5.4.1 Prove that: \[ \begin{array}{l}\frac{2 \tan \beta}{1+\tan ^{2} \beta}=\sin 2 \beta \\ \\ \frac{(1+\tan \beta)^{2}}{1+\tan ^{2} \beta}\end{array} \]

To prove the given expressions, let's start with the first one: 1. **Proving \(\frac{2 \tan \beta}{1 + \tan^2 \beta} = \sin 2\beta\)**: We know that: \[ \sin 2\beta = 2 \sin \beta \cos \beta \] Recall the identity for \(\tan \beta\): \[ \tan \beta = \frac{\sin \beta}{\cos \beta} \] In terms of sine and cosine: \[ \sin \beta = \tan \beta \cos \beta \] Therefore, \[ \sin 2\beta = 2 \sin \beta \cos \beta = 2 \left(\tan \beta \cos \beta\right) \cos \beta = 2 \tan \beta \cos^2 \beta \] Now, using the Pythagorean identity: \[ \cos^2 \beta = \frac{1}{1 + \tan^2 \beta} \] Substituting this into our equation: \[ \sin 2\beta = 2 \tan \beta \cdot \frac{1}{1 + \tan^2 \beta} \] Thus: \[ \sin 2\beta = \frac{2 \tan \beta}{1 + \tan^2 \beta} \] So, the identity is proven. --- 2. **Proving \(\frac{(1+\tan \beta)^2}{1+\tan^2 \beta}\)**: Let's simplify \(\frac{(1+\tan \beta)^2}{1+\tan^2 \beta}\). First, expand the numerator: \[ (1 + \tan \beta)^2 = 1 + 2\tan \beta + \tan^2 \beta \] Now, the expression becomes: \[ \frac{1 + 2 \tan \beta + \tan^2 \beta}{1 + \tan^2 \beta} \] Notice that we can simplify the expression. Split it up: \[ \frac{1 + \tan^2 \beta}{1 + \tan^2 \beta} + \frac{2 \tan \beta}{1 + \tan^2 \beta} = 1 + \frac{2 \tan \beta}{1 + \tan^2 \beta} \] We already showed that: \[ \frac{2 \tan \beta}{1 + \tan^2 \beta} = \sin 2 \beta \] So the entire expression becomes: \[ 1 + \sin 2\beta \] Therefore: \[ \frac{(1+\tan \beta)^2}{1+\tan^2 \beta} = 1 + \sin 2\beta \] Thus, both identities have been proven.