Proving an identity using reciprocal, quotient, or Pythagorean identities.
Sebastian Wright 
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.
$\endgroup$3 Answers
$\begingroup$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 $(1-\cos\theta)(1+\cos\theta)$?
$\endgroup$ 5 $\begingroup$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.
$\endgroup$ 1 $\begingroup$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, $\dfrac 1 {1+\sqrt 2}$
