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$\tan(2x)$ was easy, as I just had to use the identity:

$$\tan(2x) = \frac{2\tan x}{1-\tan^2x}$$

I'm having a bit of trouble with $\sec(2x)$, however!

I tried squaring it and substituting a few other identities, but I either get stuck early on or go through cycles upon cycles of identity substitution which seems unnecessarily complicated.

Thanks in advance!

tan

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    $\begingroup$ $\tan^2 x+1=\sec^2 x$. $\endgroup$ – David Mitra May 5 '16 at 9:21
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Hint:

Start from $\;\cos 2x=2\cos^2x-1$ and use $1+\tan^2x=\dfrac1{\cos^2x}$.

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  • $\begingroup$ Ah, thank you! That's cleared things up. c: $\endgroup$ – Jude Dell May 5 '16 at 10:17

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