Inverse trigonometric functions for NDA

Inverse trigonometry is a small chapter with a high trap density. Almost every NDA question on it turns on one of three things: the principal value range, the condition attached to a formula, or a quick right triangle. The formulas take an evening to learn. The conditions are what separate a correct answer from the distractor sitting next to it.

Typical question types are:

• principal values such as cos⁻¹(−1/2) or cot⁻¹(−1);
• expressions like sin⁻¹(sin 5π/6), where the answer is not 5π/6;
• sums like tan⁻¹(1/2) + tan⁻¹(1/3), or tan⁻¹ 2 + tan⁻¹ 3;
• converting sin(cos⁻¹ x) or cos(tan⁻¹ x) into algebra;
• equations such as tan⁻¹ 2x + tan⁻¹ 3x = π/4;
• domains of functions like sin⁻¹(2x − 1).

Why principal values exist

sin x takes the value 1/2 at π/6, 5π/6, 13π/6 and infinitely many other points, so it has no inverse on its whole domain. To get one, we restrict sin x to a stretch where it is one-one and takes every value from −1 to 1 exactly once. For sin that stretch is [−π/2, π/2]. The value sin⁻¹ x returns is always in that stretch, and it is called the principal value.

A memory aid: sin, tan and cosec take the "right half" around zero; cos, cot and sec take the "upper half" from 0 to π.

Negative arguments

• 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.

So sin⁻¹(−1/2) = −π/6, but cos⁻¹(−1/2) = π − π/3 = 2π/3 and cot⁻¹(−1) = π − π/4 = 3π/4. The range decides the answer.

Function of an inverse, and inverse of a function

sin(sin⁻¹ x) = x for every x in [−1, 1]. No trap there.

sin⁻¹(sin x) = x only when x is in [−π/2, π/2]. Outside that interval, find the angle inside the principal range that has the same sine. For example, sin(5π/6) = 1/2, so sin⁻¹(sin 5π/6) = π/6. Likewise cos⁻¹(cos 7π/6) = 5π/6, because cos 7π/6 = −√3/2 and the angle in [0, π] with that cosine is 5π/6. And tan⁻¹(tan 3π/4) = tan⁻¹(−1) = −π/4.

The identities

Complementary pairs:

• sin⁻¹ x + cos⁻¹ x = π/2, for −1 ≤ x ≤ 1;
• tan⁻¹ x + cot⁻¹ x = π/2, for all real x;
• sec⁻¹ x + cosec⁻¹ x = π/2, for x ≤ −1 or x ≥ 1.

Reciprocals: sin⁻¹(1/x) = cosec⁻¹ x for x ≤ −1 or x ≥ 1, and tan⁻¹(1/x) = cot⁻¹ x for x > 0. For negative x the tan⁻¹ and cot⁻¹ ranges differ, so be careful.

Sum and difference of tan⁻¹:

• tan⁻¹ x + tan⁻¹ y = tan⁻¹[(x + y)/(1 − xy)], when xy < 1;
• tan⁻¹ x + tan⁻¹ y = π + tan⁻¹[(x + y)/(1 − xy)], when x > 0, y > 0 and xy > 1;
• tan⁻¹ x − tan⁻¹ y = tan⁻¹[(x − y)/(1 + xy)], when xy > −1.

These come straight from tan(A + B) = (tan A + tan B)/(1 − tan A tan B), with the condition making sure the answer lands in the principal range.

Double-angle forms:

These are the substitutions (x = tan θ) that turn ugly inverse expressions into 2 tan⁻¹ x, and they reappear in differentiation.

The triangle method

To evaluate a ratio of an inverse, draw a right triangle. For cos(tan⁻¹ x) with x > 0, let θ = tan⁻¹ x: opposite x, adjacent 1, hypotenuse √(1 + x²). So cos θ = 1/√(1 + x²) and sin θ = x/√(1 + x²). For negative arguments, check the sign against the range: cos(sin⁻¹ x) = √(1 − x²) is never negative, because sin⁻¹ x lies in [−π/2, π/2], where cos ≥ 0.

