NDA Inverse Trigonometric Functions previous year questions with solutions

Answer: (b) Two

Method. odd/⁠supplementary rules for negative arguments

I. sin^-1(-x) = -sin^-1 x: correct (sin^-1 is odd).
II. cos^-1(-x) = π − cos^-1 x, not cos^-1 x: incorrect (e.g. cos^-1(-1) = π ≠ 0 = cos^-1 1).
III. tan^-1(-x) = -tan^-1 x, not π − tan^-1 x: incorrect (π − tan^-1 x lies outside (-π2, π2)).
IV. cot^-1(-x) = π − cot^-1 x: correct.
Two are correct.

Answer: (b) [1, 3]

Method. the argument of cos^-1 must lie in [-1, 1]

Need -1 ≤ x − 2 ≤ 1.
Add 2: 1 ≤ x ≤ 3.
Domain = [1, 3].

Answer: (c) π/⁠3

Method. sin^-1(sin θ) = θ only when θ ∈ [-π2, π2]

sin 2π3 = sin(π − π3) = sin π3 = ⁠√32.⁠
2π3 lies outside [-π2, π2], so the answer is not 2π3.
sin^-1 ⁠√32⁠ = π3.

Answer: (b) [-π2, π2]

Method. definition of the principal branch of sin^-1

sin^-1 x is defined on [-1, 1] and takes every value from sin^-1(-1) = -π2 to sin^-1 1 = π2, both ends included.
Principal value range: [-π2, π2].

Answer: (c) 3π/⁠2

Method. replace cos^-1 x by π2 − sin^-1 x

4sin^-1 x + π2 − sin^-1 x = π, so 3sin^-1 x = π2, sin^-1 x = π6.
cos^-1 x = π2 − π6 = π3.
sin^-1 x + 4cos^-1 x = π6 + 4π3 = 3π2.

Shortcut. Add the two expressions: 5(sin^-1 x + cos^-1 x) = 5π2, so the required value is 5π2 − π = 3π2.

Answer: (c) x = 1⁠√2⁠

Method. sin^-1 x + cos^-1 x = π2

cos^-1 x = sin^-1 x and their sum is π2, so sin^-1 x = π4.
x = sin π4 = 1⁠√2.⁠

Answer: (c) 1⁠√2⁠

Method. replace cos^-1 x by π2 − sin^-1 x

3sin^-1 x + π2 − sin^-1 x = π.
2sin^-1 x = π2, so sin^-1 x = π4.
x = sin π4 = 1⁠√2.⁠

Answer: (c) 3

Method. each sin^-1 is at most π2, so the maximum sum forces each to be π2

sin^-1 x, sin^-1 y, sin^-1 z ≤ π2 each, so their sum ≤ 3π2.
The sum equals 3π2 only if each equals π2, i.e. x = y = z = 1.
x¹⁰⁰⁰ + y¹⁰⁰¹ + z¹⁰⁰² = 1 + 1 + 1 = 3.

Answer: (b) unique solution

Method. sin^-1 x + cos^-1 x = π2 turns the equation into a linear pair

sin^-1 x − cos^-1 x = π6 and sin^-1 x + cos^-1 x = π2.
Adding: 2sin^-1 x = 2π3, so sin^-1 x = π3, which lies in [-π2, π2].
x = ⁠√32:⁠ exactly one solution.

Answer: (a) x ∈ R

Method. identity tan^-1 x + cot^-1 x = π2 with principal ranges

Principal ranges: tan^-1 x ∈ (-π2, π2), cot^-1 x ∈ (0, π).
Let θ = tan^-1 x; then π2 − θ ∈ (0, π) and cot(π2 − θ) = tan θ = x, so cot^-1 x = π2 − θ.
This holds for every real x (including x ≤ 0 and x = 0), so x ∈ R.

Answer: (d) 0

Method. sec^-1 x = cos^-1 1x and sin^-1 x + cos^-1 x = π2

sec^-1 54 = cos^-1 45.
sin^-1 45 + cos^-1 45 = π2.
Expression = π2 − π2 = 0.

Answer: (c) 3/⁠5

Method. right triangle with tan θ = 34

θ = tan^-1 0.75 = tan^-1 34, θ ∈ (0, π2).
Opposite 3, adjacent 4, hypotenuse 5.
sin θ = 35.

Answer: (a) 1/⁠2

Method. tan² θ2 = 1 − cos θ1 + cos θ

Let θ = cos^-1 13, so cos θ = 13.
tan² θ2 = 1 − 131 + 13 = 2/⁠34/⁠3 = 12.

