A trigonometry text like Sullivan's Algebra & Trigonometry often has a prohibition like this (Sec. 7.3):

WARNING: Be careful not to handle identities to be established as if they were conditional equations. You cannot establish an identity by such methods as adding the same expression to each side and obtaining a true statement. This practice is not allowed, because the original statement is precisely the one that you are trying to establish. You do not know until it has been established that is, in fact, true.

I'll look at the first such example in Sullivan. Establish the identity: $\csc \theta \cdot \tan \theta = \sec \theta$.

The given solution is: $\csc \theta \cdot \tan \theta = \frac{1}{\sin \theta} \cdot \frac{\sin \theta}{\cos \theta} = \frac{1}{\cos \theta} = \sec \theta$.

But let's consider using the prohibited operations that generate equivalent equations. We may write: $\csc \theta \cdot \tan \theta = \sec \theta \iff \frac{1}{\sin \theta} \cdot \frac{\sin \theta}{\cos \theta} = \frac{1}{\cos \theta} \iff \sin \theta = \sin \theta$. [Multiplying both sides by $\sin \theta \cdot \cos \theta$ in the last step.] Now, personally, this does persuade me that the original equation is an identity, because it is equivalent to an obviously-true equation (excepting any values not in the domain of one side, which are identical to the official answer above).

If we want to be completely rigorous, then I think we could just reverse the sequence of statements above. Starting with the reflexive property of equality: $\sin \theta = \sin \theta \iff \frac{1}{\sin \theta} \cdot \frac{\sin \theta}{\cos \theta} = \frac{1}{\cos \theta} \iff \csc \theta \cdot \tan \theta = \sec \theta$.

The official book proof is slightly shorter, and it can be written down directly without the reversal, but it seems mostly like a stylistic point -- fighting with students to abandon their equivalent-equation techniques is a great burden, and right now I'm not entirely seeing the necessity for it. Does anyone permit this alternative as an acceptable proof in their trigonometry class? If not, what is the compelling reason for Sullivan's dire warning?

Edit: This question presumes that any applied operations do correctly generate equivalent equations (noting gaps in the domain as required). The question is whether Sullivan's "cannot" statement is strictly true.

A trigonometry text like Sullivan's Algebra & Trigonometry often has a prohibition like this (Sec. 7.3):

WARNING: Be careful not to handle identities to be established as if they were conditional equations. You cannot establish an identity by such methods as adding the same expression to each side and obtaining a true statement. This practice is not allowed, because the original statement is precisely the one that you are trying to establish. You do not know until it has been established that is, in fact, true.

I'll look at the first such example in Sullivan. Establish the identity: $\csc \theta \cdot \tan \theta = \sec \theta$.

The given solution is: $\csc \theta \cdot \tan \theta = \frac{1}{\sin \theta} \cdot \frac{\sin \theta}{\cos \theta} = \frac{1}{\cos \theta} = \sec \theta$.

But let's consider using the prohibited operations that generate equivalent equations. We may write: $\csc \theta \cdot \tan \theta = \sec \theta \iff \frac{1}{\sin \theta} \cdot \frac{\sin \theta}{\cos \theta} = \frac{1}{\cos \theta} \iff \sin \theta = \sin \theta$. [Multiplying both sides by $\sin \theta \cdot \cos \theta$ in the last step.] Now, personally, this does persuade me that the original equation is an identity, because it is equivalent to an obviously-true equation (excepting any values not in the domain of one side, which are identical to the official answer above).

If we want to be completely rigorous, then I think we could just reverse the sequence of statements above. Starting with the reflexive property of equality: $\sin \theta = \sin \theta \iff \frac{1}{\sin \theta} \cdot \frac{\sin \theta}{\cos \theta} = \frac{1}{\cos \theta} \iff \csc \theta \cdot \tan \theta = \sec \theta$.

The official book proof is slightly shorter, and it can be written down directly without the reversal, but it seems mostly like a stylistic point -- fighting with students to abandon their equivalent-equation techniques is a great burden, and right now I'm not entirely seeing the necessity for it. Does anyone permit this alternative as an acceptable proof in their trigonometry class? If not, what is the compelling reason for Sullivan's dire warning?

