Trigonometry

How to Prove Trig Identities: A Strategy Guide

Last updated: March 2026 · Advanced
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The hardest part of proving trig identities is not the algebra. It is deciding what to do next. Students freeze because they stare at an identity and have no systematic way to attack it. Textbooks show you finished proofs — neat, polished, and obvious in hindsight — but they never explain how the author decided which step to take. This page gives you a repeatable strategy that works on any identity proof, from straightforward to intimidating.

The Golden Rule: Work on One Side Only

A trig identity proof is not the same as solving an equation. When you solve an equation, you manipulate both sides — add, subtract, multiply, divide. When you prove an identity, you must show that one side transforms into the other through valid algebraic steps.

The rule:

Pick the more complex side. Transform it step by step until it matches the simpler side. Never move terms across the equals sign.

If you move terms across the equals sign, you are assuming the identity is true (which is what you are trying to prove). That is circular reasoning. Instead, start with one side and rewrite it until it looks exactly like the other side.

How do you choose which side to work on? Pick whichever side has more terms, more fractions, or more different trig functions. That side has more “handles” you can grab to simplify.

The Strategy Toolkit

Here are six strategies for transforming trig expressions, listed in the order you should try them. When you are stuck, start at Strategy 1 and work down the list.

Strategy 1: Convert Everything to Sine and Cosine

This is your default first move. Replace every tan, cot, sec, and csc with its sine/cosine equivalent:

tan⁡θ=sin⁡θcos⁡θcot⁡θ=cos⁡θsin⁡θsec⁡θ=1cos⁡θcsc⁡θ=1sin⁡θ\tan\theta = \frac{\sin\theta}{\cos\theta} \qquad \cot\theta = \frac{\cos\theta}{\sin\theta} \qquad \sec\theta = \frac{1}{\cos\theta} \qquad \csc\theta = \frac{1}{\sin\theta}

Most identities simplify dramatically once everything is expressed in terms of just two functions. If you only remember one strategy, make it this one.

Strategy 2: Look for Pythagorean Identity Patterns

Train your eye to spot these substitutions, even when they appear in disguise:

  • sin⁡2θ+cos⁡2θ⟶1\sin^2\theta + \cos^2\theta \longrightarrow 1
  • 1−sin⁡2θ⟶cos⁡2θ1 - \sin^2\theta \longrightarrow \cos^2\theta
  • 1−cos⁡2θ⟶sin⁡2θ1 - \cos^2\theta \longrightarrow \sin^2\theta
  • sec⁡2θ−1⟶tan⁡2θ\sec^2\theta - 1 \longrightarrow \tan^2\theta
  • sec⁡2θ−tan⁡2θ⟶1\sec^2\theta - \tan^2\theta \longrightarrow 1
  • csc⁡2θ−cot⁡2θ⟶1\csc^2\theta - \cot^2\theta \longrightarrow 1

These patterns frequently appear after you have already applied Strategy 1. Once you convert to sine and cosine, squared terms often combine into a Pythagorean identity.

Strategy 3: Factor

Look for the same factoring patterns you learned in algebra:

  • Common factors: sin⁡2θcos⁡θ+sin⁡θcos⁡θ=sin⁡θcos⁡θ(sin⁡θ+1)\sin^2\theta\cos\theta + \sin\theta\cos\theta = \sin\theta\cos\theta(\sin\theta + 1)
  • Difference of squares: sin⁡2θ−cos⁡2θ=(sin⁡θ−cos⁡θ)(sin⁡θ+cos⁡θ)\sin^2\theta - \cos^2\theta = (\sin\theta - \cos\theta)(\sin\theta + \cos\theta)
  • Trinomials: sin⁡2θ+2sin⁡θ+1=(sin⁡θ+1)2\sin^2\theta + 2\sin\theta + 1 = (\sin\theta + 1)^2

Factoring is especially powerful when combined with Pythagorean substitutions, because factored forms often cancel with denominators.

