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

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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

$$
\frac{\sin 2A - \cos 2A + 1}{\sin 2A + \cos 2A + 1}=\frac{\sin A}{\cos A}.
$$
$$ \frac{\sin 2A - \cos 2A + 1}{\sin 2A + \cos 2A + 1} = \frac{\sin A}{\cos A} $$

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