
| sec⁻¹(−x) | π − sec⁻¹ x |
The first three are odd functions. The last three reflect through π because their ranges start at 0. Example: cos⁻¹(−1/2) = π − π/3 = 2π/3.
| Identity | Valid for |
|---|---|
| sin⁻¹ x + cos⁻¹ x = π/2 | −1 ≤ x ≤ 1 |
| tan⁻¹ x + cot⁻¹ x = π/2 | all real x |
| sec⁻¹ x + cosec⁻¹ x = π/2 | |x| ≥ 1 |
Reason: if θ = sin⁻¹ x then x = sin θ = cos(π/2 − θ), and π/2 − θ ∈ [0, π], the branch of cos⁻¹. So cos⁻¹ x = π/2 − θ.
| Identity | Valid for |
|---|---|
| cosec⁻¹ x = sin⁻¹(1/x) | |x| ≥ 1 |
| sec⁻¹ x = cos⁻¹(1/x) | |x| ≥ 1 |
| cot⁻¹ x = tan⁻¹(1/x) | x > 0 only |
| cot⁻¹ x = π + tan⁻¹(1/x) | x < 0 |
For x < 0, tan⁻¹(1/x) is negative but cot⁻¹ x lies in (0, π), so the plain identity fails. For example cot⁻¹(−1) = 3π/4, while tan⁻¹(−1) = −π/4.
For sine, with the restricted branch:
Similar results hold for the other functions, each on its own branch: cos⁻¹(cos x) = x for x ∈ [0, π], tan⁻¹(tan x) = x for x ∈ (−π/2, π/2), and so on.
The first rule works for every x in the domain. The second works only when x already lies in the branch.
sin⁻¹(sin x) is a zig-zag outside [−π/2, π/2], not the line y = x:
Example 6: sin⁻¹(sin 3π/5). 3π/5 is not in [−π/2, π/2], but sin(3π/5) = sin(π − 3π/5) = sin(2π/5) and 2π/5 is in the branch. So the answer is 2π/5.
Another: cos⁻¹(cos 7π/6). 7π/6 ∉ [0, π], and cos(7π/6) = cos(2π − 7π/6) = cos(5π/6), with 5π/6 ∈ [0, π]. So the answer is 5π/6.
Wrong shortcut: 'sin⁻¹(sin x) = x always'. It is false for x = 3π/5, where the answer is 2π/5.
For expressions like sin(cos⁻¹ x) or tan(sin⁻¹ x), set the inner function equal to θ, build a right triangle, and read off the trigonometric ratio. Pay attention to the sign: the branch tells you whether the angle is acute, obtuse or negative.
So sin(tan⁻¹ x) = x/√(1 + x²) and cos(tan⁻¹ x) = 1/√(1 + x²). Also sin(cos⁻¹ x) = cos(sin⁻¹ x) = √(1 − x²), since an angle in [0, π] or [−π/2, π/2] respectively has a non-negative sine or cosine.
Replace x by a trigonometric function of θ so that the expression under the inverse function becomes a single trigonometric ratio. Then the inverse and the ratio cancel, provided θ lies in the principal branch. Always state the range of θ.
| Expression | Put | Becomes |
|---|---|---|
| √(1 − x²) | x = sin θ or cos θ | cos θ or sin θ |
| √(1 + x²) | x = tan θ | sec θ |
| √(x² − 1) | x = sec θ | tan θ |
| √(a² − x²) | x = a sin θ | a cos θ |
| 2x/(1 + x²) | x = tan θ | sin 2θ |
| (1 − x²)/(1 + x²) | x = tan θ | cos 2θ |
| 2x/(1 − x²) | x = tan θ | tan 2θ |
| 3x − 4x³ | x = sin θ | sin 3θ |
| 4x³ − 3x | x = cos θ | cos 3θ |
| (3x − x³)/(1 − 3x²) | x = tan θ | tan 3θ |
Show sin⁻¹(2x√(1 − x²)) = 2 sin⁻¹ x for −1/√2 ≤ x ≤ 1/√2.
Let x = sin θ, so θ ∈ [−π/4, π/4]. Then 2x√(1 − x²) = 2 sin θ cos θ = sin 2θ, and 2θ ∈ [−π/2, π/2], so sin⁻¹(sin 2θ) = 2θ = 2 sin⁻¹ x. For 1/√2 ≤ x ≤ 1, put x = cos θ, θ ∈ [0, π/4], and the result is 2 cos⁻¹ x.
Write tan⁻¹(cos x/(1 − sin x)), −3π/2 < x < π/2, in the simplest form.
Use cos x = cos²(x/2) − sin²(x/2) and 1 − sin x = (cos(x/2) − sin(x/2))². The ratio is (cos(x/2) + sin(x/2))/(cos(x/2) − sin(x/2)) = (1 + tan(x/2))/(1 − tan(x/2)) = tan(π/4 + x/2). Since π/4 + x/2 ∈ (−π/2, π/2), the answer is π/4 + x/2.
cot⁻¹(1/√(x² − 1)), x > 1: put x = sec θ, so √(x² − 1) = tan θ and the expression is cot⁻¹(cot θ) = θ = sec⁻¹ x.
If the substitution gives the wrong interval for θ, the 'simplification' is valid only on a different range of x. Always check the interval, as in Exercise 2.2 questions 4 to 7.
Convert each term to an angle, take the tangent of the sum, and use tan(A + B) = (tan A + tan B)/(1 − tan A tan B). For the principal value of tan⁻¹ x + tan⁻¹ y:
| Condition | Result |
|---|---|
| xy < 1 | tan⁻¹((x + y)/(1 − xy)) |
| xy > 1, x, y > 0 | π + tan⁻¹((x + y)/(1 − xy)) |
| xy > 1, x, y < 0 | −π + tan⁻¹((x + y)/(1 − xy)) |
Example: tan⁻¹ 2 + tan⁻¹ 3: xy = 6 > 1 with both positive, so the sum is π + tan⁻¹(5/(1 − 6)) = π + tan⁻¹(−1) = 3π/4.
| Identity | Valid for |
|---|---|
| 2 tan⁻¹ x = sin⁻¹(2x/(1 + x²)) | |x| ≤ 1 |
| 2 tan⁻¹ x = cos⁻¹((1 − x²)/(1 + x²)) | x ≥ 0 |
| 2 tan⁻¹ x = tan⁻¹(2x/(1 − x²)) | |x| < 1 |
For results like sin⁻¹(8/17) + sin⁻¹(3/5), find the sine, cosine or tangent of the sum with the addition formulas. Then choose the inverse function whose branch contains the angle, and check the size of the sum. This is how Exercise Misc questions 3 to 7 work: for 2 sin⁻¹(3/5), tan θ = 3/4 and tan 2θ = 24/7, and 2θ is acute because 3/5 < 1/√2.
| Result | Condition |
|---|---|
| sin(sin⁻¹ x) = x | −1 ≤ x ≤ 1 |
| sin⁻¹(sin x) = x | −π/2 ≤ x ≤ π/2 |
| cos⁻¹(cos x) = x | 0 ≤ x ≤ π |
| tan⁻¹(tan x) = x | −π/2 < x < π/2 |
| sin⁻¹ x + cos⁻¹ x = π/2 | −1 ≤ x ≤ 1 |
| tan⁻¹ x + cot⁻¹ x = π/2 | x real |
| sin⁻¹(−x) = −sin⁻¹ x | −1 ≤ x ≤ 1 |
| cos⁻¹(−x) = π − cos⁻¹ x | −1 ≤ x ≤ 1 |
| 3 sin⁻¹ x = sin⁻¹(3x − 4x³) | −1/2 ≤ x ≤ 1/2 |
| 3 cos⁻¹ x = cos⁻¹(4x³ − 3x) | 1/2 ≤ x ≤ 1 |
Try each problem first, then open the solution to check your working.
Find the principal values of sin⁻¹(√3/2), cos⁻¹(−√3/2) and tan⁻¹(−1/√3).
