The Pythagorean identities in disguise

Two kinds of job:

• Simplify or prove: turn a tangle of sec, cosec and cot into something small.
• Solve: an equation has two different functions of the same angle, like $\sec^2\theta$ and $\tan\theta$. The identity turns it into a quadratic in one function.

It's the same trick as using $\sin^2\theta + \cos^2\theta \equiv 1$ at AS, now for sec/tan and cosec/cot.

Start with the one you know:

$\sin^2\theta + \cos^2\theta \equiv 1$

Divide every term by $\cos^2\theta$.

• $\dfrac{\sin^2\theta}{\cos^2\theta} = \tan^2\theta$
• $\dfrac{\cos^2\theta}{\cos^2\theta} = 1$
• $\dfrac{1}{\cos^2\theta} = \sec^2\theta$

$\tan^2\theta + 1 \equiv \sec^2\theta$

Divide every term by $\sin^2\theta$ instead.

• $\dfrac{\sin^2\theta}{\sin^2\theta} = 1$
• $\dfrac{\cos^2\theta}{\sin^2\theta} = \cot^2\theta$
• $\dfrac{1}{\sin^2\theta} = \operatorname{cosec}^2\theta$

$1 + \cot^2\theta \equiv \operatorname{cosec}^2\theta$

Check with numbers, $\theta = 45°$: $\tan 45° = 1$ and $\sec 45° = \sqrt2$. Then $1 + 1 = 2 = (\sqrt2)^2$ ✓

1. Substitute an identity so the equation has only one trig function. For example, replace $\sec^2\theta$ with $1 + \tan^2\theta$.
2. Rearrange into a quadratic in that function, and solve it.
3. Solve for the angle from each value, using the interval you're given.

The forgotten step: if you square-root something like $\tan^2\theta = 4$, there are two values, $\tan\theta = 2$ and $\tan\theta = -2$. Dropping the negative one loses half the solutions.