Edit: This question presumes that any applied operations do correctly generate equivalent equations (noting gaps in the domain as required). The question is whether Sullivan's "cannot establish an identity by such methods as adding" statement is strictly true.

A trigonometry text like Sullivan's Algebra & Trigonometry often has a prohibition like this (Sec. 7.3):

WARNING: Be careful not to handle identities to be established as if they were conditional equations. You cannot establish an identity by such methods as adding the same expression to each side and obtaining a true statement. This practice is not allowed, because the original statement is precisely the one that you are trying to establish. You do not know until it has been established that is, in fact, true.

I'll look at the first such example in Sullivan. Establish the identity: $\csc \theta \cdot \tan \theta = \sec \theta$.

The given solution is: $\csc \theta \cdot \tan \theta = \frac{1}{\sin \theta} \cdot \frac{\sin \theta}{\cos \theta} = \frac{1}{\cos \theta} = \sec \theta$.

But let's consider using the prohibited operations that generate equivalent equations. We may write: $\csc \theta \cdot \tan \theta = \sec \theta \iff \frac{1}{\sin \theta} \cdot \frac{\sin \theta}{\cos \theta} = \frac{1}{\cos \theta} \iff \sin \theta = \sin \theta$. [Multiplying both sides by $\sin \theta \cdot \cos \theta$ in the last step.] Now, personally, this does persuade me that the original equation is an identity, because it is equivalent to an obviously-true equation (excepting any values not in the domain of one side, which are identical to the official answer above).

If we want to be completely rigorous, then I think we could just reverse the sequence of statements above. Starting with the reflexive property of equality: $\sin \theta = \sin \theta \iff \frac{1}{\sin \theta} \cdot \frac{\sin \theta}{\cos \theta} = \frac{1}{\cos \theta} \iff \csc \theta \cdot \tan \theta = \sec \theta$.

The official book proof is slightly shorter, and it can be written down directly without the reversal, but it seems mostly like a stylistic point -- fighting with students to abandon their equivalent-equation techniques is a great burden, and right now I'm not entirely seeing the necessity for it. Does anyone permit this alternative as an acceptable proof in their trigonometry class? If not, what is the compelling reason for Sullivan's dire warning?

A trigonometry text like Sullivan's Algebra & Trigonometry often has a prohibition like this (Sec. 7.3):

WARNING: Be careful not to handle identities to be established as if they were conditional equations. You cannot establish an identity by such methods as adding the same expression to each side and obtaining a true statement. This practice is not allowed, because the original statement is precisely the one that you are trying to establish. You do not know until it has been established that is, in fact, true.

I'll look at the first such example in Sullivan. Establish the identity: $\csc \theta \cdot \tan \theta = \sec \theta$.

The given solution is: $\csc \theta \cdot \tan \theta = \frac{1}{\sin \theta} \cdot \frac{\sin \theta}{\cos \theta} = \frac{1}{\cos \theta} = \sec \theta$.

But let's consider using the prohibited operations that generate equivalent equations. We may write: $\csc \theta \cdot \tan \theta = \sec \theta \iff \frac{1}{\sin \theta} \cdot \frac{\sin \theta}{\cos \theta} = \frac{1}{\cos \theta} \iff \sin \theta = \sin \theta$. [Multiplying both sides by $\sin \theta \cdot \cos \theta$ in the last step.] Now, personally, this does persuade me that the original equation is an identity, because it is equivalent to an obviously-true equation (excepting any values not in the domain of one side, which are identical to the official answer above).

If we want to be completely rigorous, then I think we could just reverse the sequence of statements above. Starting with the reflexive property of equality: $\sin \theta = \sin \theta \iff \frac{1}{\sin \theta} \cdot \frac{\sin \theta}{\cos \theta} = \frac{1}{\cos \theta} \iff \csc \theta \cdot \tan \theta = \sec \theta$.

The official book proof is slightly shorter, and it can be written down directly without the reversal, but it seems mostly like a stylistic point -- fighting with students to abandon their equivalent-equation techniques is a great burden, and right now I'm not entirely seeing the necessity for it. Does anyone permit this alternative as an acceptable proof in their trigonometry class? If not, what is the compelling reason for Sullivan's dire warning?

Edit: This question presumes that any applied operations do correctly generate equivalent equations (noting gaps in the domain as required). The question is whether Sullivan's "cannot" statement is strictly true.