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Let $0 < x \leq π / 4, (\sec 2 x - \tan 2 x)$ equals

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\tan^{2} (x + π / 4)

\tan (x + π / 4)

\tan (π / 4 - x)

\tan (x - π / 4)

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detailed solution

Correct option is C

We have,

$\sec 2 x - \tan 2 x = \frac{1 - \sin 2 x}{\cos 2 x}$

$\Rightarrow \sec 2 x - \tan 2 x = - \frac{1 - \cos (\frac{π}{2} - 2 x)}{\sin (\frac{π}{2} - 2 x)}$

$\Rightarrow \sec 2 x - \tan 2 x = \frac{2 \sin^{2} (π / 4 - x)}{2 \sin (π / 4 - x) \cos (π / 4 - x)} = \tan (\frac{π}{4} - x)$

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Let 0<x≤π/4, (sec⁡2x−tan⁡2x) equals