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The identity \( \frac{\sin 2A}{1 + \cos 2A} = \tan A \) is proven true using double angle formulas.
Solution
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:
\[
\frac{\sin 2A}{1 + \cos 2A} = \tan A
\]
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Simplify this solution Bonus Knowledge
To solve the equation \( \frac{\sin 2 A}{1+\cos 2 A}=\tan A \), we can use some trigonometric identities. Recall that \( \sin 2A = 2 \sin A \cos A \) and \( \cos 2A = 2\cos^2 A - 1 \). Substituting these identities into the left side gives us: \[ \frac{2 \sin A \cos A}{1 + (2\cos^2 A - 1)} = \frac{2 \sin A \cos A}{2 \cos^2 A} = \frac{\sin A}{\cos A} = \tan A \] Both sides of the equation are indeed equal, confirming that the original expression holds true for values where \( A \) is defined! Keep in mind that \( A \) should be within the domain where the tangent function is defined (i.e., not at \( A = n\frac{\pi}{2} \) where \( n \) is an integer).