Use: Find the angle in the principal branch
sin⁻¹(√3/2) = π/3, since sin(π/3) = √3/2 and π/3 ∈ [−π/2, π/2].
cos⁻¹(−√3/2) = π − cos⁻¹(√3/2) = π − π/6 = 5π/6, which lies in [0, π].
tan⁻¹(−1/√3) = −tan⁻¹(1/√3) = −π/6.
Answer: π/3, 5π/6, −π/6
Find the value of sec⁻¹(−2) + cosec⁻¹(−2).
sec⁻¹(−2) = y gives cos y = −1/2, so y = 2π/3 ∈ [0, π] − {π/2}.
cosec⁻¹(−2) = z gives sin z = −1/2, so z = −π/6 ∈ [−π/2, π/2] − {0}.
Sum = 2π/3 − π/6 = π/2. This agrees with sec⁻¹ x + cosec⁻¹ x = π/2 for |x| ≥ 1.
Answer: π/2
Find the domain of the function f(x) = sin⁻¹(3x − 1).
Use: The input of sin⁻¹ must lie in [−1, 1]
Need −1 ≤ 3x − 1 ≤ 1. Adding 1: 0 ≤ 3x ≤ 2. Dividing by 3: 0 ≤ x ≤ 2/3.
Answer: [0, 2/3]
State whether each is defined: sin⁻¹(3/2), sec⁻¹(1/2), tan⁻¹(100), cosec⁻¹(−1). Give the value when it is.
sin⁻¹(3/2): 3/2 is outside [−1, 1], so it is not defined.
sec⁻¹(1/2): 1/2 lies in (−1, 1), outside the domain R − (−1, 1), so it is not defined.
tan⁻¹(100) is defined for every real number. It is slightly less than π/2.
cosec⁻¹(−1) is defined: sin y = −1 gives y = −π/2.
Answer: not defined; not defined; defined; −π/2
Find sin⁻¹(sin 5π/6), cos⁻¹(cos(−π/3)) and tan⁻¹(tan 5π/4).
Use: Use sin x = sin(π − x), cos(−x) = cos x, tan x = tan(x − π)
sin(5π/6) = sin(π/6), and π/6 ∈ [−π/2, π/2], so sin⁻¹(sin 5π/6) = π/6.
cos(−π/3) = cos(π/3), and π/3 ∈ [0, π], so cos⁻¹(cos(−π/3)) = π/3.
tan(5π/4) = tan(π/4), and π/4 ∈ (−π/2, π/2), so tan⁻¹(tan 5π/4) = π/4.
Answer: π/6, π/3, π/4
Find sin⁻¹(sin 7) and cos⁻¹(cos 4). Use π ≈ 3.14.
7 radians is more than 2π ≈ 6.28. sin 7 = sin(7 − 2π) with 7 − 2π ≈ 0.72 ∈ [−π/2, π/2]. So sin⁻¹(sin 7) = 7 − 2π.
4 radians is more than π. cos 4 = cos(2π − 4) with 2π − 4 ≈ 2.28 ∈ [0, π]. So cos⁻¹(cos 4) = 2π − 4.
Answer: 7 − 2π and 2π − 4
Find the values of sin(cos⁻¹(−4/5)), tan(cos⁻¹(5/13)) and cos(tan⁻¹(3/4)).
Use: Draw the right triangle
Let θ = cos⁻¹(−4/5), so θ ∈ (π/2, π] and sin θ ≥ 0. sin θ = √(1 − 16/25) = 3/5.
Let φ = cos⁻¹(5/13): sin φ = 12/13 and tan φ = 12/5.
Let ψ = tan⁻¹(3/4): the triangle has legs 3 and 4 and hypotenuse 5, so cos ψ = 4/5.
Answer: 3/5, 12/5, 4/5
Find sin(2 sin⁻¹(3/5)).
Use: sin 2θ = 2 sin θ cos θ
Let θ = sin⁻¹(3/5): sin θ = 3/5 and cos θ = 4/5 (θ is acute).
sin 2θ = 2 × 3/5 × 4/5 = 24/25.
Answer: 24/25
Simplify tan⁻¹(x/√(1 − x²)), |x| < 1.
