Let 0

Question

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

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

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(B) Given expression: $\sec 2x - 	an 2x = \frac{1 - \sin 2x}{\cos 2x}$ Using 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 $\cos x$: $= \frac{1 - 	an x}{1 + 	an x} = \frac{	an \frac{\pi}{4} - 	an x}{1 + 	an \frac{\pi}{4} 	an x}$ Using the formula $	an(A - B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$: $= 	an \left( \frac{\pi}{4} - x ight)$.

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

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