To solve the expression [tex]\(\frac{\sin 2A}{1 + \cos 2A} = \tan A\)[/tex], let's go through it step-by-step.
1. Recall the trigonometric identities:
We know the following trigonometric identities:
[tex]\[
\sin 2A = 2 \sin A \cos A
\][/tex]
and
[tex]\[
\cos 2A = 2 \cos^2 A - 1
\][/tex]
or alternatively,
[tex]\[
\cos 2A = 1 - 2 \sin^2 A
\][/tex]
2. Substitute [tex]\(\sin 2A\)[/tex]:
Substitute [tex]\(\sin 2A = 2 \sin A \cos A\)[/tex] into the numerator of the given expression.
[tex]\[
\frac{\sin 2A}{1 + \cos 2A} = \frac{2 \sin A \cos A}{1 + \cos 2A}
\][/tex]
3. Simplify the denominator:
Substitute the identity [tex]\(\cos 2A = 1 - 2 \sin^2 A\)[/tex] into the denominator.
[tex]\[
1 + \cos 2A = 1 + (1 - 2 \sin^2 A) = 2 - 2 \sin^2 A
\][/tex]
4. Simplify the expression:
Now the expression becomes:
[tex]\[
\frac{2 \sin A \cos A}{2(1 - \sin^2 A)} = \frac{2 \sin A \cos A}{2 \cos^2 A}
\][/tex]
5. Reduce the fraction:
Cancel out the common factors in the numerator and the denominator.
[tex]\[
\frac{2 \sin A \cos A}{2 \cos^2 A} = \frac{\sin A}{\cos A}
\][/tex]
6. Final simplification:
We know that [tex]\(\frac{\sin A}{\cos A} = \tan A\)[/tex]. Thus,
[tex]\[
\frac{\sin 2A}{1 + \cos 2A} = \tan A
\][/tex]
So we have shown that [tex]\(\frac{\sin 2A}{1 + \cos 2A} = \tan A\)[/tex].