Worked NDA-style MCQs

Q1. tan⁻¹(1/2) + tan⁻¹(1/3) equals:
(a) π/6 (b) π/4 (c) π/3 (d) π/2

xy = 1/6 < 1. (1/2 + 1/3)/(1 − 1/6) = (5/6)/(5/6) = 1, and tan⁻¹ 1 = π/4. Answer: (b).

Q2. tan⁻¹ 2 + tan⁻¹ 3 equals:
(a) −π/4 (b) π/4 (c) 3π/4 (d) π/2

Both positive and xy = 6 > 1, so the value is π + tan⁻¹[5/(1 − 6)] = π + tan⁻¹(−1) = π − π/4 = 3π/4. Answer: (c).

Q3. sin⁻¹(sin 5π/6) equals:
(a) 5π/6 (b) −π/6 (c) 7π/6 (d) π/6

5π/6 is outside [−π/2, π/2]. sin 5π/6 = 1/2, and the principal angle with sine 1/2 is π/6. Answer: (d).

Q4. cos(tan⁻¹(3/4)) equals:
(a) 3/5 (b) 3/4 (c) 4/5 (d) 5/4

Opposite 3, adjacent 4, hypotenuse 5. cos = 4/5. Answer: (c).

Q5. If sin⁻¹ x + sin⁻¹ y = 2π/3, then cos⁻¹ x + cos⁻¹ y equals:
(a) π/3 (b) 2π/3 (c) π/6 (d) π

cos⁻¹ x + cos⁻¹ y = (π/2 − sin⁻¹ x) + (π/2 − sin⁻¹ y) = π − 2π/3 = π/3. Answer: (a).

Q6. The solution of tan⁻¹ 2x + tan⁻¹ 3x = π/4 is:
(a) x = −1 (b) x = 1 (c) x = −1 or 1/6 (d) x = 1/6

Take tan of both sides: 5x/(1 − 6x²) = 1, so 6x² + 5x − 1 = 0, which factorises as (6x − 1)(x + 1) = 0. x = −1 makes both terms negative, so their sum cannot be π/4; reject it. x = 1/6 gives tan⁻¹(1/3) + tan⁻¹(1/2) = π/4. Answer: (d).

Practice set

1. cos⁻¹(1/2) equals: (a) π/3 (b) π/6 (c) −π/3 (d) 2π/3
2. tan⁻¹(1/4) + tan⁻¹(3/5) equals: (a) π/4 (b) π/3 (c) π/6 (d) π/2
3. cos(sin⁻¹(5/13)) equals: (a) 5/12 (b) −12/13 (c) 13/12 (d) 12/13
4. sin⁻¹(−1) equals: (a) π/2 (b) π (c) 3π/2 (d) −π/2
5. cos⁻¹(−1/2) + sin⁻¹(−1/2) equals: (a) π/2 (b) π/3 (c) 5π/6 (d) π/6
6. cot⁻¹(−1) equals: (a) −π/4 (b) π/4 (c) 3π/4 (d) 5π/4
7. sin(2 tan⁻¹(1/2)) equals: (a) 3/5 (b) 4/5 (c) 1 (d) 2/5
8. The domain of sin⁻¹(2x − 1) is: (a) [−1, 1] (b) [0, 1] (c) [−1, 0] (d) [0, 2]

Answers:

1. (a). Standard value; π/3 lies in [0, π].
2. (a). (1/4 + 3/5)/(1 − 3/20) = (17/20)/(17/20) = 1.
3. (d). Hypotenuse 13, opposite 5, adjacent 12; cos is positive in the principal range.
4. (d). −π/2 is inside [−π/2, π/2].
5. (a). 2π/3 + (−π/6) = π/2.
6. (c). π − cot⁻¹ 1 = π − π/4.
7. (b). sin 2θ = 2 tan θ/(1 + tan²θ) = 1/(5/4) = 4/5.
8. (b). −1 ≤ 2x − 1 ≤ 1 gives 0 ≤ x ≤ 1.

What to do next

• Write the range table and the three tan⁻¹ conditions on one card and revise it until you can reproduce it cold.
• Do 20 old NDA questions on this chapter. For each wrong answer, write down which range or condition you missed.
• Refresh the underlying identities in trigonometric identities, then learn the derivatives of the inverse functions in differentiation.