Answer: (d) 17

Method. double-angle for tan^-1, then tan(A − π4)

2tan^-1 12 = tan^-1 11 − 1/⁠4 = tan^-1 43.
tan(A − π4) = tan A − 11 + tan A = 43 − 11 + 43 = 1/⁠37/⁠3.
= 17.

Answer: (a) 2 - ⁠√3⁠

Method. sec^-1 x = cos^-1 1x, then tan 15°

sec^-1 2⁠√3⁠ = cos^-1 ⁠√32⁠ = π6.
tan π12 = 1 − cos π6sin π6 = 1 − ⁠√3212⁠ = 2 - ⁠√3.⁠

Answer: (b) 11/⁠24

Method. evaluate each inverse function, then the right-triangle ratios

sec^-1 2 = π3; cot² π3 = 13.
Let θ = cosec^-1 3: sin θ = 13, cos² θ = 89, so tan² θ = 1/⁠98/⁠9 = 18.
Sum = 13 + 18 = 1124.

Answer: (c) 8

Method. cot^-1 3 = tan^-1 13, double-angle, then tan(π4 − A)

cot^-1 3 = tan^-1 13; 2tan^-1 13 = tan^-1 2/⁠31 − 1/⁠9 = tan^-1 34.
tan(π4 − tan^-1 34) = 1 − 341 + 34 = 17, so cot(π4 − 2cot^-1 3) = 7.
⁠√(15⁠ + 7²) = ⁠√64⁠ = 8.

Answer: (c) 3 + ⁠√5⁠

Method. half-angle formula cot θ2 = 1 + cos θsin θ

Let θ = cos^-1 ⁠√53,⁠ θ ∈ (0, π2); cos θ = ⁠√53,⁠ sin θ = ⁠√(1⁠ −⁠ ⁠59⁠) = 23.
cot θ2 = 1 + ⁠√5323⁠ = 3 + ⁠√52.⁠
2cot θ2 = 3 + ⁠√5.⁠

Answer: (a) 1

Method. a common angle θ gives cos θ = 1p, sin θ = 1q

sec^-1 p = cosec^-1 q = θ; since p, q > 0 (and p, q ≥ 1 for the inverses to exist), θ ∈ [0, π2) ∩ (0, π2] = (0, π2).
sec θ = p and cosec θ = q, so cos θ = 1p, sin θ = 1q.
p^-2 + q^-2 = cos² θ + sin² θ = 1.

Answer: (a) 2517

Method. sin²(cos^-1 x) = 1 − x²

Let θ = cos^-1 3⁠√17,⁠ so cos² θ = 917.
sin² θ = 1 − 917 = 817.
1 + sin² θ = 2517.

Answer: (d) π3

Method. evaluate from the inside using principal values

cosec^-1 2 = sin^-1 12 = π6.
cot π6 = ⁠√3.⁠
tan^-1 ⁠√3⁠ = π3.

Answer: (b) 34

Method. tan 2θ = 2tan θ1 − tan² θ

Let θ = tan^-1 13, so tan θ = 13.
tan 2θ = 2/⁠31 − 1/⁠9 = 2/⁠38/⁠9 = 34.

Answer: (a) π4

Method. convert both to tan^-1, then the addition formula

cot^-1 9 = tan^-1 19.
cosec^-1 ⁠√414:⁠ sin θ = 4⁠√41,⁠ cos θ = 5⁠√41,⁠ so θ = tan^-1 45.
tan^-1 19 + tan^-1 45 = tan^-1 19 + 451 − 445 = tan^-1 41/⁠4541/⁠45 = tan^-1 1 = π4.

Answer: (d) π/⁠2

Method. tan^-1 k + tan^-1 1k = π2 for k > 0

Roots of x² - 4x + 1 = 0: k = 2 ± ⁠√3,⁠ both positive (product 1, sum 4).
For k > 0, tan^-1 1k = cot^-1 k, and tan^-1 k + cot^-1 k = π2.
Answer: π2.

Answer: (c) 1/⁠3

Method. tan^-1 addition formula

k + 121 − k2 = tan π4 = 1.
k + 12 = 1 − k2, so 3k2 = 12.
k = 13.

Shortcut. Check: tan^-1 13 + tan^-1 12 is the standard pair summing to π4.

Answer: (b) π4

Method. tan^-1 subtraction formula

Let t = ab; then a − ba + b = t − 1t + 1.
tan of the difference = ab − a − ba + b1 + a(a − b)b(a + b) = a(a + b) − b(a − b)b(a + b) + a(a − b) = a² + b²a² + b² = 1.
The question tacitly takes ab > -1 (as when a, b > 0); then the difference lies in (-π2, π2) and equals tan^-1 1 = π4. (For ab < -1 it would be -3π4, which is not offered.)

Shortcut. Put a = b: tan^-1 1 − tan^-1 0 = π4.

Answer: (a) 1

Method. tan^-1 addition formula

12 + x31 − x6 = tan π4 = 1 (valid as x6 < 1 for 0 < x < 6).
12 + x3 = 1 − x6, so x2 = 12.
x = 1.

Answer: (a) 617

Method. write both angles as cot^-1, then cot(A + B) = cot A cot B − 1cot A + cot B

A = sin^-1 35: sin A = 35, cos A = 45, cot A = 43. B = cot^-1 32: cot B = 32.
cot(A + B) = 43 · 32 − 143 + 32 = 2 − 117/⁠6 = 617.

Answer: (a) 3 only

Method. tan^-1 addition formula, then reject roots that make the left side negative

Taking tan: 2x + 3x1 − 6x² = 1, so 6x² + 5x − 1 = 0, i.e. (6x − 1)(x + 1) = 0, x = 16 or x = -1.
x = -1 gives tan^-1(-2) + tan^-1(-3) < 0, not π4: reject.
x = 16: tan^-1 13 + tan^-1 12 = tan^-1 5/⁠65/⁠6 = π4. (holds)
Of 8, -4, 16, -14, only 16 (value 3) is a solution.

Shortcut. For x < 0 the left side is negative, so 2 and 4 are out at once; 8 gives a sum above π4.

Answer: (b) π/⁠4

Method. tan^-1 a + tan^-1 b = tan^-1 a + b1 − ab (ab < 1)

14 · 35 = 320 < 1.
tan^-1 14 + tan^-1 35 = tan^-1 14 + 351 − 320 = tan^-1 17/⁠2017/⁠20 = tan^-1 1 = π4.

Answer: (b) π4

Method. convert sin^-1 to tan^-1, then the tan^-1 addition formula

sin^-1 35 = tan^-1 34 (sides 3, 4, 5).
tan^-1 34 + tan^-1 17 = tan^-1 34 + 171 − 328 = tan^-1 25/⁠2825/⁠28 = tan^-1 1.
= π4.

No correct option. This question is defective as printed in UPSC's paper.

Method. tan^-1 addition formula combined with the G.P. condition y² = xz

G.P.: y² = xz. A.P.: 2tan^-1 y = tan^-1 x + tan^-1 z.
If y ≠ 1: taking tan of both sides, 2y1 − y² = x + z1 − xz = x + z1 − y², so x + z = 2y; with xz = y² this gives (⁠√x⁠ - ⁠√z)²⁠ = 0, so x = y = z.
If y = 1: 2tan^-1 1 = π2 = tan^-1 x + tan^-1 1x for every x > 0, so any x, 1, 1x works, e.g. x = 2, y = 1, z = 12.
So the data allow x = y = z = 2 (xz ≠ 1) and also 2, 1, 12 (x ≠ y): neither (a) nor (b) is forced, and (c), (d) are impossible.

The conditions do not force a single option: x = y = z = 2 satisfies them with xz ≠ 1, and x = 2, y = 1, z = 1/⁠2 satisfies them with x ≠ y; the intended reading (dividing by 1 − y², i.e. assuming y ≠ 1) gives option (a).

Answer: (d) 3sin^-1(x3)

Method. triple-angle identity sin 3θ = 3sin θ − 4sin³ θ

x − 4x³27 = 3(x3) − 4(x3)³.
Put x3 = sin θ: the argument is 3sin θ − 4sin³ θ = sin 3θ.
For 3θ ∈ [-π2, π2] (|x| ≤ 32), y = sin^-1(sin 3θ) = 3θ = 3sin^-1(x3).

Answer: (c) 3⁠√(9⁠ −⁠ x²)

Method. differentiate y = 3sin^-1(x3)

y = 3sin^-1(x3) (previous item).
dydx = 3 · 1⁠√(1⁠ −⁠ ⁠x²9⁠) · 13 = 1⁠√(9⁠ −⁠ x²9).
= 3⁠√(9⁠ −⁠ x²).

Answer: (b) p − q1 + pq

Method. each term equals 2tan^-1 of its variable, then tan^-1 subtraction formula

Put p = tan α, q = tan β, x = tan γ (principal values, |p|, |q|, |x| < 1, q ≥ 0).
sin^-1 2p1 + p² = sin^-1(sin 2α) = 2tan^-1 p; cos^-1 1 − q²1 + q² = cos^-1(cos 2β) = 2tan^-1 q; tan^-1 2x1 − x² = 2tan^-1 x.
So 2tan^-1 p − 2tan^-1 q = 2tan^-1 x, i.e. tan^-1 x = tan^-1 p − tan^-1 q.
x = p − q1 + pq.