Use: Put x = sin θ
θ = sin⁻¹ x ∈ (−π/2, π/2) and √(1 − x²) = cos θ > 0. The ratio is tan θ.
tan⁻¹(tan θ) = θ because θ ∈ (−π/2, π/2).
Answer: sin⁻¹ x
Show that cos⁻¹((1 − x²)/(1 + x²)) = 2 tan⁻¹ x for x ≥ 0.
Use: Put x = tan θ
θ = tan⁻¹ x ∈ [0, π/2). (1 − tan² θ)/(1 + tan² θ) = cos 2θ with 2θ ∈ [0, π].
So cos⁻¹(cos 2θ) = 2θ = 2 tan⁻¹ x. For x < 0 the right side is negative and the left is not, so the restriction matters.
Answer: proved
Show that sin⁻¹(3/5) + sin⁻¹(5/13) = sin⁻¹(56/65).
Use: sin(A + B) = sin A cos B + cos A sin B
A = sin⁻¹(3/5): sin A = 3/5, cos A = 4/5. B = sin⁻¹(5/13): sin B = 5/13, cos B = 12/13.
sin(A + B) = (3/5)(12/13) + (4/5)(5/13) = 36/65 + 20/65 = 56/65.
A + B ≈ 0.6435 + 0.3948 = 1.04 < π/2, so A + B lies in the principal branch of sin⁻¹.
Answer: proved
Prove that tan⁻¹(1/4) + tan⁻¹(2/9) = ½ tan⁻¹(4/3).
Use: tan(A + B)
tan(tan⁻¹(1/4) + tan⁻¹(2/9)) = (1/4 + 2/9)/(1 − 2/36) = (17/36)/(34/36) = 1/2, and xy < 1, so the sum is tan⁻¹(1/2).
2 tan⁻¹(1/2) = tan⁻¹((2 × 1/2)/(1 − 1/4)) = tan⁻¹(4/3), since 1/2 < 1.
So tan⁻¹(1/2) = ½ tan⁻¹(4/3), which is the claim.
Answer: proved
Show that tan⁻¹(1/3) + tan⁻¹(1/5) + tan⁻¹(1/7) + tan⁻¹(1/8) = π/4.
Use: Combine in pairs
tan⁻¹(1/3) + tan⁻¹(1/5) = tan⁻¹((8/15)/(14/15)) = tan⁻¹(4/7).
tan⁻¹(1/7) + tan⁻¹(1/8) = tan⁻¹((15/56)/(55/56)) = tan⁻¹(3/11).
tan⁻¹(4/7) + tan⁻¹(3/11) = tan⁻¹((65/77)/(1 − 12/77)) = tan⁻¹(65/65) = tan⁻¹ 1 = π/4. All products are less than 1, so each step uses the principal value.
Answer: proved
Show that 2 tan⁻¹(1/3) + tan⁻¹(1/7) = π/4.
2 tan⁻¹(1/3) = tan⁻¹((2/3)/(1 − 1/9)) = tan⁻¹(3/4).
tan⁻¹(3/4) + tan⁻¹(1/7) = tan⁻¹((3/4 + 1/7)/(1 − 3/28)) = tan⁻¹((25/28)/(25/28)) = tan⁻¹ 1 = π/4.
Answer: proved
Solve tan⁻¹ 2x + tan⁻¹ 3x = π/4.
Use: Take tan of both sides, then check
Taking tan: (2x + 3x)/(1 − 6x²) = 1, so 5x = 1 − 6x², i.e. 6x² + 5x − 1 = 0, which factors as (6x − 1)(x + 1) = 0. So x = 1/6 or x = −1.
Check x = 1/6: tan⁻¹(1/3) + tan⁻¹(1/2) = π/4. Correct.
Check x = −1: tan⁻¹(−2) + tan⁻¹(−3) is negative (about −2.36), not π/4. Rejected.
Answer: x = 1/6
Solve sin⁻¹ x + sin⁻¹(1 − x) = cos⁻¹ x.
Use: sin⁻¹ x + cos⁻¹ x = π/2
The equation gives sin⁻¹(1 − x) = cos⁻¹ x − sin⁻¹ x = π/2 − 2 sin⁻¹ x.
