Let 0

Question

(B) Given expression: $\sec 2x - 	an 2x$$= \frac{1}{\cos 2x} - \frac{\sin 2x}{\cos 2x} = \frac{1 - \sin 2x}{\cos 2x}$Using the identities $\sin 2x =

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

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(B) Given expression: $\sec 2x - 	an 2x$$= \frac{1}{\cos 2x} - \frac{\sin 2x}{\cos 2x} = \frac{1 - \sin 2x}{\cos 2x}$Using the identities $\sin 2x = 2 \sin x \cos x$,$\cos 2x = \cos^2 x - \sin^2 x$,and $1 = \cos^2 x + \sin^2 x$:$= \frac{\cos^2 x + \sin^2 x - 2 \sin x \cos x}{\cos^2 x - \sin^2 x} = \frac{(\cos x - \sin x)^2}{(\cos x - \sin x)(\cos x + \sin x)}$$= \frac{\cos x - \sin x}{\cos x + \sin x}$Dividing numerator and denominator by $\cos x$:$= \frac{1 - 	an x}{1 + 	an x}$Since $	an \frac{\pi}{4} = 1$,we can write this as:$= \frac{	an \frac{\pi}{4} - 	an x}{1 + 	an \frac{\pi}{4} 	an x} = 	an \left( \frac{\pi}{4} - x ight)$.

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

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