Trigonometric functions are many-to-one: sin(30°) = sin(150°) = 1/2. To define an inverse, you have to artificially restrict the domain so the function becomes one-to-one. The resulting restricted inverse is called the principal value branch.
Think of it this way: if someone tells you sin θ = 1/2, there are infinitely many solutions for θ. But sin⁻¹(1/2) — arcsin of 1/2 — gives you exactly one answer: π/6. That uniqueness is only possible because we agreed to confine the output to a specific interval called the principal value range.
Here is the analogy that sticks: imagine a revolving door that can spin endlessly. Trigonometry is the full spin. Inverse trigonometry is the security guard who only lets the door open within a narrow arc — one controlled window of output.
The three functions you see most in NDA papers are:
| Function | Input domain | Principal value range |
|---|---|---|
| sin⁻¹x | [−1, 1] | [−π/2, π/2] |
| cos⁻¹x | [−1, 1] | [0, π] |
| tan⁻¹x | (−∞, ∞) | (−π/2, π/2) |
Two facts that appear in almost every NDA question on this topic:
sin⁻¹x + cos⁻¹x = π/2 for all x ∈ [−1, 1]tan⁻¹x + tan⁻¹(1/x) = π/2 when x > 0, and = −π/2 when x < 0These are not formulas to memorise in isolation — they are consequences of the geometry of the principal value ranges. Once you understand why they are true, you never confuse the signs.
For sin⁻¹: sine is increasing on [−π/2, π/2], which is the "natural" window centred at zero. It covers the full output range [−1, 1] exactly once.
For cos⁻¹: cosine is decreasing on [0, π]. We cannot use [−π/2, π/2] because cos is not one-to-one there (it equals the same value at +θ and −θ). So we shift to [0, π].
For tan⁻¹: tangent is increasing on (−π/2, π/2) and covers all of ℝ once. The endpoints are excluded because tan(±π/2) is undefined.
This difference in ranges — cos⁻¹ living on [0, π] while sin⁻¹ lives on [−π/2, π/2] — is the source of 80% of sign errors on NDA papers. Burn those ranges in.
For x ∈ [−1, 1]:
sin⁻¹x + cos⁻¹x = π/2
Why it's true: If sin⁻¹x = α, then sin α = x, which means cos(π/2 − α) = x, so cos⁻¹x = π/2 − α. Adding them: α + (π/2 − α) = π/2. It's that direct.
For x > 0: tan⁻¹x + tan⁻¹(1/x) = π/2
For x < 0: tan⁻¹x + tan⁻¹(1/x) = −π/2
Why the sign change? When x > 0, both tan⁻¹x and tan⁻¹(1/x) are positive (both live in (0, π/2)), and they are complementary. When x < 0, both values are negative (they live in (−π/2, 0)), and their sum is −π/2. Never 0, never ±π — this is a common trap.
This is the workhorse of NDA inverse trig problems. You need both forms:
Form 1 (addition):
tan⁻¹x + tan⁻¹y = tan⁻¹((x + y)/(1 − xy)), when xy < 1
When xy > 1 and x > 0, y > 0, add π to the right side.
Form 2 (subtraction):
tan⁻¹x − tan⁻¹y = tan⁻¹((x − y)/(1 + xy)), when xy > −1
These formulas derive from the standard tan(A ± B) expansion. If you know tan(A + B) = (tan A + tan B)/(1 − tan A · tan B), just let A = tan⁻¹x and B = tan⁻¹y and work backwards.
These conversions come up when a question mixes function types:
sin⁻¹x = cos⁻¹(√(1 − x²)) for x ∈ [0, 1]sin⁻¹x = tan⁻¹(x/√(1 − x²)) for x ∈ (−1, 1)cos⁻¹x = tan⁻¹(√(1 − x²)/x) for x ∈ (0, 1]The derivation for the first: if sin⁻¹x = θ, then sin θ = x, so cos θ = √(1 − x²) (positive because θ ∈ [−π/2, π/2] — and cos is non-negative there). Therefore θ = cos⁻¹(√(1 − x²)).
