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.


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 $(1-\cos\theta)(1+\cos\theta)$?

  • $\begingroup$ It is equal to $1-\cos\theta$ or $\sin^2\theta$... still deciding what to do with it though. $\endgroup$ – DMan May 1 '12 at 5:05
  • $\begingroup$ And 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? $\endgroup$ – Arturo Magidin May 1 '12 at 5:08
  • $\begingroup$ The LHS? I'm seriously at a lost here; another hint would be great. $\endgroup$ – DMan May 1 '12 at 5:12
  • $\begingroup$ Well, 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? $\endgroup$ – Arturo Magidin May 1 '12 at 5:14
  • $\begingroup$ Oh! Convert $\csc$ and $\sec$ to the reciprocals, get $\frac{\sin^2 \theta}{\cos \theta \sin \theta}$, simplify and get $\tan\$! Thanks! $\endgroup$ – DMan May 1 '12 at 5:17

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.

  • $\begingroup$ That's funny - the quick bit I saw isn't quite the same as the bit the other two saw $\endgroup$ – davidlowryduda May 1 '12 at 5:05


Multiply the numerator and denominator of the LHS by $(1-\cos \theta)$ to see what you get.


  • 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, $\dfrac 1 {1+\sqrt 2}$

  • $\begingroup$ $\frac{\tan \theta}{\sin^2 \theta}$? Is that correct? $\endgroup$ – DMan May 1 '12 at 5:09
  • $\begingroup$ @DMan You're missing a $(1-\cos \theta)$ in the numerator, and of course, the rest is fine. $\endgroup$ – user21436 May 1 '12 at 5:11
  • $\begingroup$ Oops, you are right. I understand this step as rationalizing the denominator, but I don't know how to proceed at all. $\endgroup$ – DMan May 1 '12 at 5:13
  • $\begingroup$ @DMan Now, what is $\tan \theta$? Can you write it in terms of $\sin \theta$ and $\cos \theta$? $\endgroup$ – user21436 May 1 '12 at 5:18
  • $\begingroup$ Oh I see, you are doing it a bit differently from Arturo which has mixed me up a bit. Thanks for this as well! $\endgroup$ – DMan May 1 '12 at 5:25

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