<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] One of the related questions here, seems to be more suitable for method $2$:

$$\frac{1+\sin x}{\cos x} = \frac{1+\sin x+\cos x}{1-\sin x+\cos x} (*)$$ 

Let $a =\sin x$, $b=\cos x$, then:
$(*)\Leftrightarrow \frac{1+a}{b}=\frac{1+a+b}{1-a+b}$
$\Leftrightarrow (1+a)(1-a+b) = b(1+a+b)$
$\Leftrightarrow 1-a^2+b(a+1) = b^2 + b(a+1)$
$\Leftrightarrow a^2+b^2=1$, which is provable.
For method $1$, I think you can use it to prove this simple expression:

$$a^2-b^2=(a-b)(a+b)$$

For this problem, using method $1$ or $2$ is actually the same, no method is better than the other.
For some problems when either side is hard to be made "equal" to the other side, or if you are stuck with this method because of that annoying "$\Rightarrow$", using substraction is recommended.
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