How to Prove a Trig Identity: A Step-by-Step Method

Proof 1 - moves 2 and 3: tan θ + cot θ = sec θ csc θ

Prove that tan θ + cot θ = sec θ csc θ for all θ where both sides are defined.

1. The left side has two terms and two function names, so start there. Convert both to sine and cosine (move 2).
   $$\tan\theta + \cot\theta = \dfrac{\sin\theta}{\cos\theta} + \dfrac{\cos\theta}{\sin\theta}$$
2. Combine over the common denominator sin θ cos θ (move 3).
   $$= \dfrac{\sin^{2}\theta + \cos^{2}\theta}{\sin\theta\cos\theta}$$
3. The numerator is the Pythagorean identity, so it is exactly 1 (move 4).
   $$= \dfrac{1}{\sin\theta\cos\theta}$$
4. Split the single fraction into the product of two reciprocals.
   $$= \dfrac{1}{\cos\theta}\cdot\dfrac{1}{\sin\theta} = \sec\theta\csc\theta$$

Answer. The left side has become the right side using only the quotient, Pythagorean and reciprocal identities, so the statement is proved.

Proof 2 - move 5: cos θ / (1 − sin θ) = (1 + sin θ) / cos θ

Prove that cos θ / (1 − sin θ) = (1 + sin θ) / cos θ.

1. The left has the awkward binomial denominator, so start there and multiply numerator and denominator by the conjugate 1 + sin θ (move 5).
   $$\dfrac{\cos\theta}{1 - \sin\theta} = \dfrac{\cos\theta}{1 - \sin\theta}\cdot\dfrac{1 + \sin\theta}{1 + \sin\theta}$$

   This is multiplying by 1, which is always legal on one side. It is not the same as multiplying both sides by something.

2. Expand the denominator as a difference of squares.
   $$= \dfrac{\cos\theta\,(1 + \sin\theta)}{1 - \sin^{2}\theta}$$
3. The denominator is a rearranged Pythagorean identity: 1 − sin²θ = cos²θ (move 4).
   $$= \dfrac{\cos\theta\,(1 + \sin\theta)}{\cos^{2}\theta}$$
4. Cancel a factor of cos θ.
   $$= \dfrac{1 + \sin\theta}{\cos\theta}$$

Answer. Proved. Note the temptation this identity is designed to test: cross-multiplying to get cos²θ = 1 − sin²θ looks convincing and is an illegal move on both sides at once.

Proof 3 - move 4 on a rearranged Pythagorean identity

Prove that (sec²θ − 1) / sec²θ = sin²θ.

1. Start on the left: the numerator is the rearrangement of 1 + tan²θ = sec²θ.
   $$\dfrac{\sec^{2}\theta - 1}{\sec^{2}\theta} = \dfrac{\tan^{2}\theta}{\sec^{2}\theta}$$
2. Convert both to sine and cosine (move 2).
   $$= \dfrac{\sin^{2}\theta/\cos^{2}\theta}{1/\cos^{2}\theta}$$
3. Dividing by 1/cos²θ is multiplying by cos²θ, so the cosines cancel.
   $$= \dfrac{\sin^{2}\theta}{\cos^{2}\theta}\cdot\cos^{2}\theta = \sin^{2}\theta$$

Answer. Proved. Spotting sec²θ − 1 as tan²θ turned it into a one-line cancellation.

Proof 4 - bringing in the double-angle identities

Prove that (1 − cos 2θ) / sin 2θ = tan θ.

1. The left side is messier, and it is in terms of 2θ while the right is in terms of θ. So the first job is to eliminate the double angles. Use the form of cos 2θ written in terms of sine only.
   $$\cos 2\theta = 1 - 2\sin^{2}\theta \quad\Longrightarrow\quad 1 - \cos 2\theta = 1 - (1 - 2\sin^{2}\theta) = 2\sin^{2}\theta$$

   cos 2θ has three equivalent forms. Choosing the one that leaves you with sine only is what makes this proof short.

2. Rewrite the denominator with the sine double-angle identity.
   $$\sin 2\theta = 2\sin\theta\cos\theta$$
3. Assemble and cancel the common factor 2 sin θ.
   $$\dfrac{1 - \cos 2\theta}{\sin 2\theta} = \dfrac{2\sin^{2}\theta}{2\sin\theta\cos\theta} = \dfrac{\sin\theta}{\cos\theta} = \tan\theta$$

Answer. Proved. This is the half-angle tangent identity in disguise: replacing θ with θ/2 gives tan(θ/2) = (1 − cos θ) / sin θ.

Proof 5 - a sum-to-product proof

Prove that (sin A + sin B) / (cos A + cos B) = tan((A + B)/2).

1. Both sums on the left convert directly using the sum-to-product identities.
   $$\sin A + \sin B = 2\sin\dfrac{A+B}{2}\cos\dfrac{A-B}{2}$$
2. And the same for the denominator.
   $$\cos A + \cos B = 2\cos\dfrac{A+B}{2}\cos\dfrac{A-B}{2}$$
3. Form the quotient. The factor 2 cos((A − B)/2) is common to both and cancels.
   $$\dfrac{\sin A + \sin B}{\cos A + \cos B} = \dfrac{2\sin\frac{A+B}{2}\cos\frac{A-B}{2}}{2\cos\frac{A+B}{2}\cos\frac{A-B}{2}} = \dfrac{\sin\frac{A+B}{2}}{\cos\frac{A+B}{2}}$$
4. That quotient is the definition of tangent.
   $$= \tan\dfrac{A+B}{2}$$

Answer. Proved. When an identity mixes two angles on one side and their average on the other, sum-to-product is the intended route.

Why can I not cross-multiply when proving an identity?

Cross-multiplying operates on both sides simultaneously, which presumes the two sides are already equal - the very thing you were asked to establish. The reasoning is circular. Multiplying one side by a conjugate over itself is fine, because that is multiplication by 1 applied to a single expression.

Which side should I start from?

The more complicated one: more terms, more fractions, or more different function names. Simplifying gives you far more legal moves than building up does. If the two sides are comparably messy, simplify each independently to a common third expression and state that they agree.

Is converting everything to sine and cosine always the right first step?

It is the reliable default and it will finish most first-course problems, though not always by the shortest route. When the expression already sits in one family - all secants and tangents, say - using 1 + tan²θ = sec²θ directly is usually quicker. Try the pattern first, and fall back on sin and cos if nothing appears within a step or two.

Do I need to worry about the domain in my write-up?

For a standard course proof, generally no: the identity is understood to hold wherever both sides are defined. It is worth a sentence, though, when a step cancels a factor that could be zero - for example cancelling cos θ implicitly assumes cos θ ≠ 0, which is already required for the original expression to exist.

How do I get faster at these?

Recognition, not memorisation. The five moves are quick to learn; what takes practice is seeing 1 − sin²θ and thinking cos²θ without pausing. Work through a dozen proofs writing out every step, and the substitutions start arriving before you have finished reading the problem.