a. \( \frac{\sin 2 A}{1+\cos 2 A}=\tan A \)

To prove the identity \( \frac{\sin 2A}{1 + \cos 2A} = \tan A \), we can start by using the double angle formulas for sine and cosine: \[ \sin 2A = 2 \sin A \cos A \] \[ \cos 2A = \cos^2 A - \sin^2 A = 2\cos^2 A - 1 = 1 - 2\sin^2 A \] Now, substituting these into the left-hand side of the equation: \[ \frac{\sin 2A}{1 + \cos 2A} = \frac{2 \sin A \cos A}{1 + (2 \cos^2 A - 1)} \] Simplifying the denominator: \[ 1 + \cos 2A = 1 + (2 \cos^2 A - 1) = 2 \cos^2 A \] Now we can rewrite the left-hand side: \[ \frac{2 \sin A \cos A}{2 \cos^2 A} \] We can simplify this expression: \[ = \frac{2 \sin A \cos A}{2 \cos^2 A} = \frac{\sin A}{\cos A} = \tan A \] Thus, we have shown that: \[ \frac{\sin 2A}{1 + \cos 2A} = \tan A \] Therefore, the identity is proven to be true: \[ \frac{\sin 2A}{1 + \cos 2A} = \tan A \]