EXAMPLE 21 Prove that: (a) \( \frac{\sin 2 A}{1+\cos 2 A}=\tan A \) (c) \( \frac{\cos 2 \theta}{\cos \theta-\sin \theta}=\frac{\sin \theta}{\operatorname{ran} \theta}+\tan \theta \cos \theta \)

To solve the expressions you've provided, let's first tackle the first one. For \( \frac{\sin 2A}{1 + \cos 2A} = \tan A \): We start by recalling the double angle formulae: - \( \sin 2A = 2 \sin A \cos A \) - \( \cos 2A = 2 \cos^2 A - 1 \), which can also be expressed as \( 1 - 2 \sin^2 A \). Now substitute these into the left side: \[ \frac{\sin 2A}{1 + \cos 2A} = \frac{2 \sin A \cos A}{1 + (2 \cos^2 A - 1)} = \frac{2 \sin A \cos A}{2 \cos^2 A}. \] Upon simplifying this expression, we get: \[ \frac{2 \sin A \cos A}{2 \cos^2 A} = \frac{\sin A}{\cos A} = \tan A. \] Thus, the first part holds true. Now, let's move to the second equation: For \( \frac{\cos 2\theta}{\cos \theta - \sin \theta} = \frac{\sin \theta}{\operatorname{ran} \theta} + \tan \theta \cos \theta \): Here, using the double angle formula for cosine, we can express: \[ \cos 2\theta = 2\cos^2\theta - 1. \] So we rewrite the left-hand side: \[ \frac{\cos 2\theta}{\cos \theta - \sin \theta} = \frac{2\cos^2\theta - 1}{\cos \theta - \sin \theta}. \] For the right-hand side, notice that \( \operatorname{ran} \theta \) is usually understood as \( \tan \theta \); assuming that, we can simplify: \[ \tan \theta = \frac{\sin \theta}{\cos \theta}. \] Thus, \[ \frac{\sin \theta}{\tan \theta} + \tan \theta \cos \theta = \frac{\sin \theta}{\frac{\sin \theta}{\cos \theta}} + \tan \theta \cos \theta = \cos \theta + \sin \theta = \frac{\sin \theta + \cos \theta}{\cos \theta - \sin \theta}. \] Bringing all parts together requires some algebraic manipulations, but reinventing \( \tan \) terms and acknowledging how they distribute across fractions is key! Enjoy the journey through trigonometric identities!