Miscellaneous Exercise answers: Inverse Trigonometric Functions

Open NCERT p. 31Matches NCERT’s answer

1. 13π/6 = 2π + π/6, so cos(13π/6) = cos(π/6) because cosine repeats every 2π.
2. π/6 lies in [0, π], the range of cos⁻¹.
3. So cos⁻¹(cos(13π/6)) = π/6.

Answerπ/6

Watch this explained “More than a whole turn away”, 15:41 into When the two cancellation rules hold, and the angles where the second one fails

Open NCERT p. 31Matches NCERT’s answer

1. 7π/6 = π + π/6, so tan(7π/6) = tan(π/6) because tangent repeats every π.
2. π/6 lies in (−π/2, π/2), the range of tan⁻¹.
3. So tan⁻¹(tan(7π/6)) = π/6.

Answerπ/6

Watch this explained “More than a whole turn away”, 15:41 into When the two cancellation rules hold, and the angles where the second one fails

Open NCERT p. 31One way to think about it

1. Let A = sin⁻¹(3/5), so sinA = 3/5 and, from a 3-4-5 triangle, cosA = 4/5, tanA = 3/4.
2. Use tan(2A) = 2tanA/(1 − tan²A) = 2(3/4)/(1 − 9/16) = (3/2)/(7/16) = 24/7.
3. Since A ≈ 36.9°, 2A ≈ 73.7° lies in (−90°, 90°), so 2A is itself the principal value tan⁻¹(24/7).
4. So 2sin⁻¹(3/5) = 2A = tan⁻¹(24/7), as required.

In shortProved: 2sin⁻¹(3/5) = tan⁻¹(24/7).

Watch this explained “Adding two of them”, 16:37 into Substituting an angle for the variable to collapse a messy expression

Open NCERT p. 31One way to think about it

1. Let A = sin⁻¹(8/17). Then sin A = 8/17 and A lies between 0 and π/2. An 8-15-17 right triangle gives cos A = 15/17, so tan A = 8/15.
2. Let B = sin⁻¹(3/5). Then sin B = 3/5 and B lies between 0 and π/2. A 3-4-5 right triangle gives cos B = 4/5, so tan B = 3/4.
3. tan(A + B) = (tan A + tan B)/(1 − tan A·tan B) = (8/15 + 3/4)/(1 − (8/15)(3/4)).
4. Top: 8/15 + 3/4 = 32/60 + 45/60 = 77/60. Bottom: 1 − 24/60 = 36/60. So tan(A + B) = (77/60) ÷ (36/60) = 77/36.
5. Range check: 8/17 and 3/5 are both less than 1/√2 (about 0.707), so A and B are each less than π/4. So A + B lies between 0 and π/2, inside the range (−π/2, π/2) of tan⁻¹.
6. So A + B = tan⁻¹(77/36), that is, sin⁻¹(8/17) + sin⁻¹(3/5) = tan⁻¹(77/36).

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

Watch this explained “Adding two of them”, 16:37 into Substituting an angle for the variable to collapse a messy expression

Open NCERT p. 31One way to think about it

1. Let A = cos⁻¹(4/5). Since 4/5 > 0, A lies between 0 and π/2. A 3-4-5 right triangle gives sin A = 3/5.
2. Let B = cos⁻¹(12/13). Since 12/13 > 0, B lies between 0 and π/2. A 5-12-13 right triangle gives sin B = 5/13.
3. 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.
4. Range check: A and B are each between 0 and π/2, so A + B lies between 0 and π, the range of cos⁻¹.
5. So A + B = cos⁻¹(33/65), that is, cos⁻¹(4/5) + cos⁻¹(12/13) = cos⁻¹(33/65).

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

Watch this explained “Adding two of them”, 16:37 into Substituting an angle for the variable to collapse a messy expression

Open NCERT p. 31One way to think about it

1. Let A = cos⁻¹(12/13). Then A lies between 0 and π/2, cos A = 12/13, and a 5-12-13 right triangle gives sin A = 5/13.
2. Let B = sin⁻¹(3/5). Then B lies between 0 and π/2, sin B = 3/5, and a 3-4-5 right triangle gives cos B = 4/5.
3. 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.
4. Range check: sin A = 5/13 and sin B = 3/5 are both less than 1/√2 (about 0.707), so A and B are each less than π/4. So A + B lies between 0 and π/2, inside the range [−π/2, π/2] of sin⁻¹.
5. So A + B = sin⁻¹(56/65), that is, cos⁻¹(12/13) + sin⁻¹(3/5) = sin⁻¹(56/65).

