For example:
$\color{red}{\text{Show that}}$$$\color{red}{\frac{4\cos(2x)}{1+\cos(2x)}=4-2\sec^2(x)}$$
En la escuela secundaria, mi profesor de matemáticas me dijo
Para probar la igualdad de la ecuación; se comienza en un lado y manipular algebraicamente hasta que sea igual para el otro lado.
Así que a partir de la LHS: $$\frac{4\cos(2x)}{1+\cos(2x)}=\frac{4(2\cos^2(x)-1)}{2\cos^2(x)}=\frac{2(2\cos^2(x)-1)}{\cos^2(x)}=\frac{4\cos^2(x)-2}{\cos^2(x)}=4-2\sec^2(x)$$ $\gran\fbox{}$
En la Universidad, mi Matemáticas Análisis de la maestra me dice
Para demostrar que un enunciado es verdadero, usted debe no usar lo que usted está tratando de demostrar.
Así que, usando el mismo ejemplo de antes:
LHS = $$\frac{4\cos(2x)}{1+\cos(2x)}=\frac{4(2\cos^2(x)-1)}{2\cos^2(x)}=\frac{2(2\cos^2(x)-1)}{\cos^2(x)}=\frac{2\Big(2\cos^2(x)-\left[\sin^2(x)+\cos^2(x)\right]\Big)}{\cos^2(x)}=\frac{2(\cos^2(x)-\sin^2(x))}{\cos^2(x)}=\bbox[yellow]{2-2\tan^2(x)}$$
RHS =$$4-2\sec^2(x)=4-2(1+\tan^2(x))=\bbox[yellow]{2-2\tan^2(x)}$$
So I have shown that the two sides of the equality in $\color{red}{\rm{rojo}}$ are equal to the same highlighted expression. But is this a sufficient proof?
Since I used both sides of the equality (which is effectively; using what I was trying to prove
) to show that $$\color{red}{\frac{4\cos(2x)}{1+\cos(2x)}=4-2\sec^2(x)}$$
One of the reasons why I am asking this question is because I have a bounty question which is suffering from the exact same issue that this post is about.
EDIT:
Comments and answers below seem to indicate that you can use both sides to prove equality. So does this mean that my high school maths teacher was wrong?
$$\bbox[#AFF]{\text{Suppose we have an identity instead of an equality:}}$$ $$\bbox[#AFF]{\text{Is it plausible to manipulate both sides of an identity to prove the identity holds?}}$$
Gracias.