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Pure · Trigonometry

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The sec/cosec/cot identities

The two new Pythagorean identities 1 + tan²θ ≡ sec²θ and 1 + cot²θ ≡ cosec²θ — deriving them from sin²+cos²≡1, and using them to prove identities, simplify expressions and turn equations into a solvable quadratic.

Pure · Trigonometry

The sec/cosec/cot identities

The two new Pythagorean identities 1 + tan²θ ≡ sec²θ and 1 + cot²θ ≡ cosec²θ — deriving them from sin²+cos²≡1, and using them to prove identities, simplify expressions and turn equations into a solvable quadratic.

Why it works

These two identities aren't new facts — they're sin⁡2θ+cos⁡2θ≡1\sin^2\theta + \cos^2\theta \equiv 1 in disguise, divided through by cos⁡2θ\cos^2\theta and then by sin⁡2θ\sin^2\theta.

Divide sin⁡2θ+cos⁡2θ≡1\sin^2\theta + \cos^2\theta \equiv 1 by cos⁡2θ\cos^2\theta: sin⁡2θcos⁡2θ+1=1cos⁡2θ ⟹ tan⁡2θ+1≡sec⁡2θ.\frac{\sin^2\theta}{\cos^2\theta} + 1 = \frac{1}{\cos^2\theta} \ \Longrightarrow\ \tan^2\theta + 1 \equiv \sec^2\theta.

Divide it by sin⁡2θ\sin^2\theta: 1+cos⁡2θsin⁡2θ=1sin⁡2θ ⟹ 1+cot⁡2θ≡cosec⁡2θ.1 + \frac{\cos^2\theta}{\sin^2\theta} = \frac{1}{\sin^2\theta} \ \Longrightarrow\ 1 + \cot^2\theta \equiv \operatorname{cosec}^2\theta.

So the toolkit is now: 1+tan⁡2θ≡sec⁡2θ,1+cot⁡2θ≡cosec⁡2θ,1 + \tan^2\theta \equiv \sec^2\theta, \qquad 1 + \cot^2\theta \equiv \operatorname{cosec}^2\theta, together with the rearrangements you'll actually reach for: tan⁡2θ≡sec⁡2θ−1,cot⁡2θ≡cosec⁡2θ−1.\tan^2\theta \equiv \sec^2\theta - 1, \qquad \cot^2\theta \equiv \operatorname{cosec}^2\theta - 1.

Three jobs they do.
  1. Prove an identity — work one side, convert to sin⁡/cos⁡\sin/\cos if stuck, and
look for a sec⁡2−tan⁡2\sec^2 - \tan^2 or sin⁡2+cos⁡2\sin^2 + \cos^2 to collapse.
  1. Simplify — e.g. $\dfrac{\sec^2\theta - 1}{\sec^2\theta} =
\dfrac{\tan^2\theta}{\sec^2\theta} = \sin^2\theta$.
  1. Solve equations — when an equation mixes sec⁡θ\sec\theta with tan⁡2θ\tan^2\theta
(or cosec⁡θ\operatorname{cosec}\theta with cot⁡2θ\cot^2\theta), use the identity to write everything in one function. You almost always land on a quadratic.

The recurring trap on solving: a reciprocal can't be small. After factorising, reject any branch giving ∣sec⁡θ∣<1|\sec\theta| < 1 or ∣cosec⁡θ∣<1|\operatorname{cosec}\theta| < 1 — those have no solutions.