trig identity: which is equivalent to cot(x) / sin(2x) ?

[hugs4trees]

Can someone please help me with this question and explain how they did it.

Problem #1: Which of the following is equal to \(\displaystyle \, \dfrac{\cot(x)}{\sin(2x)}\)?

. . .(i) \(\displaystyle \, \dfrac{1}{1\, -\, \cos(2x)}\). . .(ii) \(\displaystyle \, \left(\dfrac{1}{2}\right)\, \left(1\, +\, \tan^2(x)\right)\). . .(iii) \(\displaystyle \, \dfrac{1}{2\, \left(1\, -\, \cos^2(x)\right)}\)

(A) (iii) only . . . (B) (i) and (ii) only . . . (C) all of them . . . (D) (ii) and (iii) only
(E) (i) and (iii) only . . . (G) (ii) only . . . (H) none of them

I always get stuck with identities.
Any tips on how to make identities easier would be appreciated!!

 

[ksdhart]

Unfortunately, there aren't really too many tips I can give for trig identities. They frustrated me at first, too. The best method is just to practice, practice, practice. If you haven't already you're definitely want to memorize the core (sometimes called "fundamental") identities. These include, double and half angle identities, sin²(x) + cos²(x) = 1, tan(x) = sin(x)/cos(x), etc. Also note that even though the identities specifically say x, they also work for 2x, 3x, (1/7)x, ...

Now for this particular problem, my strategy would be to start by simplifying the given form a bit. If possible, I like to rewrite the problem such that only sines and cosines remain. When you do that, what do you end up with? After that, work with the given answers and try to work your way toward the original form. Like for answer i, you have an identity for cos(2x), so what does 1 - [that expression] equal? And the reciprocal of that? Then apply the same process for answers ii and iii.​