Take sine of both sides: 1 − x = cos(2 sin⁻¹ x) = 1 − 2x², so 2x² − x = 0 and x = 0 or x = 1/2.
Check x = 0: 0 + π/2 = π/2 = cos⁻¹ 0. Correct. Check x = 1/2: π/6 + π/6 = π/3 = cos⁻¹(1/2). Correct.
Answer: x = 0 or x = 1/2
Solve tan⁻¹((x − 1)/(x − 2)) + tan⁻¹((x + 1)/(x + 2)) = π/4.
Take tan of both sides: [(x − 1)(x + 2) + (x + 1)(x − 2)] / [(x − 2)(x + 2) − (x − 1)(x + 1)] = 1.
The top is (x² + x − 2) + (x² − x − 2) = 2x² − 4. The bottom is (x² − 4) − (x² − 1) = −3.
So 2x² − 4 = −3, giving x² = 1/2 and x = ±1/√2.
Both values satisfy the original equation (checked by substitution).
Answer: x = ±1/√2
If sin⁻¹ x + sin⁻¹ y = 2π/3, find cos⁻¹ x + cos⁻¹ y.
Use: cos⁻¹ t = π/2 − sin⁻¹ t
cos⁻¹ x + cos⁻¹ y = (π/2 − sin⁻¹ x) + (π/2 − sin⁻¹ y) = π − (sin⁻¹ x + sin⁻¹ y) = π − 2π/3 = π/3.
Answer: π/3
Solve cos⁻¹ x + sin⁻¹(x/2) = π/6.
Try x = 1: cos⁻¹ 1 = 0 and sin⁻¹(1/2) = π/6, so the sum is π/6.
No other root: on [−1, 1], cos⁻¹ x falls from π to 0 faster than sin⁻¹(x/2) rises, so the sum is strictly decreasing (its derivative is −1/√(1 − x²) + 1/√(4 − x²) < 0). A strictly decreasing function takes each value once.
Answer: x = 1
Find the range of f(x) = cos⁻¹ x − sin⁻¹ x.
Use: Use cos⁻¹ x = π/2 − sin⁻¹ x
f(x) = π/2 − 2 sin⁻¹ x for x ∈ [−1, 1]. Since sin⁻¹ x runs over [−π/2, π/2], 2 sin⁻¹ x runs over [−π, π].
So f(x) runs from π/2 − π = −π/2 (at x = 1) to π/2 + π = 3π/2 (at x = −1).
Answer: [−π/2, 3π/2]
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Answers to every exercise question of the NCERT chapter. Try each one first, then tap Show answer.
Let sin⁻¹(−1/2) = y, so sin y = −1/2 = sin(−π/6). The principal branch of sin⁻¹ is [−π/2, π/2] and −π/6 lies in it.
Answer: −π/6
cos y = √3/2 = cos(π/6) and π/6 ∈ [0, π].
Answer: π/6
cosec y = 2 gives sin y = 1/2, and the principal branch is [−π/2, π/2] − {0}; y = π/6 works.
Answer: π/6
tan y = −√3 = tan(−π/3) and −π/3 ∈ (−π/2, π/2).
Answer: −π/3
cos y = −1/2 = cos(π − π/3) = cos(2π/3) and 2π/3 ∈ [0, π].
Answer: 2π/3
tan(−π/4) = −1 and −π/4 ∈ (−π/2, π/2).
Answer: −π/4
sec y = 2/√3 gives cos y = √3/2, so y = π/6, which lies in [0, π] − {π/2}.
Answer: π/6
cot y = √3 = cot(π/6) and π/6 ∈ (0, π).
Answer: π/6
cos y = −1/√2 = cos(π − π/4) = cos(3π/4) and 3π/4 ∈ [0, π].
Answer: 3π/4
cosec y = −√2 gives sin y = −1/√2 = sin(−π/4), and −π/4 ∈ [−π/2, π/2] − {0}.
Answer: −π/4
tan⁻¹(1) = π/4, cos⁻¹(−1/2) = 2π/3, sin⁻¹(−1/2) = −π/6.
Sum = π/4 + 2π/3 − π/6 = (3π + 8π − 2π)/12 = 9π/12.
