Why is it that when proving trig identities, one must work both sides independently? Suppose that you have to prove the trig identity:

$$\frac{\sin\theta - \sin^3\theta}{\cos^2\theta}=\sin\theta$$

I have always been told that I should manipulate the left and right sides of the equation separately, until I have transformed them each into something identical. So I would do:

$$\frac{\sin\theta - \sin^3\theta}{\cos^2\theta}$$
$$=\frac{\sin\theta(1 - \sin^2\theta)}{\cos^2\theta}$$
$$=\frac{\sin\theta(\cos^2\theta)}{\cos^2\theta}$$
$$=\sin\theta$$

And then, since the left side equals the right side, I have proved the identity. My problem is: why can't I manipulate the entire equation? In this situation it probably won't make things any easier, but for certain identities, I can see ways to "prove" the identity by manipulating the entire equation, but cannot prove it by keeping both sides isolated.

I understand, of course, that I can't simply assume the identity is true. If I assume a false statement, and then derive from it a true statement, I still haven't proved the original statement. However, why can't I do this:

$$\frac{\sin\theta - \sin^3\theta}{\cos^2\theta}\not=\sin\theta$$
$$\sin\theta - \sin^3\theta\not=(\sin\theta)(\cos^2\theta)$$
$$\sin\theta(1 - \sin^2\theta)\not=(\sin\theta)(\cos^2\theta)$$
$$(\sin\theta)(\cos^2\theta)\not=(\sin\theta)(\cos^2\theta)$$

Since the last statement is obviously false, is this not a proof by contradiction that the first statement is false, and thus the identity is true?

Or, why can't I take the identity equation, manipulate it, arrive at $(\sin\theta)(\cos^2\theta)=(\sin\theta)(\cos^2\theta)$, and then work backwards to arrive at the trig identity. Now, I start with a statement which is obviously true, and derive another statement (the identity) which must also be true - isn't that correct?

Another argument that I have heard for keeping the two sides isolated is that manipulating an equation allows you to do things that are not always valid in every case. But the same is true when manipulating just one side of the equation. In my first proof, the step

$$\frac{\sin\theta(\cos^2\theta)}{\cos^2\theta}$$
$$=\sin\theta$$

is not valid when theta is $\pi/2$, for example, because then it constitutes division by zero.
Why can't I manipulate the entire equation?


You can. The analytical method for proving an identity consists of starting with the
identity you want to prove, in the present case
$$
\begin{equation}
\frac{\sin \theta -\sin ^{3}\theta }{\cos ^{2}\theta }=\sin \theta,\qquad \cos \theta \neq 0
\tag{1}
\end{equation}
$$
and establish a sequence of identities so that each one is a consequence of
the next one. For the identity $(1)$ to be true is enough that the following
holds
$$
\begin{equation}
\sin \theta -\sin ^{3}\theta =\sin \theta \cos ^{2}\theta   \tag{2}
\end{equation}
$$
or this equivalent one
$$
\begin{equation}
\sin \theta \left( 1-\sin ^{2}\theta \right) =\sin \theta \cos ^{2}\theta  
\tag{3}
\end{equation}
$$
or finally this last one 
$$
\begin{equation}
\sin \theta \cos ^{2}\theta =\sin \theta \cos ^{2}\theta   \tag{4}
\end{equation}
$$

Since $(4)$ is true so is $(1)$.

The book indicated below illustrates this method with the following identity
$$
\frac{1+\sin a}{\cos a}=\frac{\cos a}{1-\sin a}\qquad a\neq (2k+1)\frac{\pi 
}{2}
$$

It is enough that the following holds
$$
(1+\sin a)(1-\sin a)=\cos a\cos a
$$

or
$$
1-\sin ^{2}a=\cos ^{2}a,
$$

which is true if 
$$
1=\cos ^{2}a+\sin ^{2}a
$$

is true. Since this was proven to be true, all the previous indentities
hold, and so does the first identity.




Reference: J. Calado, Compêndio de Trigonometria, Empresa Literária
Fluminense, Lisbon, pp. 90-91, 1967.
I think it can be done.  We are proving that LHS =RHS,  as by assuming so,  you arrive at a Universal truth. 

Similarly,  by assuming it's not true,  we arrive at a contradiction ,  which is called,  proof by contradiction.