2 sin⁻¹x = sin⁻¹(2x√(1 − x²)) for x ∈ [−1/√2, 1/√2]2 cos⁻¹x = cos⁻¹(2x² − 1) for x ∈ [0, 1]2 tan⁻¹x = tan⁻¹(2x/(1 − x²)) for |x| < 1These surface occasionally in simplification questions. Recognise the form 2x√(1 − x²) as the signal to consolidate two inverse trig terms into one.
When you see tan⁻¹k + tan⁻¹(1/k), the answer is π/2 · sign(k). Positive k gives π/2, negative k gives −π/2. No formula lookup needed.
Micro-example: tan⁻¹(2 + √3) + tan⁻¹(1/(2 + √3)) — since 2 + √3 > 0, the answer is π/2 immediately.
Standard method: expand each term, check product, apply formula → 40 seconds. This pattern: 5 seconds.
sin⁻¹x + cos⁻¹x = π/2 always. So whenever you see both together in a question, replace the pair with π/2 and continue.
Micro-example: cos⁻¹x = sin⁻¹x → both are equal → their sum is π/2 → each equals π/4 → x = sin(π/4) = 1/√2. Three steps total.
Without the identity: you would set up cos⁻¹x = sin⁻¹x, try to equate arguments, likely make a range error. 5 steps, 60+ seconds. With the identity: 3 steps, 15 seconds.
In tan⁻¹(a/b) − tan⁻¹((a−b)/(a+b)), substitute the arctan subtraction formula. Notice that (x − y)/(1 + xy) simplifies to exactly 1 — giving tan⁻¹(1) = π/4 — whenever the arguments have this (a/b) and ((a−b)/(a+b)) structure. You can verify mentally: the denominator 1 + xy always equals the numerator x − y for this pair.
Standard method: full algebraic expansion → 90 seconds. Pattern recognition: 20 seconds.
Before applying tan⁻¹x + tan⁻¹y = tan⁻¹((x+y)/(1−xy)), check the product xy:
xy < 1: use the formula directly.xy > 1 and both positive: add π to the result.xy > 1 and both negative: subtract π from the result.This three-branch test prevents the most common error on NDA arctan addition questions. Write "check xy first" at the top of your rough work.
Without the test: wrong sign on final answer in roughly 1 in 3 problems. With the test: zero conditional errors.
If a question says "k is a root of [quadratic]", extract what you need from Vieta's formulas before computing anything. Sum of roots = −(coefficient of x)/leading coefficient, product of roots = constant/leading coefficient.
Micro-example: roots of x² − 4x + 1 = 0 have product = 1 and sum = 4. Both roots are positive (sum positive, product positive). Knowing both roots are positive is sufficient to conclude tan⁻¹k + tan⁻¹(1/k) = π/2.
You never need to find k = 2 ± √3 explicitly. Standard method (solve quadratic, substitute): 75 seconds. Vieta + sign test: 20 seconds.
When you encounter an inverse trig question in the exam hall, run this decision tree in order:
Step 1 — Identify the structure. Is it a pair (f⁻¹(x) + g⁻¹(x))? Is it a subtraction? Is it an equation to solve for x?
Step 2 — Check for pair identities first. If you see sin⁻¹x + cos⁻¹x, write π/2 and move on. If you see tan⁻¹x + tan⁻¹(1/x), check sign of x and write ±π/2.
Step 3 — If it is an arctan addition/subtraction, check whether xy < 1 or xy > 1, then apply the addition formula. Write out numerator and denominator of (x ± y)/(1 ∓ xy) step by step — algebraic slips here are the main time-wasters.
Step 4 — If solving for x, isolate one inverse trig term, apply the direct trig function to both sides, and check that the answer lies within the valid input domain (e.g., x ∈ [−1, 1] for arcsin/arccos).
Step 5 — Range-check your answer. If the question is multiple choice and your computed value is, say, 3π/4, check whether that is achievable for the function in question. tan⁻¹ cannot output 3π/4 — it lives strictly inside (−π/2, π/2). This eliminates wrong answers without full computation.