Strategy 4: Combine Fractions

If you see multiple fractions, find a common denominator and combine them into a single fraction. This frequently reveals a Pythagorean identity hiding in the numerator.

Strategy 5: Multiply by a Conjugate

When you are stuck with an expression like (1−sin⁡θ)(1 - \sin\theta) or (1+cos⁡θ)(1 + \cos\theta) in a denominator, multiply the numerator and denominator by the conjugate — the same expression with the opposite sign:

11−sin⁡θ⋅1+sin⁡θ1+sin⁡θ=1+sin⁡θ1−sin⁡2θ=1+sin⁡θcos⁡2θ\frac{1}{1 - \sin\theta} \cdot \frac{1 + \sin\theta}{1 + \sin\theta} = \frac{1 + \sin\theta}{1 - \sin^2\theta} = \frac{1 + \sin\theta}{\cos^2\theta}

The conjugate creates a difference of squares in the denominator, which almost always simplifies via a Pythagorean identity.

Strategy 6: Split a Single Fraction

If you have a fraction with a sum or difference in the numerator, try splitting it into separate fractions:

a+bc=ac+bc\frac{a + b}{c} = \frac{a}{c} + \frac{b}{c}

Each resulting fraction may simplify into a recognizable trig function. For example:

sin⁡θ+cos⁡θcos⁡θ=sin⁡θcos⁡θ+cos⁡θcos⁡θ=tan⁡θ+1\frac{\sin\theta + \cos\theta}{\cos\theta} = \frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\cos\theta} = \tan\theta + 1

The Decision Flowchart

When you sit down with a proof and do not know where to start, run through this checklist:

  1. See tan, cot, sec, or csc? Convert to sin/cos (Strategy 1)
  2. See squared terms? Look for Pythagorean patterns (Strategy 2)
  3. See a factorable expression? Factor it (Strategy 3)
  4. Multiple fractions? Combine them over a common denominator (Strategy 4)
  5. Stuck on (1±trig function)(1 \pm \text{trig function}) in a denominator? Multiply by the conjugate (Strategy 5)
  6. Complex single fraction? Split it apart (Strategy 6)

You will often use multiple strategies in a single proof. The flowchart tells you which one to try first.

Worked Examples

Example 1: Prove that sin⁡θcsc⁡θ+cos⁡θsec⁡θ=1\dfrac{\sin\theta}{\csc\theta} + \dfrac{\cos\theta}{\sec\theta} = 1

Strategy: Convert to sin/cos (Strategy 1).

Work on the left-hand side. Replace csc and sec with their definitions:

LHS=sin⁡θ1/sin⁡θ+cos⁡θ1/cos⁡θ\text{LHS} = \frac{\sin\theta}{1/\sin\theta} + \frac{\cos\theta}{1/\cos\theta}

Dividing by a fraction is multiplying by its reciprocal:

=sin⁡θ⋅sin⁡θ+cos⁡θ⋅cos⁡θ=sin⁡2θ+cos⁡2θ= \sin\theta \cdot \sin\theta + \cos\theta \cdot \cos\theta = \sin^2\theta + \cos^2\theta

Apply the Pythagorean identity:

=1=RHS✓= 1 = \text{RHS} \checkmark

Example 2: Prove that 1sin⁡θ−cos⁡2θsin⁡θ=sin⁡θ\dfrac{1}{\sin\theta} - \dfrac{\cos^2\theta}{\sin\theta} = \sin\theta

Strategy: Combine fractions (Strategy 4), then Pythagorean identity (Strategy 2).