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

Watch this explained “Adding two of them”, 16:37 into Substituting an angle for the variable to collapse a messy expression

Open NCERT p. 31One way to think about it

1. Let A = sin⁻¹(5/13). Then A lies between 0 and π/2, and a 5-12-13 right triangle gives cos A = 12/13, so tan A = 5/12.
2. Let B = cos⁻¹(3/5). Then B lies between 0 and π/2, and a 3-4-5 right triangle gives sin B = 4/5, so tan B = 4/3.
3. tan(A + B) = (tan A + tan B)/(1 − tan A·tan B) = (5/12 + 4/3)/(1 − (5/12)(4/3)).
4. Top: 5/12 + 16/12 = 21/12 = 7/4. Bottom: 1 − 20/36 = 16/36 = 4/9. So tan(A + B) = (7/4) ÷ (4/9) = 63/16.
5. Range check: tan A·tan B = 5/9, which is less than 1. So tan A < 1/tan B = tan(π/2 − B), which means A < π/2 − B, that is, A + B < π/2. So A + B lies between 0 and π/2, inside the range of tan⁻¹.
6. So A + B = tan⁻¹(63/16), that is, tan⁻¹(63/16) = sin⁻¹(5/13) + cos⁻¹(3/5).

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

Watch this explained “Adding two of them”, 16:37 into Substituting an angle for the variable to collapse a messy expression

Open NCERT p. 31One way to think about it

1. To prove: tan⁻¹ √x = (1/2) cos⁻¹((1 − x)/(1 + x)), x ∈ [0, 1].
2. Put √x = tanθ, so x = tan²θ, with θ ∈ [0, π/4] because x ∈ [0, 1].
3. Then (1−x)/(1+x) = (1 − tan²θ)/(1 + tan²θ) = cos(2θ).
4. So cos⁻¹[(1−x)/(1+x)] = 2θ, because 2θ ∈ [0, π/2] lies inside cos⁻¹'s range [0, π].
5. Divide by 2: (1/2)cos⁻¹[(1−x)/(1+x)] = θ = tan⁻¹√x, as required.

In shortProved: tan⁻¹√x = (1/2)cos⁻¹[(1−x)/(1+x)].

Watch the lesson Substituting an angle for the variable to collapse a messy expression

Open NCERT p. 31One way to think about it

1. To prove: cot⁻¹((√(1 + sin x) + √(1 − sin x))/(√(1 + sin x) − √(1 − sin x))) = x/2, x ∈ (0, π/4).
2. Since x ∈ (0, π/4), x/2 ∈ (0, π/8), so both cos(x/2) and sin(x/2) are positive and cos(x/2) > sin(x/2).
3. Write 1 + sinx = (cos(x/2) + sin(x/2))² and 1 − sinx = (cos(x/2) − sin(x/2))², so their square roots are cos(x/2)+sin(x/2) and cos(x/2)−sin(x/2).
4. Adding gives √(1+sinx) + √(1−sinx) = 2cos(x/2); subtracting gives √(1+sinx) − √(1−sinx) = 2sin(x/2).
5. The ratio is 2cos(x/2)/2sin(x/2) = cot(x/2).
6. cot⁻¹[cot(x/2)] = x/2, because x/2 ∈ (0, π/8) lies inside cot⁻¹'s range (0, π).

In shortProved: the expression equals x/2.

Watch this explained “The one with no root in it”, 8:20 into Substituting an angle for the variable to collapse a messy expression

Open NCERT p. 31One way to think about it

1. To prove: tan⁻¹((√(1 + x) − √(1 − x))/(√(1 + x) + √(1 − x))) = π/4 − (1/2) cos⁻¹x, −1/√2 ≤ x ≤ 1.
2. Put x = cos 2θ with 2θ between 0 and π, so θ = (1/2)cos⁻¹x. Since −1/√2 ≤ x ≤ 1, 2θ runs from 0 to 3π/4, so θ lies between 0 and 3π/8. There cos θ > 0 and sin θ ≥ 0.
3. 1 + x = 1 + cos 2θ = 2cos²θ and 1 − x = 1 − cos 2θ = 2sin²θ. So √(1 + x) = √2·cos θ and √(1 − x) = √2·sin θ.
4. The fraction becomes (√2 cos θ − √2 sin θ)/(√2 cos θ + √2 sin θ) = (cos θ − sin θ)/(cos θ + sin θ).
5. Divide top and bottom by cos θ: (1 − tan θ)/(1 + tan θ) = tan(π/4 − θ), because tan(π/4) = 1.
6. θ between 0 and 3π/8 puts π/4 − θ between −π/8 and π/4, inside the range (−π/2, π/2) of tan⁻¹. So the left side is tan⁻¹[tan(π/4 − θ)] = π/4 − θ.
7. Since θ = (1/2)cos⁻¹x, the left side equals π/4 − (1/2)cos⁻¹x, the right side.

