An identity is a mathematical sentence that is always true.
(A bit more precisely—everywhere it is defined, it is true.)
To verify (or prove) an identity means to prove that the sentence is always true.
This section discusses two common approaches for verifying (proving) identities.
Fundamental Trigonometric Identities talks about identities in general, and their usefulness.
It also introduces three important trigonometric identities (listed below).
Review this earlier section as needed.
Verifying identities is all about renaming—to show that two (initially differentlooking) expressions are actually the same.
You'll use:
Start with: $\,\sin^2 + \cos^2 = 1\,$ (For simplicity, inputs are not shown.) 

divide both sides by $\,\cos^2\,$: $$ \begin{gather} \cssId{s30}{\frac{\sin^2 + \cos^2}{\cos^2} = \frac{1}{\cos^2}}\cr\cr \cssId{s31}{\frac{\sin^2}{\cos^2} + \frac{\cos^2}{\cos^2} = \frac{1}{\cos^2}}\cr\cr \cssId{s32}{\left(\frac{\sin}{\cos}\right)^2 + 1 = \left(\frac{1}{\cos}\right)^2}\cr\cr\cr \cssId{s33}{\color{red}{\tan^2 + 1 = \sec^2}} \end{gather} $$  divide both sides by $\,\sin^2\,$: $$ \begin{gather} \cssId{s35}{\frac{\sin^2 + \cos^2}{\sin^2} = \frac{1}{\sin^2}}\cr\cr \cssId{s36}{\frac{\sin^2}{\sin^2} + \frac{\cos^2}{\sin^2} = \frac{1}{\sin^2}}\cr\cr \cssId{s37}{1 + \left(\frac{\cos}{\sin}\right)^2 = \left(\frac{1}{\sin}\right)^2}\cr\cr\cr \cssId{s38}{\color{red}{1 + \cot^2 = \csc^2}} \end{gather} $$ 
Equations of the form:  
COMPLICATED = SIMPLER  COMPL1 = COMPL2 
Often, one side of a (suspected) identity is more complicated than the other.
In such cases, it is usually easiest to:

Often, both sides of a (suspected) identity are similarly complicated. In such cases, you can often work with each side separately, renaming each side as the same simpler expression: $$ \cssId{s83}{ \begin{align} \text{COMPL1} &= \ \text{<rename>}\cr &= \ \ldots\cr &= \ \color{red}{\text{SIMPLER}}\cr\cr \text{COMPL2} &= \ \text{<rename>}\cr &= \ \ldots\cr &= \ \color{red}{\text{SIMPLER}} \end{align}} $$ The two $\,\color{red}{\text{SIMPLER}}\,$ expressions are the same! (Sometimes, when renaming one side, you inadvertently end up with the other side. If so—you're done!) 
Notes:
(In the examples and exercises, only one correct approach is shown.) 
$$ \begin{alignat}{2} &\cssId{s126}{\text{LHS}} \cssId{s127}{= \frac{1 + \cos t}{\cos t}} &\qquad\qquad& \cssId{s128}{\text{(start with LHS)}} \cr\cr &\qquad\cssId{s129}{= \ \frac{1}{\cos t} + \frac{\cos t}{\cos t}} &&\cssId{s130}{\text{(algebra)}}\cr\cr &\qquad\cssId{s131}{= \ \color{red}{\sec t + 1}} &&\cssId{s132}{\text{(definition of secant)}}\cr \end{alignat} $$  $$ \begin{alignat}{2} &\cssId{s133}{\text{RHS}} \cssId{s134}{= \frac{\tan^2 t}{\sec t  1}} &\qquad\qquad& \cssId{s135}{\text{(start with RHS)}} \cr\cr &\qquad\cssId{s136}{= \ \frac{\tan^2 t}{\sec t  1}\cdot\frac{\sec t + 1}{\sec t + 1}} &&\cssId{s137}{\text{(conjugate technique)}}\cr\cr &\qquad\cssId{s138}{= \ \frac{\tan^2 t(\sec t + 1)}{\sec^2 t  1}} &&\cssId{s139}{\text{(FOIL)}}\cr\cr &\qquad\cssId{s140}{= \ \frac{\tan^2 t(\sec t + 1)}{\tan^2 t}} &&\cssId{s141}{\text{(alternate Pythagorean Identity)}}\cr\cr &\qquad\cssId{s142}{= \ \color{red}{\sec t + 1}} &&\cssId{s143}{\text{(cancel)}}\cr \end{alignat} $$ 
Notice that both the LHS and RHS have been renamed as the same expression, $\,\sec t + 1\,$.

For many years, I worked in a math lab.
Any student could come in to get help.
I have a recollection of a time when I learned an important lesson.
I was helping someone verify an identity—but nothing was working!
We tried oodles of things, for a very long time.
Finally, the thought occurred to me that perhaps it wasn't an identity after all—maybe it was just a mistake.
Indeed, it was a typo in the book.
If you ever find yourself in the same situation, check that you're really working with an identity.
If the LHS and RHS are always equal then they will have precisely the same graphs.
At WolframAlpha, if you type in two commaseparated expressions, then both will be graphed in the same coordinate system.
If you graph both sides of an identity, then it will look like a single graph, because the graphs will overlap perfectly!
For example, cutandpaste the following into WolframAlpha:

The blue and red graphs overlap perfectly, giving the appearance of a single graph. This graphically shows that ‘$\,\displaystyle \tan x + \frac{\cos x}{1 + \sin x} = \sec x\,$’ is an identity. 
If you prefer, you can get an equivalent equation with zero on one side:
$$\cssId{s163}{\tan x + \frac{\cos x}{1 + \sin x}  \sec x = 0}$$
Since this is an identity, the left side graphs as the zero function. Not much to see! 
If the graphs don't match up (and you've input both sides correctly), then you're not working with an identity.
To prove that a sentence is not an identity, you need only provide a single number for which the sentence is false.
For example, ‘$\,\displaystyle \tan x + \frac{\cos x}{1 + \sin x} = 1 + \sec x\,$’ is not an identity:
On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. 
PROBLEM TYPES:
