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So I have an establishing the identity problem that I'm trying to figure out.

$\frac{\sin\theta + \cos\theta} {\sec\theta + \csc\theta} = \cos(\theta) \sin(\theta) $

I'm being told that the first step is to multiply the right side of the identity by

$\frac{\sec\theta + \csc\theta} {\sec\theta + csc\theta} $

because "since this fraction is equivalent to 1, the product is unchanged."

Why are we doing this?

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  • $\begingroup$ Notice the edit $\endgroup$
    – user3370603
    Apr 30, 2015 at 2:46
  • $\begingroup$ Because it works quickly. In some ways, it is not a good way to proceed. More natural for a student is to work with the LHS, expressing it as $\frac{\sin\theta+\cos\theta}{\frac{1}{\cos\theta}+\frac{1}{\sin\theta}}$. Then a little algebra takes care of things. Uglier, for sure, but less "magical." $\endgroup$
    – André Nicolas
    Apr 30, 2015 at 2:52
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    $\begingroup$ Uglier is in the eye of the beholder. There is a simple "reason" for the identity, that $\sin\theta\cos\theta(\sec\theta+\csc\theta)=\sin\theta+\cos\theta$. But as I mention in a comment below, that particular step is "not allowed." $\endgroup$
    – André Nicolas
    Apr 30, 2015 at 3:02
  • $\begingroup$ I keep thinking this question has been answered before on MSE, but I only find related questions, not the same one. The identity $\frac{\sec\theta+\csc\theta}{\tan\theta + \cot\theta} = \sin\theta + \cos\theta$ (math.stackexchange.com/q/171675/139123) is curiously related to this one by the identity $\frac{1}{\tan\theta + \cot\theta} = \sin\theta\cos\theta$. $\endgroup$
    – David K
    Apr 30, 2015 at 3:39

2 Answers 2

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If you multiply the right side by $\frac{\sec\theta + \csc\theta}{sec\theta+\csc\theta}$ then you will get $\cos\theta \sin\theta(\frac{\sec\theta + \csc\theta}{\sec\theta+\csc\theta})$. But $\sec\theta = \frac{1}{\cos\theta}$ and $\csc\theta = \frac{1}{\sin\theta}$ so the top of the fraction will simplify to $\sin\theta + \cos\theta$. This is the easiest way to solve the problem in my opinion.

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  • $\begingroup$ I wouldn't necessarily think of it as a first step either, but if I were solving the problem I might start with the right hand side as it is simpler to work with. At that point, multiplying by $\frac{\sec\theta + \csc\theta}{\sec\theta + \csc\theta}$ is the logical next step, and from there the problem just works itself out. $\endgroup$
    – Danny
    Apr 30, 2015 at 3:02
  • $\begingroup$ This approach makes perfect sense. We would like to multiply the RHS by a fraction whose value is $1$ and whose denominator is the same as the denominator of the LHS. $\endgroup$
    – John Joy
    Apr 30, 2015 at 12:39
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I don't know why, but it seems much harder that way. This is a better way:

$$\frac{\sin \theta + \cos\theta}{\sec \theta + \csc\theta} = \frac{\sin \theta + \cos\theta}{\frac{1}{\cos\theta}+\frac{1}{\sin\theta}} = \frac{\sin \theta + \cos\theta}{\frac{\sin \theta + \cos\theta}{\cos\theta \sin\theta}} = \sin\theta \cos\theta$$. This way is much faster and cleaner.

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  • $\begingroup$ Yes, that is how I would have answered it. I don't know where the pulled that from and why that was recommended as the first step. $\endgroup$
    – user3370603
    Apr 30, 2015 at 2:53
  • $\begingroup$ @user3370603 are you sure you read the directions right? Cause there might be special instructions or something. $\endgroup$
    – SalmonKiller
    Apr 30, 2015 at 2:53
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    $\begingroup$ In HS trig identities, there is a convention that one does not bother about the issue of reversibility, nor does one bother with where the functions involved are defined. As is often the case, the identity is not really an identity, for the right side is defined everywhere and the left side is not. $\endgroup$
    – André Nicolas
    Apr 30, 2015 at 3:07
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    $\begingroup$ If $\sec\theta+\csc\theta \neq 0,$ then $\frac{a}{\sec\theta+\csc\theta} = b$ is equivalent to $a = (\sec\theta+\csc\theta)b,$ that is, the multiplication on both sides is reversible. If $\sec\theta+\csc\theta = 0$ then the LHS of the "identity" is undefined, so no proof will work in that case. $\endgroup$
    – David K
    Apr 30, 2015 at 3:33
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    $\begingroup$ @SalmonKiller That is precisely the point. We have to assume $\sec\theta + \csc\theta \neq 0$, so multiplication by that quantity is reversible (that is, division by that quantity is permitted). $\endgroup$
    – David K
    Apr 30, 2015 at 3:48

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