In shortProved: the left side equals π/4 − (1/2)cos⁻¹x.

Watch the lesson Substituting an angle for the variable to collapse a messy expression

Open NCERT p. 31Matches NCERT’s answer

1. The right side contains cosec x, so sin x ≠ 0. Then cos x cannot be 1 or −1, so −1 < cos x < 1.
2. Let θ = tan⁻¹(cos x). Because −1 < cos x < 1, θ lies between −π/4 and π/4, so 2θ lies between −π/2 and π/2 — the same interval in which tan⁻¹ gives its values. So the two sides are equal exactly when their tangents are equal.
3. Tangent of the left side: tan 2θ = 2tan θ/(1 − tan²θ) = 2cos x/(1 − cos²x) = 2cos x/sin²x.
4. Tangent of the right side: tan(tan⁻¹(2 cosec x)) = 2 cosec x = 2/sin x.
5. Set them equal: 2cos x/sin²x = 2/sin x. Multiply both sides by sin²x/2 (allowed, since sin x ≠ 0): cos x = sin x, so tan x = 1.
6. tan x = 1 when x = π/4, and again every π after that: x = nπ + π/4, where n is any integer. At all of these sin x = ±1/√2, which is not 0, so none has to be thrown out.
7. The question gives no interval, so every one of these is a solution, not only π/4. Check x = 5π/4: cos x = −1/√2, so the left side is 2tan⁻¹(−1/√2) ≈ −1.2310; sin x = −1/√2, so the right side is tan⁻¹(−2√2) ≈ −1.2310. They agree.

Answerx = nπ + π/4, where n is any integer (for example π/4, 5π/4, −3π/4)

Watch this explained “Solving, and rejecting”, 19:36 into Substituting an angle for the variable to collapse a messy expression

Open NCERT p. 31Matches NCERT’s answer

1. Solve: tan⁻¹((1 − x)/(1 + x)) = (1/2) tan⁻¹x, (x > 0).
2. Let x = tan θ, so θ = tan⁻¹x. Since x > 0, θ lies between 0 and π/2.
3. (1 − x)/(1 + x) = (1 − tan θ)/(1 + tan θ) = tan(π/4 − θ), because tan(π/4) = 1.
4. θ between 0 and π/2 puts π/4 − θ between −π/4 and π/4, inside the range of tan⁻¹. So the left side is π/4 − θ.
5. The equation becomes π/4 − θ = θ/2, so (3/2)θ = π/4 and θ = π/6.
6. x = tan(π/6) = 1/√3, which is positive as required.

Answerx = 1/√3

Watch this explained “Solving, and rejecting”, 19:36 into Substituting an angle for the variable to collapse a messy expression

Open NCERT p. 31Matches NCERT’s answer

1. Let θ = tan⁻¹x, so tanθ = x, with θ ∈ (−π/2, π/2).
2. Draw a right triangle with opposite side x and adjacent side 1, so the hypotenuse is √(1+x²).
3. sinθ = opposite/hypotenuse = x/√(1+x²), option (D).

Answerx/√(1+x²), option (D)

Watch this explained “Name it as an angle”, 1:28 into Substituting an angle for the variable to collapse a messy expression

Open NCERT p. 31Matches NCERT’s answer

1. Both sin⁻¹(1 − x) and sin⁻¹x must exist, so x and 1 − x both lie between −1 and 1. That means 0 ≤ x ≤ 1.
2. Rearrange: sin⁻¹(1 − x) = π/2 + 2sin⁻¹x. Take sine of both sides: 1 − x = sin(π/2 + 2sin⁻¹x) = cos(2sin⁻¹x).
3. With θ = sin⁻¹x, cos 2θ = 1 − 2sin²θ = 1 − 2x². So 1 − x = 1 − 2x², which gives 2x² − x = 0, x(2x − 1) = 0, so x = 0 or x = 1/2.
4. Taking sine of both sides can let in extra answers, because different angles can have the same sine. So check each one in the original equation.
5. x = 0: sin⁻¹(1) − 2sin⁻¹(0) = π/2 − 0 = π/2. True.
6. x = 1/2: sin⁻¹(1/2) − 2sin⁻¹(1/2) = π/6 − π/3 = −π/6, not π/2. Rejected.
7. Only x = 0 works, option (C).

Answerx = 0, option (C)

Watch this explained “Solving, and rejecting”, 19:36 into Substituting an angle for the variable to collapse a messy expression