Answer: 3π/4
cos⁻¹(1/2) = π/3 and sin⁻¹(1/2) = π/6.
Sum = π/3 + 2(π/6) = π/3 + π/3.
Answer: 2π/3
The principal value branch of sin⁻¹ has range [−π/2, π/2], and the end points are included because sin(±π/2) = ±1.
Answer: (B) −π/2 ≤ y ≤ π/2
tan⁻¹ √3 = π/3. sec⁻¹(−2) = y gives cos y = −1/2, so y = 2π/3 (in [0, π] − {π/2}).
Difference = π/3 − 2π/3 = −π/3.
Answer: (B) −π/3
Let x = sin θ. Then θ = sin⁻¹ x, and x ∈ [−1/2, 1/2] gives θ ∈ [−π/6, π/6], so 3θ ∈ [−π/2, π/2].
3x − 4x³ = 3 sin θ − 4 sin³ θ = sin 3θ.
So sin⁻¹(3x − 4x³) = sin⁻¹(sin 3θ) = 3θ, because 3θ lies in the principal branch [−π/2, π/2].
Hence sin⁻¹(3x − 4x³) = 3 sin⁻¹ x.
Answer: proved
Let x = cos θ. Then θ = cos⁻¹ x, and x ∈ [1/2, 1] gives θ ∈ [0, π/3], so 3θ ∈ [0, π].
4x³ − 3x = 4 cos³ θ − 3 cos θ = cos 3θ.
So cos⁻¹(4x³ − 3x) = cos⁻¹(cos 3θ) = 3θ, because 3θ lies in [0, π].
Hence cos⁻¹(4x³ − 3x) = 3 cos⁻¹ x.
Answer: proved
Put x = tan θ, so θ = tan⁻¹ x ∈ (−π/2, π/2), θ ≠ 0. Then √(1 + x²) = sec θ.
(sec θ − 1)/tan θ = (1 − cos θ)/sin θ = 2 sin²(θ/2) / (2 sin(θ/2) cos(θ/2)) = tan(θ/2).
Since θ/2 ∈ (−π/4, π/4), tan⁻¹(tan(θ/2)) = θ/2.
Answer: ½ tan⁻¹ x
(1 − cos x)/(1 + cos x) = 2 sin²(x/2) / 2 cos²(x/2) = tan²(x/2).
For 0 < x < π, x/2 ∈ (0, π/2), so tan(x/2) > 0 and the square root is tan(x/2).
tan⁻¹(tan(x/2)) = x/2, since x/2 ∈ (−π/2, π/2).
Answer: x/2
For cos x ≠ 0, divide the numerator and denominator by cos x: (1 − tan x)/(1 + tan x) = tan(π/4 − x). The point x = π/2 is checked separately below.
For −π/4 < x < 3π/4 we get π/4 − x ∈ (−π/2, π/2), the principal branch of tan⁻¹.
So tan⁻¹(tan(π/4 − x)) = π/4 − x.
Check at x = π/2, where cos x = 0: the expression is (−1)/(1) = −1 and tan⁻¹(−1) = −π/4 = π/4 − π/2.
Answer: π/4 − x
Put x = a sin θ, so θ = sin⁻¹(x/a) ∈ (−π/2, π/2) (|x| < a).
√(a² − x²) = a cos θ, with cos θ > 0 on this interval.
x/√(a² − x²) = tan θ, and tan⁻¹(tan θ) = θ.
Answer: sin⁻¹(x/a)
Put x = a tan θ, so θ = tan⁻¹(x/a). From −a/√3 < x < a/√3 we get θ ∈ (−π/6, π/6).
(3a²x − x³)/(a³ − 3ax²) = (3 tan θ − tan³ θ)/(1 − 3 tan² θ) = tan 3θ.
Here 3θ ∈ (−π/2, π/2), so tan⁻¹(tan 3θ) = 3θ.
Answer: 3 tan⁻¹(x/a)
sin⁻¹(1/2) = π/6, so 2 sin⁻¹(1/2) = π/3.
2 cos(π/3) = 2 × 1/2 = 1.
tan⁻¹(1) = π/4.
