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## Main Question or Discussion Point

Sorry for the dumb question by my high school teachers never explained this well and I'm an engineering student which means I don't do many proofs...

Let us say that I have to prove that the following is an identity:

$$\frac { cosx-siny }{ cosy-sinx } =\frac { cosy+sinx }{ cosx+siny }$$

My high school teachers said that I would have to set the left side of the equation to "LHS" and the right side of the equation to "RHS". Then manipulate the variables "LHS" and "RHS" until LHS=RHS. The rationale given by my high school teachers for the reason why I can't just work on it in the original form is because we don't know yet if LHS=RHS.

However, what if we make a proposition or we suppose that:

$$\frac { cosx-siny }{ cosy-sinx } =\frac { cosy+sinx }{ cosx+siny }$$

is true.

Let's say it is reduced to something like: 0=1. Can we say that by proof of contradiction, the original equation is not an identity?

Let's say it is reduced to something like: 1=1. Can we say that because the original equation is reduced to an identity, the original equation is an identity?

Is my choice of words correct?

So to recap, the below would be accepted as proofs by a university professor, right?

Question: Prove the following is an identity: ##\frac { cosx-siny }{ cosy-sinx } =\frac { cosy+sinx }{ cosx+siny } ##

Answer Case A: Suppose that ##\frac { cosx-siny }{ cosy-sinx } =\frac { cosy+sinx }{ cosx+siny }## is an identity.

##\frac { cosx-siny }{ cosy-sinx } =\frac { cosy+sinx }{ cosx+siny }##

##{ sin }^{ 2 }x+{ cos }^{ 2 }x={ sin }^{ 2 }y+{ cos }^{ 2 }y##

1=1

Therefore, because the original equation is reduced to an identity, ##\frac { cosx-siny }{ cosy-sinx } =\frac { cosy+sinx }{ cosx+siny }## is an identity.

Answer Case B: Suppose that ##\frac { cosx-siny }{ cosy-sinx } =\frac { cosy+sinx }{ cosx+siny }## is an identity.

##\frac { cosx-siny }{ cosy-sinx } =\frac { cosy+sinx }{ cosx+siny }## ... (in this case it would reduce into 1=1, but let's just pretend that doens't happen) 0=1 Therefore, because our original assumption led to a contradiction, our assumption is incorrect, and by proof by contradiction, ##\frac { cosx-siny }{ cosy-sinx } =\frac { cosy+sinx }{ cosx+siny }## is not an identity.

Let us say that I have to prove that the following is an identity:

$$\frac { cosx-siny }{ cosy-sinx } =\frac { cosy+sinx }{ cosx+siny }$$

My high school teachers said that I would have to set the left side of the equation to "LHS" and the right side of the equation to "RHS". Then manipulate the variables "LHS" and "RHS" until LHS=RHS. The rationale given by my high school teachers for the reason why I can't just work on it in the original form is because we don't know yet if LHS=RHS.

However, what if we make a proposition or we suppose that:

$$\frac { cosx-siny }{ cosy-sinx } =\frac { cosy+sinx }{ cosx+siny }$$

is true.

Let's say it is reduced to something like: 0=1. Can we say that by proof of contradiction, the original equation is not an identity?

Let's say it is reduced to something like: 1=1. Can we say that because the original equation is reduced to an identity, the original equation is an identity?

Is my choice of words correct?

So to recap, the below would be accepted as proofs by a university professor, right?

Question: Prove the following is an identity: ##\frac { cosx-siny }{ cosy-sinx } =\frac { cosy+sinx }{ cosx+siny } ##

Answer Case A: Suppose that ##\frac { cosx-siny }{ cosy-sinx } =\frac { cosy+sinx }{ cosx+siny }## is an identity.

##\frac { cosx-siny }{ cosy-sinx } =\frac { cosy+sinx }{ cosx+siny }##

##{ sin }^{ 2 }x+{ cos }^{ 2 }x={ sin }^{ 2 }y+{ cos }^{ 2 }y##

1=1

Therefore, because the original equation is reduced to an identity, ##\frac { cosx-siny }{ cosy-sinx } =\frac { cosy+sinx }{ cosx+siny }## is an identity.

Answer Case B: Suppose that ##\frac { cosx-siny }{ cosy-sinx } =\frac { cosy+sinx }{ cosx+siny }## is an identity.

##\frac { cosx-siny }{ cosy-sinx } =\frac { cosy+sinx }{ cosx+siny }## ... (in this case it would reduce into 1=1, but let's just pretend that doens't happen) 0=1 Therefore, because our original assumption led to a contradiction, our assumption is incorrect, and by proof by contradiction, ##\frac { cosx-siny }{ cosy-sinx } =\frac { cosy+sinx }{ cosx+siny }## is not an identity.

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