Proving an identity using reciprocal, quotient, or Pythagorean identities. 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.
 A: 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 $(1-\cos\theta)(1+\cos\theta)$?
A: 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.
A: 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, $\dfrac 1 {1+\sqrt 2}$