Answer: π/4
Put x = tan α and y = tan β. Since |x| < 1, α ∈ (−π/4, π/4); since y > 0, β ∈ (0, π/2).
2x/(1 + x²) = sin 2α with 2α ∈ (−π/2, π/2), so sin⁻¹(2x/(1 + x²)) = 2α.
(1 − y²)/(1 + y²) = cos 2β with 2β ∈ (0, π), so cos⁻¹((1 − y²)/(1 + y²)) = 2β.
The expression is tan ½(2α + 2β) = tan(α + β) = (tan α + tan β)/(1 − tan α tan β) = (x + y)/(1 − xy), and xy < 1 keeps the denominator positive.
Answer: (x + y)/(1 − xy)
2π/3 ∉ [−π/2, π/2]. But sin(2π/3) = sin(π − 2π/3) = sin(π/3), and π/3 ∈ [−π/2, π/2].
So sin⁻¹(sin 2π/3) = sin⁻¹(sin π/3) = π/3.
Answer: π/3
3π/4 ∉ (−π/2, π/2). tan(3π/4) = tan(3π/4 − π) = tan(−π/4) = −1.
tan⁻¹(−1) = −π/4.
Answer: −π/4
Let A = sin⁻¹(3/5): sin A = 3/5, cos A = 4/5, so tan A = 3/4.
cot⁻¹(3/2) = tan⁻¹(2/3), so tan B = 2/3 for B = cot⁻¹(3/2).
tan(A + B) = (3/4 + 2/3)/(1 − (3/4)(2/3)) = (17/12)/(1/2) = 17/6.
Answer: 17/6
7π/6 ∉ [0, π]. cos(7π/6) = cos(2π − 7π/6) = cos(5π/6), and 5π/6 ∈ [0, π].
Answer: (B) 5π/6
sin⁻¹(−1/2) = −π/6, so the angle is π/3 + π/6 = π/2.
sin(π/2) = 1.
Answer: (D) 1
tan⁻¹ √3 = π/3. cot⁻¹(−√3) = y with cot y = −√3 and y ∈ (0, π), so y = 5π/6.
π/3 − 5π/6 = −π/2.
Answer: (B) −π/2
13π/6 = 2π + π/6, so cos(13π/6) = cos(π/6), and π/6 ∈ [0, π].
Answer: π/6
tan(7π/6) = tan(π + π/6) = tan(π/6), and π/6 ∈ (−π/2, π/2).
Answer: π/6
Let θ = sin⁻¹(3/5). Then sin θ = 3/5, cos θ = 4/5, tan θ = 3/4, and θ < π/4 because 3/5 < 1/√2.
tan 2θ = 2 tan θ/(1 − tan² θ) = (3/2)/(1 − 9/16) = (3/2)/(7/16) = 24/7.
Since 2θ ∈ (0, π/2), 2θ = tan⁻¹(24/7).
Answer: proved
sin⁻¹(8/17) = A has tan A = 8/15. sin⁻¹(3/5) = B has tan B = 3/4.
tan(A + B) = (8/15 + 3/4)/(1 − (8/15)(3/4)) = (77/60)/(3/5) = 77/36.
Both A and B are acute and A + B ≈ 1.13 < π/2, so A + B = tan⁻¹(77/36).
Answer: proved
Let A = cos⁻¹(4/5) and B = cos⁻¹(12/13). Then sin A = 3/5 and sin B = 5/13.
cos(A + B) = cos A cos B − sin A sin B = (4/5)(12/13) − (3/5)(5/13) = 48/65 − 15/65 = 33/65.
A + B ∈ (0, π), the range of cos⁻¹, so A + B = cos⁻¹(33/65).
Answer: proved
Let A = cos⁻¹(12/13) and B = sin⁻¹(3/5). Then sin A = 5/13 and cos B = 4/5.
sin(A + B) = sin A cos B + cos A sin B = (5/13)(4/5) + (12/13)(3/5) = 20/65 + 36/65 = 56/65.
A and B are acute and A + B ≈ 1.03 < π/2, so A + B = sin⁻¹(56/65).
