Simplifying shows what an expression equals. Verifying asks a yes-or-no question: are two given expressions equal for every $x$? You will use the same fundamental identities and the same algebra as with rational expressions. What is new is the logic of a proof.
Key idea
To prove an identity, start with one side (usually the more complicated one) and rewrite it, one justified step at a time, until it becomes the other side. Do not do the same thing to both sides.
Why it works
Operating on both sides assumes the two sides are already equal, which is what you are trying to show. Such a chain only proves "if the identity is true, then something else is true." It can even "prove" false statements: start from $-1 = 1$, square both sides, and you get $1 = 1$, which is true. Rewriting one side never assumes anything, so when it reaches the other side, the equality is proved.
- Rewrite in terms of sine and cosine.
- Combine fractions over a common denominator, or split one fraction into two.
- Factor, especially differences of squares like $$1 - \sin^2 x = (1 - \sin x)(1 + \sin x).$$
- Multiply the top and bottom by a conjugate, such as $1 + \cos x$.
- Keep an eye on the target side: it tells you what to aim for.
Worked examples
Worked example: Combine over a common denominator
Verify $\sin x\tan x + \cos x = \sec x$.
- Start with the left side and write it in sines and cosines: $\sin x \cdot \dfrac{\sin x}{\cos x} + \cos x$.
- Common denominator: $\dfrac{\sin^2 x}{\cos x} + \dfrac{\cos^2 x}{\cos x} = \dfrac{\sin^2 x + \cos^2 x}{\cos x}$.
- Pythagorean identity: $= \dfrac{1}{\cos x} = \sec x$, the right side.
Answer: Verified
Worked example: Use a conjugate
Verify $\dfrac{\sin x}{1 - \cos x} = \dfrac{1 + \cos x}{\sin x}$.
- Start with the left side. Multiply top and bottom by $1 + \cos x$ (multiplying by $1$ changes nothing): $\dfrac{\sin x(1 + \cos x)}{(1 - \cos x)(1 + \cos x)}$.
- The bottom is $1 - \cos^2 x = \sin^2 x$: $\dfrac{\sin x(1 + \cos x)}{\sin^2 x}$.
- Cancel one factor of $\sin x$: $\dfrac{1 + \cos x}{\sin x}$, the right side.
Answer: Verified
Worked example: Two fractions into one
Verify $\tan x + \cot x = \sec x\csc x$.
- Left side: $\dfrac{\sin x}{\cos x} + \dfrac{\cos x}{\sin x} = \dfrac{\sin^2 x + \cos^2 x}{\sin x\cos x}$.
- $$= \dfrac{1}{\sin x\cos x} = \dfrac{1}{\cos x} \cdot \dfrac{1}{\sin x} = \sec x\csc x.$$
Answer: Verified
Common mistake
Treating the identity as an equation: multiplying both sides by $\cos x$, or squaring both sides, then arriving at something true like $1 = 1$. That assumes what you set out to prove. Work on one side only. If you want to use both sides, transform each side separately until they meet in the middle.
One value of $x$ where the two sides differ is enough to show an equation is not an identity. For example, $\sin x + \cos x = 1$ fails at $x = \frac{\pi}{4}$, where the left side is $\sqrt{2}$.