The left-hand side already has a common denominator of sin⁡θ\sin\theta:

LHS=1−cos⁡2θsin⁡θ\text{LHS} = \frac{1 - \cos^2\theta}{\sin\theta}

Recognize the Pythagorean identity: 1−cos⁡2θ=sin⁡2θ1 - \cos^2\theta = \sin^2\theta:

=sin⁡2θsin⁡θ=sin⁡θ=RHS✓= \frac{\sin^2\theta}{\sin\theta} = \sin\theta = \text{RHS} \checkmark

Example 3: Prove that 11−sin⁡θ=sec⁡2θ+sec⁡θtan⁡θ\dfrac{1}{1 - \sin\theta} = \sec^2\theta + \sec\theta\tan\theta

Strategy: Work on the right-hand side (it is more complex). Convert to sin/cos (Strategy 1), then combine fractions (Strategy 4).

RHS=sec⁡2θ+sec⁡θtan⁡θ=1cos⁡2θ+1cos⁡θ⋅sin⁡θcos⁡θ\text{RHS} = \sec^2\theta + \sec\theta\tan\theta = \frac{1}{\cos^2\theta} + \frac{1}{\cos\theta} \cdot \frac{\sin\theta}{\cos\theta}

=1cos⁡2θ+sin⁡θcos⁡2θ=1+sin⁡θcos⁡2θ= \frac{1}{\cos^2\theta} + \frac{\sin\theta}{\cos^2\theta} = \frac{1 + \sin\theta}{\cos^2\theta}

Replace cos⁡2θ\cos^2\theta with 1−sin⁡2θ1 - \sin^2\theta (Pythagorean identity):

=1+sin⁡θ1−sin⁡2θ= \frac{1 + \sin\theta}{1 - \sin^2\theta}

Factor the denominator as a difference of squares:

=1+sin⁡θ(1−sin⁡θ)(1+sin⁡θ)= \frac{1 + \sin\theta}{(1 - \sin\theta)(1 + \sin\theta)}

Cancel the common factor of (1+sin⁡θ)(1 + \sin\theta):

=11−sin⁡θ=LHS✓= \frac{1}{1 - \sin\theta} = \text{LHS} \checkmark

Example 4: Prove that cos⁡θ1+sin⁡θ+1+sin⁡θcos⁡θ=2sec⁡θ\dfrac{\cos\theta}{1 + \sin\theta} + \dfrac{1 + \sin\theta}{\cos\theta} = 2\sec\theta

Strategy: Combine fractions over a common denominator (Strategy 4), then simplify.

Work on the left-hand side. The common denominator is (1+sin⁡θ)cos⁡θ(1 + \sin\theta)\cos\theta:

LHS=cos⁡2θ+(1+sin⁡θ)2(1+sin⁡θ)cos⁡θ\text{LHS} = \frac{\cos^2\theta + (1 + \sin\theta)^2}{(1 + \sin\theta)\cos\theta}

Expand the numerator:

=cos⁡2θ+1+2sin⁡θ+sin⁡2θ(1+sin⁡θ)cos⁡θ= \frac{\cos^2\theta + 1 + 2\sin\theta + \sin^2\theta}{(1 + \sin\theta)\cos\theta}

Group cos⁡2θ+sin⁡2θ=1\cos^2\theta + \sin^2\theta = 1 (Pythagorean identity):

=1+1+2sin⁡θ(1+sin⁡θ)cos⁡θ=2+2sin⁡θ(1+sin⁡θ)cos⁡θ= \frac{1 + 1 + 2\sin\theta}{(1 + \sin\theta)\cos\theta} = \frac{2 + 2\sin\theta}{(1 + \sin\theta)\cos\theta}

Factor the numerator:

=2(1+sin⁡θ)(1+sin⁡θ)cos⁡θ= \frac{2(1 + \sin\theta)}{(1 + \sin\theta)\cos\theta}

Cancel the common factor of (1+sin⁡θ)(1 + \sin\theta):

=2cos⁡θ=2sec⁡θ=RHS✓= \frac{2}{\cos\theta} = 2\sec\theta = \text{RHS} \checkmark

What If You Are Stuck?

