I've been trying to prove this for a while, to no avail. I am only allowed to use pythagorean, quotient, and reciprocal identities: $$\frac{\tan \theta}{1 + \cos \theta} = \sec \theta \csc\theta(1-\cos \theta)$$ I've tried converting $\tan \theta$ to $\frac{\sin \theta}{\cos \theta}$ and such, but could only get it simplified down to $\frac{\tan \theta}{\cos \theta + 1}$ on the LHS. As for the right, I tried a common denominator and ended up with $$\frac{1-\cos \theta}{\cos \theta \sin \theta}$$ but couldn't see how I could go further from there.
Proving an identity using reciprocal, quotient, or Pythagorean identities.
3 Answers
Hint: $$\sec\theta\csc\theta(1-\cos\theta) = \sec\theta\csc\theta(1-\cos\theta)\frac{1+\cos\theta}{1+\cos\theta}.$$ How much is $(1-\cos\theta)(1+\cos\theta)$?
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0It is equal to $1-\cos\theta$ or $\sin^2\theta$... still deciding what to do with it though. – 2012-05-01
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0And so you would have $$\frac{\sec\theta\csc\theta\sin^2\theta}{1+\cos\theta}.$$ If only you could show that $\sec\theta\csc\theta\sin^2\theta$ is equal to... what was it we wanted to get? – 2012-05-01
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0The LHS? I'm seriously at a lost here; another hint would be great. – 2012-05-01
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0Well, if the numerator here, which is $\sec\theta\csc\theta\sin^2\theta$ were equal tot he numerator on the left hand side, which is $\tan\theta$, then we'd be done, no? We started with the right hand side, multiplied by $1$, and have done nothing but simple manipulation and used the pythagorean identity (to go from $1-\cos^2\theta$ to $\sin^2\theta$), and we already have the denominator right. So... *is* the numerator we got equal to the numerator on the left hand side? – 2012-05-01
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0Oh! Convert $\csc$ and $\sec$ to the reciprocals, get $\frac{\sin^2 \theta}{\cos \theta \sin \theta}$, simplify and get $\tan\$! Thanks! – 2012-05-01
I happen to have a very large algebra stick. Teddy once told me to write quickly a carry a big algebra stick - this advice got me through many a test.
$\displaystyle \frac{\sin \theta}{\cos \theta (1 + \cos \theta)} = \frac{1 - \cos \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta} - \frac{1}{\sin \theta}$
Throw the $\sin$ term to the other side, and we'll check for equality.
$\displaystyle \frac{\sin \theta}{\cos \theta + \cos^2 \theta} + \frac{1}{\sin \theta} = \frac{\sin^2 \theta + \cos \theta + \cos ^2 \theta}{\sin \theta \cos \theta (1 + \cos \theta)} = \frac{1 + \cos \theta}{\sin \theta \cos \theta (1 + \cos \theta)} = \frac{1}{\cos \theta \sin \theta}$
Which is what we wanted. And everything is reversible.
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0That's funny - the quick bit I saw isn't quite the same as the bit the other two saw – 2012-05-01
Hint
Multiply the numerator and denominator of the LHS by $(1-\cos \theta)$ to see what you get.
P.S.
This step is not quite magical as you see a $(1-\cos \theta)$ on the RHS. But, don't worry, you'd start thinking along these lines with practice.
And, you may want to compare this with the method you used the rationalise the denominator of, say, $\dfrac 1 {1+\sqrt 2}$
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0$\frac{\tan \theta}{\sin^2 \theta}$? Is that correct? – 2012-05-01
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0@DMan You're missing a $(1-\cos \theta)$ in the numerator, and of course, the rest is fine. – 2012-05-01
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0Oops, you are right. I understand this step as rationalizing the denominator, but I don't know how to proceed at all. – 2012-05-01
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0@DMan Now, what is $\tan \theta$? Can you write it in terms of $\sin \theta$ and $\cos \theta$? – 2012-05-01
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0Oh I see, you are doing it a bit differently from Arturo which has mixed me up a bit. Thanks for this as well! – 2012-05-01
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0@DMan Yes I was. I started out from LHS while he started out from RHS. :) – 2012-05-01