TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-02

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Derivatives and fundamental periods of tangent, cotangent, secant, and cosecant

Statement

On their natural domains, $(\tan x)^{'} = \sec^{2} x, (\cot x)^{'} = - \csc^{2} x, (\sec x)^{'} = \sec x \tan x, (\csc x)^{'} = - \csc x \cot x .$ Tangent and cotangent have least positive period $π$ ; secant and cosecant have least positive period $2 π$ .

Facts & Assumptions

Given: A real $x$ in the domain of the function under discussion.

[L1]

The quotient definitions and exact excluded points are those of Tangent, cotangent, secant, and cosecant on their exact natural domains.

[L2]

$\sin^{'} = \cos$ , $\cos^{'} = - \sin$ , and $\sin^{2} + \cos^{2} = 1$ (The derivatives of sine and cosine are cosine and minus sine, Parity and the Pythagorean identity for sine and cosine).

[L3]

Quotient and reciprocal differentiation rules apply, and the zero sets and periods of sine/cosine are known (Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , The zero sets of sine and cosine and the least positive common period 2 pi).

Proof

technique · direct

1.1L1L2L3

Quotient and reciprocal differentiation applied to [L1] gives the four displayed derivatives after using $\sin^{2} + \cos^{2} = 1$ .

1.2L1L3

Shifting by $π$ negates both sine and cosine, so their quotients have period $π$ , while their reciprocals change sign and therefore have period $2 π$ .

2.1step 1.2L2L3∎

The zero-set and quarter-turn values rule out a smaller positive period in each case.

Derivatives and fundamental periods of tangent, cotangent, secant, and cosecant

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources