Why a Familiar Algebra Step Needs Care
Cross-multiplying and canceling are useful algebraic moves, but each has a condition. Trigonometric functions take the value zero at particular angles, and expressions such as \(\tan x\) or \(\frac{1}{\sin x}\) are undefined at particular inputs. If a factor is zero, multiplying or canceling can change which statements are equivalent—or even whether an expression is defined.
When verifying an identity, the goal is to show that the two expressions agree at every input where both are defined. An algebraic transformation must preserve that domain. An equation obtained after multiplying by a factor may be true at inputs where the original expression was undefined, or canceling a factor may discard inputs that satisfy the original equation.
Multiplication Is Not Always Reversible
For real numbers \(A\), \(B\), and \(C\), if \(A=B\), then \(AC=BC\). This implication holds even when \(C=0\). But the reverse implication requires \(C\ne 0\): only then can we divide both sides of \(AC=BC\) by \(C\) and recover \(A=B\).
If \(C=0\), the multiplied equality becomes \(0=0\), regardless of whether \(A=B\). Thus, multiplying both sides by a possibly zero factor can produce an equality that does not carry the original information. Canceling has the same condition in reverse: cancel a factor only on inputs where it is nonzero.
Cross-multiplication follows the same rule. If \(B\ne 0\) and \(D\ne 0\), then \(\frac{A}{B}=\frac{C}{D}\) is equivalent to \(AD=BC\). The nonzero conditions are part of the equivalence; the cross-product equation alone does not record them.
A Complete Justification for the Nonzero Condition
Here is why the condition matters. Suppose \(B\ne 0\) and \(D\ne 0\). Starting with the fraction equality, multiply both sides by \(BD\), which is nonzero, to obtain \(AD=BC\). Conversely, start from \(AD=BC\) and divide both sides by \(BD\). Since \(B\) and \(D\) are nonzero, this division is allowed, and the result is \(\frac{A}{B}=\frac{C}{D}\).
Without the restrictions, the cross-product equation might hold even though one or both fractions do not exist. For trigonometric expressions, those restrictions often involve \(\sin x\) or \(\cos x\). The safe procedure is therefore to identify where the original expressions are defined first, and only then cross-multiply.
For each denominator, state which inputs make it nonzero.
Cross-multiply or cancel only while keeping the nonzero conditions in force.
Do not treat inputs excluded from the original expressions as valid inputs to the original statement.
Worked Example: A Cross-Product Holds Where the Original Expressions Do Not Exist
Worked Example: A Cross-Product Holds Where the Original Expressions Do Not Exist
Consider the equation \(\tan x=\sec x\). Determine whether its cross-multiplied form can be used without retaining a domain restriction.
By the quotient and reciprocal identities, \(\tan x=\frac{\sin x}{\cos x}\) and \(\sec x=\frac{1}{\cos x}\). Both sides are defined only when \(\cos x\ne 0\). On that domain, multiplication by \(\cos x\) gives the necessary equation \(\sin x=1\).
Now check \(x=\frac{\pi}{2}\). The cross-product equation holds because \(\sin\left(\frac{\pi}{2}\right)=1\). But \(\cos\left(\frac{\pi}{2}\right)=0\), so neither \(\tan\left(\frac{\pi}{2}\right)\) nor \(\sec\left(\frac{\pi}{2}\right)\) is defined. Thus \(x=\frac{\pi}{2}\) cannot be treated as an input satisfying the original equation.
Check: Substitution into the cross-product equation gives \(1=1\); substitution into the original expressions requires division by \(\cos\left(\frac{\pi}{2}\right)=0\), so the original equation is undefined there. The cross-product equation is not a replacement for the original equation unless the condition \(\cos x\ne 0\) is retained.
A Safe Identity Verification
For an identity involving quotients, it is useful to state the common domain explicitly. The following example uses the Pythagorean identity \(\sin^2 x+\cos^2 x=1\). It shows how cross-multiplication is safe when its denominator restrictions are preserved.
Worked Example: Verify a Quotient Identity on Its Common Domain
Verify that \(\frac{\sin x}{1-\cos x}=\frac{1+\cos x}{\sin x}\) where both sides are defined.
The left side requires \(1-\cos x\ne 0\), or \(\cos x\ne 1\). The right side requires \(\sin x\ne 0\). For real \(x\), \(\sin x=0\) occurs at integer multiples of \(\pi\), and at those inputs \(\cos x\) is either \(1\) or \(-1\). Thus every input where the right side is defined also has \(\cos x\ne 1\), and the common domain is precisely the inputs with \(\sin x\ne 0\).
