LemmaStatement: AI-adaptedProof: AI-adaptedverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)

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Reciprocal positive arguments of the principal inverse tangent

Physical scope

Pure trigonometric identity for the principal inverse tangent; it is used only to rewrite a formal model limit.

Statement

For every $y > 0$ ,

$\arctan y + \arctan (1 / y) = \frac{π}{2},$

where $\arctan$ takes values in $(- π / 2, π / 2)$ .

Facts & Assumptions

Given: A positive real $y$ and the principal inverse-tangent convention.

[F1]

Tangent is a continuous strictly increasing bijection from $(- π / 2, π / 2)$ onto $R$ , and its inverse obeys $\tan (\arctan z) = z$ and $\arctan (\tan u) = u$ on that interval (The principal inverse tangent $\arctan : R \to (- π / 2, π / 2)$ , Tangent is a continuous strictly increasing bijection from $(- π / 2, π / 2)$ onto $R$ ).

[F2]

Tangent is $\sin u / \cos u$ on the domain where cosine is nonzero (Tangent, cotangent, secant, and cosecant on their exact natural domains).

[F3]

The values at zero, quarter-turn shifts, and sine-cosine parity identities hold (The derivatives of sine and cosine are cosine and minus sine, Quarter-turn values and shifts by pi/2 and pi, Pythagorean and parity identities for all six trigonometric functions on their natural domains).

[F4]

Cosine is positive on $[0, π / 2)$ and tangent is defined on $(- π / 2, π / 2)$ (Signs, monotonicity intervals, and ranges of sine and cosine, Tangent, cotangent, secant, and cosecant on their exact natural domains).

Proof

technique · use the complementary-angle tangent identity and uniqueness of the principal inverse

1.1givenF1F2F3algebra

Put $u = \arctan y$ . By [F3], $\tan 0 = 0$ ; [F1] therefore gives $\arctan 0 = 0$ . Since $y > 0$ and the inverse in [F1] is strictly increasing, $u > 0$ , while its range gives $u < π / 2$ .

2.1step 1.1F2F3F4givenalgebra

The angle $π / 2 - u$ lies in $(0, π / 2)$ . The shift and parity identities in [F3] give $\sin (π / 2 - u) = \cos u$ and $\cos (π / 2 - u) = \sin u$ . By [F4], $\cos u > 0$ ; because $\tan u = y > 0$ and $\tan u = \sin u / \cos u$ , also $\sin u > 0$ . Thus tangent is defined at both $u$ and $π / 2 - u$ , and [F2] gives $\tan (π / 2 - u) = \cos u / \sin u = 1 / y$ .

3.1step 1.1step 2.1F1givenalgebra∎

Since $π / 2 - u \in (- π / 2, π / 2)$ , the inverse identity in [F1] yields $\arctan (1 / y) = π / 2 - u$ . Substituting $u = \arctan y$ proves the formula.

Reciprocal positive arguments of the principal inverse tangent

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources