Conceptual Idea of Functional Reduction I have a doubt regarding functional reduction(In mathematics, reduction refers to the rewriting of an expression into a simpler form). 
I have a function, let me call it as $f(x)$ and I have rewritten it to a simpler form let me call it as $g(x)$. Can we say both $f(x)$ and $g(x)$ are equivalent? From a rough over view, we may argue that both render same result for the values of $x$.  
Assume I have a function  $f(x) = \frac{x^2-9}{x-3}$ . Clearly it is not defined at $x=3$. But if I rewrite it as $g(x) = x+3$ you can say that it is defined on $x=3$. Since $g(x)$ and $f(x)$ both  don't agree on common domain  and range can we say both are equivalent? Is it conceptually fair to argue such point?  Then what is the point of reduction if we gain some domain which was not the original property of the function before reduction.? 

Equal functions agree on domain and values.
When one proves identities, one only requires the (restricted) functions to agree on their mutual domain, the intersection of their domains.  If we do not make this relaxation, many identities become false.  For instance, "$1 = \frac{\sin z}{\sin z}$" is undefined on the right for $z$ an integer multiple of $\pi$.  If we use ideas from calculus, we can "fix this" -- the right-hand side has finite limits which equal the left-hand side at the points where the right-hand side is undefined.
You are right to keep the change of domain in mind.  Although the above is what we do for identities, it is not what we do when solving equations.  Consider $\sin x \cos x = \sin x$.  If we "divide both sides by $\sin x$", we do exactly the sort of manipulation we were describing above.  However, the integer multiples of $\pi$ that we remove from the domains of both sides actually contain solutions, which our reduced equation has "forgotten".  So there are two ways to go.
First, we can choose to not reduce our domain by neither multiplying nor dividing both sides by something that can ever be zero...
\begin{align}
\sin x \cos x &= \sin x  \\
\sin x \cos x - \sin x &= 0  \\
\sin x (\cos x - 1) &= 0  \\
\sin x = 0 \text{ or }& \cos x = 1  \\
x = \sin^{-1}(0) + 2 \pi k \text{ or }& x = \pi - \sin^{-1}(0) + 2 \pi k \\
    \text{ or }& x = \pm \cos^{-1}(1) + 2 \pi k, \text{ for any integer $k$}  \\
x = 2 \pi k \text{ or }& x = \pi + 2 \pi k \\
    \text{ or }& x = \pm \pi/2 + 2 \pi k, \text{ for any integer $k$}
\end{align}
Second, we can allow ourselves to alter the domain, but must then check for solutions in the set we removed.  To reduce writing, we reuse the particular solution sets found above.
\begin{align}
\sin x \cos x &= \sin x  \\
\cos x &= 1 &&\text{ -- Check } \sin x = 0 \text{ in } \sin x \cos x = \sin x \text{.} \\
x &= \pm \pi/2 + 2 \pi k, \text{ for any integer $k$}
\end{align}
and we check
\begin{align}
\sin x &= 0  \\
x &= 2 \pi k \text{ or } x = \pi + 2 \pi k, \text{ for any integer $k$}.
\end{align}\begin{align}
x &= 2 \pi k: & \sin (2 \pi k) \cos (2 \pi k) &\overset{?}{=} \sin (2 \pi k)  \\
    && 0 \cdot 1 &\overset{\checkmark}{=} 0  \text{ and }\\
x &= \pi + 2 \pi k: & \sin (\pi + 2 \pi k) \cos (\pi + 2 \pi k) &\overset{?}{=} \sin (\pi + 2 \pi k)  \\
    && 0 \cdot -1 &\overset{\checkmark}{=} 0  \text{.}
\end{align}
So the solution set is $x = \pm \pi/2 + 2 \pi k$, $x = 2 \pi k$, or $x = \pi + 2 \pi k$, for any integer $k$.  Notice that this second method has significantly more cognitive load for both the writer and the reader -- goals are stored for later use.  I usually recommend the first method because of this.
