Trigonometric identities --- working on both sides of the equation at once When solving trigonometric identities, you aren't allowed to work on both sides of the equation at once. The reason for this is that if you do arrive at a valid conclusion, it doesn't provide the validity of the initial equation - it just proves that if the initial equation is true, you can arrive at a valid equation.
I have a number of questions about this:
1) Why can't adding or subtracting to both sides by allowed? Regardless of the relation between the sides ($=$,$<$,$>$,$≤$ or $≥$), adding and subtracting doesn't change the relation. Unlike dividing or multiplying by a negative number (which inverses the sign), adding or subtracting doesn't change the sign.
2) If you do work on both sides, and you do arrive at a valid equation (i.e. the Pythagorean identity), can't you prove the initial equation by working backwards? Through this reasoning if you reach a valid equation by working on both sides, the initial equation is valid.
For example:
Let $x$ represent a trigonometric identity you are checking the validity of. Let $y$ represent a proven identity such as the Pythagorean identity. 
Let's you say you do the following:
i) Divide both sides of $x$ by $a$
ii) Multiply both sides of $x$ by $b$
iii) Add $c$ to both sides of $x$
iiii) You arrive at equation $y$
My math teacher would argue this doesn't prove the validity of $x$ - it simply proves that if $x$ is true, you can arrive at $y$. However, if you start at $y$, and do the above steps backward (Subtract $c$, divide by $b$ and multiply by $a$) won't you arrive at $x$? Therefore if you do arrive at $y$ through working on both sides of $x$, shouldn't $x$ be valid since you can arrive at $x$ by working backwards starting at $y$?
3) Some trigonometric identities are extremely complicated and take a while to solve. How do you tell the difference between you not being able to find the proof and when the equation is not true? Because it would be a waste of time trying to find the proof of an trigonometric identity that is invalid.
 A: Suppose you write
$$
A=B
$$
Then you do something to both sides and get
$$
C=D.
$$
For example:
$$
3=-3.
$$
Square both sides:
$$
9=9.
$$
Lo and behold, this is TRUE!
So you see the logical fallacy.
There is nothing wrong with doing this when you're trying to figure out how to prove the identity.  But once you're there, you haven't yet got a proof.  One way to make it a proof is to write
$$
A = C = D = B.
$$
Then you've got it.
Another way is this: at each step where you did something to both sides, write an explanation of why it is that IF the lower equality is true, THEN so is the one above it.  Very often that is trivial and doesn't require you to say much.  But it should be a valid explanation of that point.
A: I disagree with Michael Hardy: He assumes something that the OP doesn't
1)

Reversible steps theorem (a stronger condition of property of equality):
$ A = B  \Rightarrow f(A) = f(B)$ iff $ f^{-1}(x)$ exists for all $A, B  \in f^{-1}(y)$

$f(x) = x^2$ does not have a (global) inverse and thus doesn't follow the theorem above. OP only mentioned steps, addition, subtraction, multiplication and division, that are reversible.
Therefore, it is valid to manipulate both sides of the trig identity as long as you know what your doing i.e. every step is reversible. The reason students aren't encouraged to do this is that most high school students do not grasp the depth of real function theory to understand these theorems.


*you can also manipulate both sides of the equation via substitution. Substitution is reversible and obeys the properties of equality.

e.g.

$ \frac{1 - cos^{2}(x)}{sin(x)} = cos(x)tan(x)$
we may substitute in $1-cos^2(x) = sin^2(x)$ and $tan(x) = \frac{sin(x)}{cos(x)}$
$ \frac{sin^2(x)}{sin(x)} = cos(x)\frac{sin(x)}{cos(x)} \implies sin(x) = sin(x)$

