When tackling trig identities, which side is the best to start from? I'm currently in Y13. I wanted to know which side of a trig identity is typically best to work from (since I know you have to work from one side to the other). Here's an example of an identity I just proved:
$$\tan(\frac{\pi}{4}-\frac{x}{2}) = \sec(x)-\tan(x)$$
I figured out the proof after a false start by working from both ends, then putting the steps together. But working from the RHS is a lot more straightforward than the LHS; the latter requires you to deduce that:
$$\frac{1-\tan(\frac{x}{2})}{1+\tan(\frac{x}{2})} = \frac{(1-\tan(\frac{x}{2}))^2}{1-\tan^2(\frac{x}{2})}$$
which is fine, but not immediately obvious. Even if you continue working in this direction, the next steps are not very reassuring either. I was trying to multiply by $\frac{1+\tan(\frac{x}{2})}{1+\tan(\frac{x}{2})}$, which wasn't very helpful.
So I was wondering: are there any giveaway signs that one side of the identity will be easier to work from than the other? In hindsight, for this one, the double angles on the RHS were much easier to manipulate than the angle addition on the LHS. Is there a sort of hierarchy anyone has learnt from their experience?
 A: Welcome to Math Stack Exchange! Thanks for the nicely worded and well-formatted question. In general, I'd say that it is better to start from the side that looks "weirder." I would argue that the side that looks more "natural" (in your example, $\sec(x)-\tan(x)$ looks more "natural") is harder to work with. Sometimes the weirder side has an obvious first step (like multiplying by a conjugate or using a common trig identity). Starting from the more natural looking side might involve lots of creativity (like adding and subtracting an obscure term). This is by no means a set-in-stone rule, and should only be a guideline. As you discovered in your example, sometimes it is best to work on both sides to reach a common answer.
The beauty to math is that there are often multiple correct approaches, and each one is valuable in its own regard.
A: It is rather simple thinking of the addition formulae and the half-angle formulæ:
Set $t=\tan \frac x2$. By the addition formula for the tangent, you have, for $t\ne -1$,
$$\tan\Bigl(\frac{\pi}{4}-\frac{x}{2}\Bigr) = \frac{1-t}{1+t} \qquad (\tan \frac\pi 4=1!)$$
On the other hand, if $t\ne\pm 1$,
$$\sec(x)-\tan(x)=\frac{1+t^2}{1-t^2}-\frac{2t}{1-t^2}=\frac{(1-t)^2}{1-t^2}=\frac{1-t}{1+t}$$
A: Changing sign of x on both sides produces reciprocals on each side. So here tackling both sides at the same time is helpful once the identities are recognized.
$$\tan(\frac{\pi}{4}-\frac{x}{2}) = \sec(x)-\tan(x)$$
$$\tan(\frac{\pi}{4}+\frac{x}{2}) = \sec(x)+\tan(x)$$
Make a substitution such that z is brought in using symmetry wrt argument $\pi/2$ on  LHS.
Product of LHS is $ \tan z\cdot \cot z =1 $ and right hand side product is also $ \sec^2(x)-\tan^2(x)=1$
However there is no hard and fast rule. Better to tackle all that appears to accept simplification from known identities.
A: For a question like this where one angle is double another, start by writing both sides as functions of $t:=\tan\frac{x}{2}$ in separate calculations.
A: For this proof, it may not matter that much. Continue working from the LHS,
$$ \frac{(1-\tan\frac{x}{2})^2}{1-\tan^2\frac{x}{2}}$$
$$= \frac{(\cos\frac x2-\sin\frac{x}{2})^2}{\cos^2\frac x2-\sin^2\frac{x}{2}}
=\frac{1-2\sin\frac x2\cos\frac x2}{\cos x} 
= \frac{ 1-\sin x}{\cos x} = \sec x -\tan x$$
