<s>[INST] Validity of proving identities by showing LHS-RHS =0 or using reversible steps? When proving $\mathrm{LHS}=\mathrm{RHS}$, the most common way of doing it is by manipulating it in such a way to show that $\mathrm{LHS}$ equals to some expression which equals to $\mathrm{RHS}$. But what about these methods:
Method 1:
Showing that $\mathrm{LHS}-\mathrm{RHS}=0$
For instance, if we are required to prove
$x^2\cos^2(x)+\sin^2(x) = x^2 - x^2\sin^2(x) + 1 - \cos^2(x)$
We instead show that $\mathrm{LHS}-\mathrm{RHS}=0$ as follows:
$\mathrm{LHS} - \mathrm{RHS} = x^2\cos^2(x)+\sin^2(x) - x^2 + x^2\sin^2(x) - 1 + \cos^2(x)$
$ = x^2(\cos^2(x) + \sin^2(x)) - x^2 + (\sin^2(x) + \cos^2(x))  - 1$
$= x^2 - x^2 + 1 - 1 = 0$
Method 2: Showing that $\mathrm{LHS}=\mathrm{RHS}$ is equivalent with another equation which is true (taking care that we can always reverse the steps and showing it by putting $\iff$):
$x^2\cos^2(x)+\sin^2(x) = x^2 - x^2\sin^2(x) + 1 - \cos^2(x)$
$ \iff  x^2\cos^2(x)+x^2\sin^2(x) - x^2 = 1 - \sin^2(x) - \cos^2(x)$
$ \iff x^2 - x^2 = 1 - 1$
$ \iff 0 = 0$
Are these two methods of proving valid? Are there any cases where we can make fallacious argument by using these methods? Are any one of them better than the other?
 [/INST] This question works both from LHS $\rightarrow$ RHS and RHS $\rightarrow$ LHS to prove it.
</s>