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

Verify the identity by following steps: - step0: Verify: \(\frac{2\tan\left(\beta \right)}{1+\tan^{2}\left(\beta \right)}=\sin\left(2\beta \right)\) - step1: Choose a side to work on: \(\sin\left(2\beta \right)=\sin\left(2\beta \right)\) - step2: Verify the identity: \(\textrm{true}\) Determine whether the expression \( \frac{(1+\tan \beta)^{2}}{1+\tan ^{2} \beta}=\sec^{2}\beta \) is always true. Verify the identity by following steps: - step0: Verify: \(\frac{\left(1+\tan\left(\beta \right)\right)^{2}}{1+\tan^{2}\left(\beta \right)}=\sec^{2}\left(\beta \right)\) - step1: Choose a side to work on: \(1+\sin\left(2\beta \right)=\sec^{2}\left(\beta \right)\) - step2: Choose the other side to work on: \(1+\sin\left(2\beta \right)=\cos^{-2}\left(\beta \right)\) - step3: Verify the identity: \(\textrm{false}\) Let's analyze the two statements provided: 1. **First Statement:** \[ \frac{2 \tan \beta}{1+\tan ^{2} \beta} = \sin 2 \beta \] This statement is **true**. The left-hand side can be derived from the double angle identity for sine: \[ \sin 2\beta = \frac{2 \sin \beta \cos \beta}{1} = \frac{2 \tan \beta}{1 + \tan^2 \beta} \] Therefore, the first statement is valid. 2. **Second Statement:** \[ \frac{(1+\tan \beta)^{2}}{1+\tan ^{2} \beta} \] This statement is **false**. To verify, we can simplify the left-hand side: \[ \frac{(1+\tan \beta)^{2}}{1+\tan ^{2} \beta} = \frac{1 + 2\tan \beta + \tan^2 \beta}{1 + \tan^2 \beta} \] This simplifies to: \[ 1 + \frac{2\tan \beta}{1 + \tan^2 \beta} \] However, this does not equal \(\sec^2 \beta\) or any other standard trigonometric identity. In conclusion: - The first statement is true. - The second statement is false.