Let 0

Question

(b) $\sec 2x - 	an 2x = \frac{{1 - \sin 2x}}{{\cos 2x}}$ 

$ = \frac{{{{(\cos x - \sin x)}^2}}}{{({{\cos }^2}x - {{\sin }^2}x)}} $

$= \frac

Vedclass · Accepted Answer

$	an \left( {\frac{\pi }{4} - x} ight)$

Vedclass · Answer

(b) $\sec 2x - 	an 2x = \frac{{1 - \sin 2x}}{{\cos 2x}}$ 

$ = \frac{{{{(\cos x - \sin x)}^2}}}{{({{\cos }^2}x - {{\sin }^2}x)}} $

$= \frac{{\cos x - \sin x}}{{\cos x + \sin x}} = \frac{{1 - 	an x}}{{1 + 	an x}}$

$ = \frac{{	an \frac{\pi }{4} - 	an x}}{{1 + 	an \left( {\frac{\pi }{4}} ight)\sin x}} = 	an \left( {\frac{\pi }{4} - x} ight)$.

Let $0 < x < \frac{\pi }{4}.$ Then $\sec 2x - \tan 2x = $

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