Let 0<x<π4, then sec2x−tan2x is equal to

∵0<x<π4 ∴sec2x−tan2x=1−sin2xcos2x=1−cos(π2−2x)sin(π2−2x)=sin(π4−x)cos(π4−x)=tan(π4−x)

Let 0<x<π4, then sec2x−tan2x is equal to

∵0<x<π4 ∴sec2x−tan2x=1−sin2xcos2x=1−cos(π2−2x)sin(π2−2x)=sin(π4−x)cos(π4−x)=tan(π4−x)

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Let $0 < x < \frac{π}{4},$ then $\sec 2 x - \tan 2 x$ is equal to

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$\tan (x - \frac{π}{4})$

$\tan (\frac{π}{4} - x)$

$\tan (x + \frac{π}{4})$

$\tan^{2} (x + \frac{π}{4})$

answer is B.

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Detailed Solution

$∵ 0 < x < \frac{π}{4}$
$∴ \sec 2 x - \tan 2 x = \frac{1 - \sin 2 x}{\cos 2 x} = \frac{1 - \cos (\frac{π}{2} - 2 x)}{\sin (\frac{π}{2} - 2 x)} = \frac{\sin (\frac{π}{4} - x)}{\cos (\frac{π}{4} - x)} = \tan (\frac{π}{4} - x)$

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Let $0 < x < \frac{π}{4},$ then $\sec 2 x - \tan 2 x$ is equal to