Why this question: Tests whether you connect a quadratic's roots to the tan pair identity — a two-step insight that most students miss because they rush to solve the quadratic.
Solving path: Read "k is a root of x² − 4x + 1 = 0". By Vieta's, product of roots = 1. If k is one root, the other root is 1/k. Sum of roots = 4 > 0, product = 1 > 0, so both roots are positive. For any k > 0: tan⁻¹k + tan⁻¹(1/k) = π/2. Answer: π/2. Time: under 20 seconds.
Why this question: Tests the arctan subtraction formula in reverse — you're given the sum equals π/4 and asked to find k. Clean two-step algebra.
Solving path: tan⁻¹k = π/4 − tan⁻¹(1/2). Take tan of both sides: k = tan(π/4 − tan⁻¹(1/2)). Use tan(A − B) = (tan A − tan B)/(1 + tan A · tan B) with A = π/4 (so tan A = 1) and B = tan⁻¹(1/2) (so tan B = 1/2): k = (1 − 1/2)/(1 + 1 · 1/2) = (1/2)/(3/2) = 1/3. Answer: 1/3.
Why this question: Uses the complementary pair identity sin⁻¹x + cos⁻¹x = π/2 directly. If you know this identity, the problem takes 10 seconds. If you don't, it looks like a transcendental equation with no clean solution.
Solving path: Given cos⁻¹x = sin⁻¹x. We know sin⁻¹x + cos⁻¹x = π/2 for all valid x. Substituting cos⁻¹x = sin⁻¹x: 2 sin⁻¹x = π/2, so sin⁻¹x = π/4, giving x = sin(π/4) = 1/√2. Answer: x = 1/√2.
Why this question: A pure arctan subtraction test. The structure tan⁻¹(a/b) − tan⁻¹((a−b)/(a+b)) is a disguised constant — once you apply the subtraction formula, the numerator and denominator of the combined fraction are always equal regardless of a and b.
Solving path: Let x = a/b, y = (a−b)/(a+b). Apply tan⁻¹x − tan⁻¹y = tan⁻¹((x−y)/(1+xy)).
Compute x − y:
a/b − (a−b)/(a+b) = [a(a+b) − b(a−b)] / [b(a+b)] = [a² + ab − ab + b²] / [b(a+b)] = (a²+b²)/[b(a+b)]
Compute 1 + xy:
1 + (a/b) · (a−b)/(a+b) = [b(a+b) + a(a−b)] / [b(a+b)] = [ab + b² + a² − ab] / [b(a+b)] = (a²+b²)/[b(a+b)]
So (x−y)/(1+xy) = 1, and tan⁻¹(1) = π/4. Answer: π/4.
Forgetting the sign condition on the tan pair identity. tan⁻¹x + tan⁻¹(1/x) = π/2 only when x > 0. When x < 0, the answer is −π/2, not π/2. This exact trap appears in NDA options.
Applying the arctan addition formula when xy > 1 without the ±π correction. If x = 2 and y = 3, then xy = 6 > 1. The raw formula gives tan⁻¹((5)/(−5)) = tan⁻¹(−1) = −π/4, but the correct answer is −π/4 + π = 3π/4. Missing the correction costs the mark.
Confusing the ranges of sin⁻¹ and cos⁻¹. sin⁻¹(−1/2) = −π/6, but cos⁻¹(−1/2) = 2π/3 — not −π/3. The negative sign behaves differently because cos⁻¹ lives on [0, π], where all outputs are non-negative.
Solving cos⁻¹x = sin⁻¹x by squaring both sides. Squaring introduces extraneous solutions. The correct route is the complementary identity → 2 sin⁻¹x = π/2.
Treating sin⁻¹(sin θ) = θ as always true. It is only true when θ ∈ [−π/2, π/2]. For θ = 5π/6, sin⁻¹(sin(5π/6)) = π/6, not 5π/6. NDA options routinely include both to catch this.
Not checking the domain of the final answer. After solving for x, verify x ∈ [−1, 1] for arcsin/arccos and any real number for arctan. An answer like x = 2 from an arcsin problem is immediately invalid — but it will appear as a distractor option.