Another Verify the identity: $\sec^2 \frac{x}{2} = \frac{2}{1+\cos x}$ Another Verify the identity that I can't get:
$$\sec^2 \frac{x}{2} = \frac{2}{1+\cos x}$$
$$ = \frac{1 + \left(\frac{1}{\cos x}\right)}{2}$$
$$ = \frac{\cos x + 1}{2 \cos x}$$

Again, both ways are possible : 
Going backwards (which is simpler, coincidentally)
$$\frac{2}{1 + \cos x} = \frac{2}{1 + (2\cos^2\frac{x}{2} - 1)}\\
= \frac{2}{2\cos^2\frac{x}{2}}\\
= \sec^2\frac{x}{2}$$

Going forward, start by rewriting as:
$$\sec^2\frac{x}{2} = \frac{1}{\cos^2\frac{x}{2}}$$
By the double angle formula for $\cos$, $\cos 2\theta = 2\cos^2\theta - 1$ we have (by letting $\theta = \frac{x}{2}$):
$$\cos x = 2\cos^2\frac{x}{2} - 1$$
Rearrange to get
$$\cos^2\frac{x}{2} = \frac{1 + \cos x}{2}$$
Simply substitute this back, to deduce
$$\sec^2\frac{x}{2} = \frac{1}{\frac{1 + \cos x}{2}} = \frac{2}{1 + \cos x}$$

It's probably worth mentioning that both approaches are equivalent, differing only by which direction you choose to prove the identity.