On this common domain, both denominators are nonzero, so cross-multiplication is reversible:
Expand the right side using the difference of squares, then use the Pythagorean identity:
The cross-product equation is therefore true, so the original expressions are equal wherever both are defined. As a numerical check, take \(x=\frac{\pi}{2}\). The left side is \(\frac{1}{1-0}=1\), and the right side is \(\frac{1+0}{1}=1\).
Check: The proof only cross-multiplied after establishing that both denominators were nonzero. At \(x=\pi\), for instance, the right side is undefined because \(\sin\pi=0\); the identity makes no claim that the two original expressions are equal there.
Canceling a Zero Trigonometric Factor Can Lose Cases
Cancellation can fail in the other direction: rather than allowing an input that was originally excluded, it can discard an input that satisfied the original equation. Factoring first does not authorize cancellation of a factor that might be zero.
Worked Example: Cancellation Discards a Valid Input
Consider the equation \(\sin x\cos x=\sin x\). Factor the difference between its sides:
One might try to cancel \(\sin x\) and conclude that \(\cos x-1=0\), or \(\cos x=1\). That cancellation is permitted only when \(\sin x\ne 0\). If \(\sin x=0\), the factored equation is already true, whatever the value of \(\cos x-1\).
Check \(x=\pi\). Then \(\sin\pi=0\) and \(\cos\pi=-1\). The original left side is \(0\cdot(-1)=0\), and the original right side is \(0\), so the equation holds. But the canceled conclusion \(\cos\pi=1\) is false. This confirms that canceling \(\sin x\) without checking whether it is zero loses a valid case.
Check: At \(x=\pi\), the factored left side is \(0\cdot(-2)=0\), as required. The canceled equation would read \(-1=1\), which fails. The correct conclusion is that the factored equation requires at least one factor to be zero; neither factor may simply be discarded without a nonzero condition.
Keep the Domain When Simplifying
A common source of confusion is that simplifying an expression can make it look defined at more inputs than the original expression. For example, \(\frac{\sin x}{\sin x}\) simplifies to \(1\) whenever \(\sin x\ne 0\). But the original quotient remains undefined where \(\sin x=0\). The simplified expression does not restore those excluded inputs.
This distinction matters when verifying an identity. The claim is about the expressions as originally written, not about an informal extension of a simplified expression. Keep a record of any restrictions introduced by denominators or by cancellation, even if those factors disappear later in the algebra.
| Algebraic move | What is always valid? | What must be checked? |
|---|---|---|
| Multiply both sides by \(C\) | \(A=B\) implies \(AC=BC\) | To reverse the implication, require \(C\ne 0\) |
| Cancel a common factor \(C\) | Cancellation preserves equality where \(C\ne 0\) | Check whether the original equation also has cases with \(C=0\) |
| Cross-multiply fractions | The cross-product equality follows on the fractions’ common domain | Both original denominators must be nonzero |
A Practical Domain Check
Before cross-multiplying or canceling in a trigonometric statement, use this short check:
- Write down the inputs excluded by every denominator in the original expressions.
- Identify any factor you plan to cancel, and state where that factor is nonzero.
- Use equivalence only while its conditions hold; otherwise mark the step as a one-way implication.
- When an algebraic result includes an input excluded at the start, do not count it as valid for the original expressions.
- After simplifying, compare the final domain with the original domain and keep all original exclusions.
These checks do not prevent cross-multiplication or cancellation. They make clear when those moves preserve the original statement and when they do not. For identity verification, the safest habit is to work on the common domain and leave each nonzero condition visible in the reasoning.
Check Your Understanding
For each question, identify the nonzero condition that matters before deciding whether an algebraic step is reversible.
- Why does \(AC=BC\) imply \(A=B\) when \(C\ne 0\), but not necessarily when \(C=0\)?
- What domain restriction must be retained when cross-multiplying an equation involving \(\tan x\) and \(\sec x\)?
- In the identity \(\frac{\sin x}{1-\cos x}=\frac{1+\cos x}{\sin x}\), which original denominators must be nonzero?
- Why does canceling \(\sin x\) from \(\sin x\cos x=\sin x\) fail to preserve every valid input?
- If a simplified expression is defined at an input where the original quotient was undefined, does the simplification make the original expression defined there? Explain.
