Inverse Trigonometric Simplification
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Direct answer
The addition formula for tan⁻¹ is the most quoted line of the chapter and the least checked: tan⁻¹x + tan⁻¹y = tan⁻¹[(x + y)/(1 − xy)] is true as written only when xy < 1 — cross that threshold and a π must be added or subtracted, which is exactly where answer options diverge. Simplification in this chapter means collapsing sums and multiples of inverse functions into single ones with the branch condition doing the real work: the double-angle family 2tan⁻¹x = sin⁻¹[2x/(1 + x²)] = tan⁻¹[2x/(1 − x²)] for |x| < 1, the complementary collapse tan⁻¹x + tan⁻¹(1/x) = π/2 for x > 0, and a range audit at the end of every reduction.
What you must remember
- Addition with conditions: xy < 1 → formula as written; xy > 1 with x, y > 0 → add π; xy > 1 with x, y < 0 → subtract π; xy = 1 with x, y > 0 → the sum is π/2 outright.
- Subtraction: tan⁻¹x − tan⁻¹y = tan⁻¹[(x − y)/(1 + xy)], valid when xy > −1; for x > 0 > y with xy < −1, add π.
- Double-angle set: 2tan⁻¹x = sin⁻¹[2x/(1 + x²)] for |x| ≤ 1, = tan⁻¹[2x/(1 − x²)] for |x| < 1, and = cos⁻¹[(1 − x²)/(1 + x²)] for x ≥ 0.
- Triple angle: 3tan⁻¹x = tan⁻¹[(3x − x³)/(1 − 3x²)] for x² < 1/3.
- Half-angle gift: tan⁻¹[(√(1 + x²) − 1)/x] = ½tan⁻¹x for x ≠ 0 — the standard rearrangement of the double-angle identity.
- Complementary collapse: tan⁻¹x + tan⁻¹(1/x) = π/2 for x > 0 and −π/2 for x < 0; sin⁻¹x + cos⁻¹x = π/2 on [−1, 1].
- Range audit: after every collapse, confirm the result sits in the correct principal branch — a tan⁻¹ output must remain in (−π/2, π/2).
Branch decisions in practice
Evaluate tan⁻¹(1/2) + tan⁻¹(1/3): here xy = 1/6 < 1, so the formula applies directly, giving tan⁻¹[(5/6)/(5/6)] = tan⁻¹1 = π/4. Next, tan⁻¹1 + tan⁻¹2: xy = 2 > 1 with both arguments positive, so the sum lies in (π/2, π) and equals π + tan⁻¹[3/(1 − 2)] = π + tan⁻¹(−3) = π − tan⁻¹3. Numerically: 0.785 + 1.107 ≈ 1.893, and π − 1.249 ≈ 1.893 ✓ — the π correction is not cosmetic, it is the answer. Third, tan⁻¹(−2) + tan⁻¹(−3): xy = 6 > 1 with both negative, so subtract π: the bare formula gives tan⁻¹[(−5)/(−5)] = tan⁻¹1 = π/4, and the true sum is π/4 − π = −3π/4.
Three evaluations, three different branch decisions — positive-small, positive-large, negative-large — and that spread is precisely what a multiple-correct option set tests. Notice also the clean one-liners these formulas generate: 2tan⁻¹(1/3) = tan⁻¹[(2/3)/(8/9)] = tan⁻¹(3/4), a Main favourite.
The condition examiners test
Advanced sets options that differ only by the added π or by a π/4-type shift, so the branch check is the entire question; Main uses the clean pairs (1/2, 1/3) and (1, 2) and single collapses like 2tan⁻¹(1/3). The standing traps: applying the addition formula with xy > 1 and reporting the arctangent of a negative number — a point in the wrong quadrant; using 2tan⁻¹x = tan⁻¹[2x/(1 − x²)] for |x| > 1, where the two sides differ by π and the sin⁻¹ form is the safe one; and mixing degrees into identities that assume radian measure throughout. The discipline that resolves all of them: before quoting any collapsed value, place it in its principal branch and ask whether the original sum could actually live there.
Frequently asked questions
What is tan⁻¹(1/2) + tan⁻¹(1/3)?
π/4 — since xy = 1/6 < 1, the addition formula applies directly and gives tan⁻¹1.
Why does tan⁻¹1 + tan⁻¹2 equal π − tan⁻¹3 and not tan⁻¹(−3)?
xy = 2 > 1 with both arguments positive places the sum in (π/2, π); tan⁻¹(−3) alone sits in the wrong quadrant.
For which x does 2tan⁻¹x = tan⁻¹[2x/(1 − x²)] hold?
For |x| < 1; beyond that the two sides differ by π, while the sin⁻¹ form 2tan⁻¹x = sin⁻¹[2x/(1 + x²)] holds for |x| ≤ 1.
What is tan⁻¹x + tan⁻¹(1/x)?
π/2 for x > 0 and −π/2 for x < 0 — a complementary collapse, never the addition formula.
How do I simplify tan⁻¹[(√(1 + x²) − 1)/x]?
It equals ½tan⁻¹x for x ≠ 0 — the half-angle rearrangement of the double-angle identity.