Sec Tan Expression 66E767

1. **State the problem:** We want to simplify the expression $$\sec 2x - \tan 2x$$ for $0 < x < \frac{\pi}{4}$ and identify which of the given options it equals. 2. **Recall the definitions:** $$\sec \theta = \frac{1}{\cos \theta}$$ $$\tan \theta = \frac{\sin \theta}{\cos \theta}$$ 3. **Rewrite the expression:** $$\sec 2x - \tan 2x = \frac{1}{\cos 2x} - \frac{\sin 2x}{\cos 2x} = \frac{1 - \sin 2x}{\cos 2x}$$ 4. **Use the Pythagorean identity:** $$1 - \sin^{2} 2x = \cos^{2} 2x$$ 5. **Multiply numerator and denominator by the conjugate of numerator's denominator:** $$\frac{1 - \sin 2x}{\cos 2x} \times \frac{1 + \sin 2x}{1 + \sin 2x} = \frac{(1 - \sin 2x)(1 + \sin 2x)}{\cos 2x (1 + \sin 2x)} = \frac{1 - \sin^{2} 2x}{\cos 2x (1 + \sin 2x)}$$ 6. **Simplify numerator using identity:** $$\frac{\cos^{2} 2x}{\cos 2x (1 + \sin 2x)} = \frac{\cancel{\cos 2x} \cos 2x}{\cancel{\cos 2x} (1 + \sin 2x)} = \frac{\cos 2x}{1 + \sin 2x}$$ 7. **Recall the tangent half-angle identity:** $$\tan\left(\frac{\pi}{4} - x\right) = \frac{\sin\left(\frac{\pi}{4} - x\right)}{\cos\left(\frac{\pi}{4} - x\right)} = \frac{\cos 2x}{1 + \sin 2x}$$ 8. **Conclusion:** $$\sec 2x - \tan 2x = \tan\left(\frac{\pi}{4} - x\right)$$ **Final answer:** $\boxed{\tan\left(\frac{\pi}{4} - x\right)}$