6. Prove that: \( \frac{\sin 2 \mathrm{~A}-\cos 2 \mathrm{~A}+1}{\sin 2 \mathrm{~A}+\cos 2 \mathrm{~A}+1}=\frac{\sin \mathrm{A}}{\cos \mathrm{A}} \)

We start with the expression \[ \frac{\sin 2A - \cos 2A + 1}{\sin 2A + \cos 2A + 1}. \] A fruitful way to prove the identity is to express the double-angle functions in terms of the tangent of \(A\). Recall that \[ \sin 2A=\frac{2\tan A}{1+\tan^2A} \quad \text{and} \quad \cos 2A=\frac{1-\tan^2A}{1+\tan^2A}. \] Substitute these into the given expression: \[ \frac{\frac{2\tan A}{1+\tan^2A}-\frac{1-\tan^2A}{1+\tan^2A}+1}{\frac{2\tan A}{1+\tan^2A}+\frac{1-\tan^2A}{1+\tan^2A}+1}. \] Since all terms have the common denominator \(1+\tan^2 A\), we can combine them. Write the numerator and denominator over the common denominator: \[ \frac{\frac{2\tan A - (1-\tan^2 A) + (1+\tan^2 A)}{1+\tan^2A}}{\frac{2\tan A + (1-\tan^2 A) + (1+\tan^2A)}{1+\tan^2A}}. \] Notice that the \(1+\tan^2A\) cancels in the fraction: \[ \frac{2\tan A - (1-\tan^2A) + (1+\tan^2A)}{2\tan A + (1-\tan^2A) + (1+\tan^2A)}. \] Now, simplify the numerator: \[ 2\tan A - (1-\tan^2A) + (1+\tan^2A) = 2\tan A - 1 + \tan^2A + 1 + \tan^2A = 2\tan A + 2\tan^2A. \] And simplify the denominator: \[ 2\tan A + (1-\tan^2A) + (1+\tan^2A) = 2\tan A + 1 - \tan^2A + 1 + \tan^2A = 2\tan A + 2. \] Thus, the expression becomes \[ \frac{2\tan A + 2\tan^2A}{2\tan A + 2}. \] Factor \(2\) out of the numerator and denominator: \[ \frac{2(\tan A+\tan^2A)}{2(\tan A+1)}=\frac{\tan A+\tan^2A}{\tan A+1}. \] Factor \(\tan A\) from the numerator: \[ \frac{\tan A (1+\tan A)}{\tan A+1}. \] Since \((1+\tan A)\) appears in both numerator and denominator (provided \(\tan A \neq -1\)), they cancel, leaving \[ \tan A. \] But by definition \[ \tan A=\frac{\sin A}{\cos A}. \] Thus, we have shown that \[ \frac{\sin 2A - \cos 2A + 1}{\sin 2A + \cos 2A + 1}=\frac{\sin A}{\cos A}. \]