Another Verify the identity: $\sec^2 \frac{x}{2} = \frac{2}{1+\cos x}$

Question

Another Verify the identity that I can't get:

$$\sec^2 \frac{x}{2} = \frac{2}{1+\cos x}$$

$$ = \frac{1 + \left(\frac{1}{\cos x}\right)}{2}$$

$$ = \frac{\cos x + 1}{2 \cos x}$$

Answer 1

Again, both ways are possible :

Going backwards (which is simpler, coincidentally)

$$\frac{2}{1 + \cos x} = \frac{2}{1 + (2\cos^2\frac{x}{2} - 1)}\\ = \frac{2}{2\cos^2\frac{x}{2}}\\ = \sec^2\frac{x}{2}$$

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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

$$\sec^2\frac{x}{2} = \frac{1}{\frac{1 + \cos x}{2}} = \frac{2}{1 + \cos x}$$

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