# Proving identities using Pythagorean, Reciprocal and Quotient

Back again, with one last identity that I cannot solve:

$$\frac{\cos \theta}{\csc \theta - 2 \sin \theta} = \frac{\tan\theta}{1-\tan^2\theta}$$.

The simplest I could get the left side to, if at all simpler, is $$\frac{\cos\theta}{\csc^2\theta-2}$$

As for the right side, it has me stumped, especially since the denominator is so close to a identity yet so far away. I've tried rationalizing the denominators (both sides) to little success. From my last question, where multiplying by a '1' worked, I didn't see the association here.

Thanks!

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 You could try writing each side in terms of sin and cos. – Mark Bennet May 1 '12 at 6:09 When you wrote $\dfrac{\cos\theta}{\csc^2\theta-2}$ you probably meant $\dfrac{\cot\theta}{\csc^2\theta-2}$ – Henry May 1 '12 at 6:41

You'll want to rewrite $\csc\theta=\frac{1}{\sin\theta}$, to make things easier. Then

$\cfrac{\cos\theta}{\csc\theta-2\sin\theta}$

$\cfrac{\cos\theta\sin\theta}{1-2\sin^2\theta}$

$\cfrac{\cos\theta\sin\theta}{\cos^2\theta-\sin^2\theta}$

Dividing top and bottom by $\cos^2\theta$, you'll get exactly the right hand side.

To make it even simpler, you'll generally want to rewrite things in terms of sines and cosines--this won't always be the easiest approach, but at least if you're limited to reciprocal, quotient, and Pythagorean identities, you should get started that way. In this case, the left hand side becomes

$\cfrac{\cos\theta}{\frac{1}{\sin\theta}-2\sin\theta}=\cfrac{\cos\theta\sin\theta}{1-2\sin^2\theta}$

and the right hand side becomes

$\cfrac{\frac{\sin\theta}{\cos\theta}}{1-\frac{\sin^2\theta}{\cos^2\theta}}=\cfrac{\cos\theta\sin\theta}{\cos^2\theta-\sin^2\theta}$.

The only thing different now is the denominator. Thus, all that remains to do is figure out how to show that $1-2\sin^2\theta=\cos^2\theta-\sin^2\theta$, using only the permitted types of identity. In this case, gathering like terms may be useful (bring all your trigonometric terms to the same side of the equation to see why it is true). Once you see that, you should be able to proceed from one side to the other, step-by-step.

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If you use \dfrac instead of \frac, you'll get visibly bigger expressions. – The Chaz 2.0 May 1 '12 at 6:11
Well, using double dollar signs for displayed equations instead of single dollar signs for inline equations would be a start. – joriki May 1 '12 at 6:12
True enough. I think \cfrac also does the trick. – Cameron Buie May 1 '12 at 6:13
Could you explain how the denominator of step 2 transforms to the squared cos subtracted by sin? – DMan May 1 '12 at 6:26
Use the Pythagorean identity on the 1 in the denominator from step 2. – Cameron Buie May 1 '12 at 7:01

(Hint)$$\dfrac{\cos\theta\dfrac{\sin \theta}{\cos^2\theta}}{(\csc \theta - 2 \sin \theta )\dfrac{\sin \theta }{\cos ^2 \theta}} = \dfrac{\tan \theta}{(1 - 2\sin^2\theta)\dfrac{1}{\cos^2 \theta}} = \dfrac{\tan \theta}{(\cos^2 \theta - \sin^2\theta)\dfrac{1}{\cos^2 \theta}}$$

Added (inspired by joriki's observation of the double tangent)

Since the right hand side is $$\dfrac{1}{2}\tan (2\theta) = \dfrac{\sin(2 \theta)}{2\cos(2\theta)} = \dfrac{\sin\theta \cos \theta}{1 - 2\sin^2 \theta}$$ we divide numerator and denominator by $\sin \theta$ to get the left hand side.

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HINT:

\begin{align*} \frac{\tan\theta}{1-\tan^2\theta}&=\frac{\sin\theta\cos\theta}{\cos^2\theta-\sin^2\theta} \end{align*}

\begin{align*} \frac{\cos\theta}{\csc\theta-2\sin\theta}&=\frac{\sin\theta\cos\theta}{1-2\sin^2\theta} \end{align*}

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Write everything in terms of sines and cosines, then multiply through by the inner denominators so that you have a single fraction on both sides; those two fractions should turn out to be the same. By the way, note that what they turn out to be is $\frac12\tan(2\theta)$.

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 It's true that they both end up equal to $\frac{1}{2}\tan(2\theta)$, but given that only reciprocal, quotient, and Pythagorean identities are allowed as tools, it isn't necessarily useful. They may not even have covered any double angle identities, yet. Still, this approach is ideal. – Cameron Buie May 1 '12 at 8:28 @CameronBuie: Yes, that was just a side remark. – joriki May 1 '12 at 10:03