Answer: proved
sin⁻¹(5/13) = A has tan A = 5/12, and cos⁻¹(3/5) = B has tan B = 4/3.
tan(A + B) = (5/12 + 4/3)/(1 − (5/12)(4/3)) = (21/12)/(4/9) = 63/16.
Both angles are acute and tan(A + B) > 0 with 1 − tan A tan B > 0, so A + B ∈ (0, π/2) and A + B = tan⁻¹(63/16).
Answer: proved
Put √x = tan θ, so θ = tan⁻¹ √x ∈ [0, π/4] for x ∈ [0, 1]. Then x = tan² θ.
(1 − x)/(1 + x) = (1 − tan² θ)/(1 + tan² θ) = cos 2θ, and 2θ ∈ [0, π/2] ⊂ [0, π].
So cos⁻¹((1 − x)/(1 + x)) = 2θ, and ½ cos⁻¹((1 − x)/(1 + x)) = θ = tan⁻¹ √x.
Answer: proved
1 + sin x = (cos(x/2) + sin(x/2))² and 1 − sin x = (cos(x/2) − sin(x/2))².
For x ∈ (0, π/4), cos(x/2) > sin(x/2), so √(1 + sin x) = cos(x/2) + sin(x/2) and √(1 − sin x) = cos(x/2) − sin(x/2).
Numerator = 2 cos(x/2) and denominator = 2 sin(x/2), so the fraction is cot(x/2).
Since x/2 ∈ (0, π/8) ⊂ (0, π), cot⁻¹(cot(x/2)) = x/2.
Answer: proved
Put x = cos 2θ, so 2θ = cos⁻¹ x ∈ [0, 3π/4] and θ ∈ [0, 3π/8].
√(1 + x) = √(2 cos² θ) = √2 cos θ and √(1 − x) = √2 sin θ (both non-negative for θ ∈ [0, π/2]).
The fraction is (cos θ − sin θ)/(cos θ + sin θ) = (1 − tan θ)/(1 + tan θ) = tan(π/4 − θ).
Here π/4 − θ ∈ [−π/8, π/4] ⊂ (−π/2, π/2), so the value is π/4 − θ = π/4 − ½ cos⁻¹ x.
Answer: proved
Take tan of both sides. The left side gives tan(2 tan⁻¹(cos x)) = 2 cos x/(1 − cos² x) = 2 cos x/sin² x. The right side gives 2/sin x.
So 2 cos x/sin² x = 2/sin x, which gives cos x = sin x (sin x ≠ 0), that is tan x = 1.
So x = π/4 (and in general x = nπ + π/4).
Check at x = π/4: 2 tan⁻¹(1/√2) = tan⁻¹(2√2), because 2(1/√2)/(1 − 1/2) = 2√2, and tan⁻¹(2 cosec(π/4)) = tan⁻¹(2√2). Equal.
Answer: x = π/4
For x > 0, tan⁻¹((1 − x)/(1 + x)) = tan⁻¹ 1 − tan⁻¹ x = π/4 − tan⁻¹ x.
So π/4 − tan⁻¹ x = ½ tan⁻¹ x, which gives (3/2) tan⁻¹ x = π/4 and tan⁻¹ x = π/6.
Hence x = tan(π/6) = 1/√3.
Answer: x = 1/√3
Let θ = tan⁻¹ x, so tan θ = x. In a right triangle with legs x and 1, the hypotenuse is √(1 + x²), so sin θ = x/√(1 + x²).
Answer: (D) x/√(1 + x²)
Let sin⁻¹ x = θ, so sin⁻¹(1 − x) = π/2 + 2θ. This must lie in [−π/2, π/2], so θ ∈ [−π/2, 0], i.e. x ≤ 0.
1 − x = sin(π/2 + 2θ) = cos 2θ = 1 − 2 sin² θ = 1 − 2x². So 2x² − x = 0 and x = 0 or x = 1/2.
x = 1/2 gives θ = π/6 > 0, so it is rejected (it gives sin⁻¹(1/2) − 2 sin⁻¹(1/2) = −π/6 ≠ π/2). x = 0 gives sin⁻¹ 1 − 0 = π/2.
Answer: (C) 0