Even experienced students get stuck on identity proofs. Here is what to do:

  • Try the other side. If you have been working on the left-hand side for several steps without progress, switch to the right-hand side. Sometimes one direction is much easier than the other.
  • Write out every step. Do not skip algebra. The step you skip is often where the simplification happens.
  • Go back to sin and cos. If you have tried multiple strategies and nothing works, erase your work and convert the entire original expression to sine and cosine from scratch. This fresh start with the simplest building blocks often reveals a path you missed.
  • Look at the target. Study what the simpler side looks like. If the target has cos⁡θ\cos\theta in the denominator, you probably need to create a cos⁡θ\cos\theta denominator somewhere in your work. Let the destination guide your steps.

Common Mistakes

  1. Moving terms across the equals sign. This is the most common error. Writing steps like “subtract sin⁡θ\sin\theta from both sides” treats the identity as an equation to solve. In a proof, you work on one side only.

  2. Giving up too early. Most proofs take 4 to 8 algebraic steps. If you have only done two steps, you are not stuck — you are just getting started. Keep applying strategies from the toolkit.

  3. Not converting to sine and cosine first. This is the single most common missed opportunity. Strategy 1 clears the path for everything else.

  4. Working both sides toward a middle point. Some textbooks accept “meet in the middle” proofs, but most instructors and standardized tests require one-sided proofs. Get in the habit of transforming one side completely into the other.

Practice Problems

Test your strategy skills with these proofs. Each one indicates which strategy to try first — but the real exercise is figuring out the strategy before you peek.

Problem 1: Prove that tan⁡θ⋅cos⁡θ=sin⁡θ\tan\theta \cdot \cos\theta = \sin\theta.

Strategy: Convert to sin/cos (Strategy 1).

LHS=tan⁡θ⋅cos⁡θ=sin⁡θcos⁡θ⋅cos⁡θ=sin⁡θ=RHS✓\text{LHS} = \tan\theta \cdot \cos\theta = \frac{\sin\theta}{\cos\theta} \cdot \cos\theta = \sin\theta = \text{RHS} \checkmark

Problem 2: Prove that sec⁡2θ−tan⁡2θ=1\sec^2\theta - \tan^2\theta = 1.

Strategy: Pythagorean identity (Strategy 2).

The identity tan⁡2θ+1=sec⁡2θ\tan^2\theta + 1 = \sec^2\theta rearranges directly to:

sec⁡2θ−tan⁡2θ=1\sec^2\theta - \tan^2\theta = 1

Alternatively, convert to sin/cos:

LHS=1cos⁡2θ−sin⁡2θcos⁡2θ=1−sin⁡2θcos⁡2θ=cos⁡2θcos⁡2θ=1=RHS✓\text{LHS} = \frac{1}{\cos^2\theta} - \frac{\sin^2\theta}{\cos^2\theta} = \frac{1 - \sin^2\theta}{\cos^2\theta} = \frac{\cos^2\theta}{\cos^2\theta} = 1 = \text{RHS} \checkmark

Problem 3: Prove that (sin⁡θ+cos⁡θ)2=1+2sin⁡θcos⁡θ(\sin\theta + \cos\theta)^2 = 1 + 2\sin\theta\cos\theta.

Strategy: Expand (related to Strategy 3 — factoring in reverse).

LHS=sin⁡2θ+2sin⁡θcos⁡θ+cos⁡2θ\text{LHS} = \sin^2\theta + 2\sin\theta\cos\theta + \cos^2\theta

Apply the Pythagorean identity to sin⁡2θ+cos⁡2θ\sin^2\theta + \cos^2\theta:

=1+2sin⁡θcos⁡θ=RHS✓= 1 + 2\sin\theta\cos\theta = \text{RHS} \checkmark

Problem 4: Prove that cos⁡θ1−sin⁡θ=1+sin⁡θcos⁡θ\dfrac{\cos\theta}{1 - \sin\theta} = \dfrac{1 + \sin\theta}{\cos\theta}.

