Q. (cosx * secx) / cotx = tanx

A. cosx secx = tanx cotx
cosx (1/cosx) = tanx (1/tanx)
1 = 1

He was marked wrong for the following reason:
"You cannot multiply both sides by anything in an identity. My instructions have been very clear that you with one side only and make it look like the other. If you do anything to both sides, you are assuming that the statement you are starting with is true. We do not know that this is true until we prove it, which is in verifying the identity."

We were discussing this afterwards, and he argued that:
"Yes, I may be working backward, but if the identity is true, I can apply algebraic manipulation to it and it will remain an identity (equality). If it is not, in fact, an identity, whatever algebraic manipulation I do, assuming it is done correctly, will not make it'show as an equality.
Therefore, if, after I manipulate it, I can show an equality, then the identity must be true."

It feels wrong to me, but I can't articulate why. If an equation is not known to be true, say, x=x, then we can't *assume* that algebraic properties hold true and that 2x=2x. However, his logic seems clear also, and I can't come up with an equation that proves him wrong.

Anyone have any input here?

1 = 2, since on multiplying both sides by 0 we get an identity

7 = -7, since on squaring both sides we get an identity.

Those are, of course, absurd examples. But more subtle things can happen and trip you up. The sort of argument your friend used is fine if each of the functions applied to both sides is injective (note: neither multiplication by 0 nor squaring is an injective function, as the above examples illustrate). So, if you want to do your working that way around, you have to think carefully about whether what you are doing is injective.

Well, your friend multiplied through by cot x. This is an injective operation provided that cot x != 0. And if cot x = 0, the question makes no sense. So your friend seems to have been OK (though the question should have really specified cot x != 0).

But the reason that an argument is normally written out the other way around is that you don't have to mess about checking whether things are injective.

I hope that helped.

Thankyou

sfwc
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The very first question to ask yourself is, What are the values of x for which the above makes sense?

The left-hand side is defined for those x, for which sec(x) and cot(x) are defined and for which cot(x) is not zero. It is not defined for p/2 + kp and kp, where k is an integer. The right hand side, on the other hand, is defined for all real x other than p/2 + kp. Therefore, the left- and the right-hand sides of Q are defined for different sets of x! Without some elaboration the two expressions can't be said to be equal. Strictly speaking, one can't say that the identity Q holds without, say, explaining why sin(x)/sin(x) = 1 for x = 0.

I have a feeling that somehow your question assumes that this issue should be skimmed. Let's continue under this assumption.

>
>A. cosx secx = tanx cotx
>cosx (1/cosx) = tanx (1/tanx)
>1 = 1
>
>He was marked wrong for the following reason:
>"You cannot multiply both sides by anything in an identity.

As a matter of course, this assertion does not make sense. It's rather the other way around: both sides of an identity can be multiplied by anything without disturbing the identity.

If 5 = 5, then 5×7 = 5×7. If a = 5, then a×7 = 5×7. There is no question about that.

>My instructions have been very clear that you start with one side
>only and make it look like the other.

This is a more or less standard approach to proving identities. But it's in no way the only one.

It is totally legitimate to do what your friend did, with one caveat: after getting 1 = 1, he had to retrace his steps and verify that on every step an identitiy has been replaced with an equivalent identity, or establish the conditions under which the equivalency holds. So

1 = 1

Does it follow that

cosx (1/cosx) = tanx (1/tanx)?

It does, provided we have removed the x's mentioned above from consideration. How? For all x, except the above,

cos(x)/cos(x) = 1 and also 1 = tan(x)/tan(x)

therefore,

cos(x)/cos(x) = tan(x)/tan(x),

or, which is the same,

cos(x)×sec(x) = tan(x)×cot(x).

Since x's, where cot(x) = 0 have been excluded from the consideration anyway, we may divide both sides of this identity by cot(x) to get

cos(x)×sec(x) / cot(x) = tan(x).

Q.E.D.

>We were discussing this afterwards, and he argued that:
>"Yes, I may be working backward, but if the identity is
>true, I can apply algebraic manipulation to it and it will
>remain an identity (equality). If it is not, in fact, an
>identity, whatever algebraic manipulation I do, assuming it
>is done correctly, will not make it show as an equality.
>Therefore, if, after I manipulate it, I can show an
>equality, then the identity must be true."

This may or may not be correct. As a rule, falsity implies anything. In particular, starting with a falsity it is possible to derive a correct result. For example,

since

a = b and c = d implies ac = bd,

we may algebraicly derive 1 = 1 from 1 = -1, which would not make the latter correct.