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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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 quic$k$ 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@DMan Yes I was. I started out from LHS while he started out from RHS. :) – 2012-05-01