Strategy: Multiply by the conjugate (Strategy 5).

Work on the left-hand side. Multiply numerator and denominator by the conjugate (1+sin⁡θ)(1 + \sin\theta):

LHS=cos⁡θ1−sin⁡θ⋅1+sin⁡θ1+sin⁡θ=cos⁡θ(1+sin⁡θ)1−sin⁡2θ\text{LHS} = \frac{\cos\theta}{1 - \sin\theta} \cdot \frac{1 + \sin\theta}{1 + \sin\theta} = \frac{\cos\theta(1 + \sin\theta)}{1 - \sin^2\theta}

Apply the Pythagorean identity: 1−sin⁡2θ=cos⁡2θ1 - \sin^2\theta = \cos^2\theta:

=cos⁡θ(1+sin⁡θ)cos⁡2θ=1+sin⁡θcos⁡θ=RHS✓= \frac{\cos\theta(1 + \sin\theta)}{\cos^2\theta} = \frac{1 + \sin\theta}{\cos\theta} = \text{RHS} \checkmark

Problem 5: Prove that cot⁡2θ1+csc⁡θ=csc⁡θ−1\dfrac{\cot^2\theta}{1 + \csc\theta} = \csc\theta - 1.

Strategy: Convert to sin/cos (Strategy 1), then factor (Strategy 3).

Convert the left-hand side:

LHS=cos⁡2θ/sin⁡2θ1+1/sin⁡θ=cos⁡2θ/sin⁡2θ(sin⁡θ+1)/sin⁡θ\text{LHS} = \frac{\cos^2\theta/\sin^2\theta}{1 + 1/\sin\theta} = \frac{\cos^2\theta/\sin^2\theta}{(\sin\theta + 1)/\sin\theta}

Dividing by a fraction is multiplying by its reciprocal:

=cos⁡2θsin⁡2θ⋅sin⁡θsin⁡θ+1=cos⁡2θsin⁡θ(sin⁡θ+1)= \frac{\cos^2\theta}{\sin^2\theta} \cdot \frac{\sin\theta}{\sin\theta + 1} = \frac{\cos^2\theta}{\sin\theta(\sin\theta + 1)}

Replace cos⁡2θ\cos^2\theta with 1−sin⁡2θ1 - \sin^2\theta (Pythagorean identity) and factor:

=1−sin⁡2θsin⁡θ(sin⁡θ+1)=(1−sin⁡θ)(1+sin⁡θ)sin⁡θ(1+sin⁡θ)= \frac{1 - \sin^2\theta}{\sin\theta(\sin\theta + 1)} = \frac{(1 - \sin\theta)(1 + \sin\theta)}{\sin\theta(1 + \sin\theta)}

Cancel the common factor of (1+sin⁡θ)(1 + \sin\theta):

=1−sin⁡θsin⁡θ=1sin⁡θ−sin⁡θsin⁡θ=csc⁡θ−1=RHS✓= \frac{1 - \sin\theta}{\sin\theta} = \frac{1}{\sin\theta} - \frac{\sin\theta}{\sin\theta} = \csc\theta - 1 = \text{RHS} \checkmark

Key Takeaways

  • Work on one side only — never move terms across the equals sign in an identity proof
  • Convert to sine and cosine first — this is your most powerful default strategy
  • Use the decision flowchart — when stuck, run through the six strategies in order
  • Pythagorean identities hide everywhere — train your eye to spot sin⁡2θ+cos⁡2θ\sin^2\theta + \cos^2\theta and its rearranged forms
  • Conjugate multiplication unlocks denominators of the form (1±trig)(1 \pm \text{trig})
  • Most proofs take 4 to 8 steps — if you have only written two lines, keep going
  • When truly stuck, switch sides or restart entirely in sine and cosine

Return to Trigonometry for more topics in this section.

Last updated: March 28, 2026