Certainly! Let's solve the equation \(\frac{1 - \cos 2\theta}{\sin 2\theta} = \tan \theta\) step by step.
### Step 1: Recall Trigonometric Identities
First, recall the double-angle identities for sine and cosine:
1. \(\cos 2\theta = 1 - 2\sin^2 \theta\)
2. \(\sin 2\theta = 2 \sin \theta \cos \theta\)
3. \(\tan \theta = \frac{\sin \theta}{\cos \theta}\)
### Step 2: Simplify the Left-Hand Side
Evaluate the left-hand side \(\frac{1 - \cos 2\theta}{\sin 2\theta}\) using the identities:
[tex]\[
\frac{1 - \cos 2\theta}{\sin 2\theta}
\][/tex]
Substitute \(\cos 2\theta = 1 - 2\sin^2 \theta\):
[tex]\[
\frac{1 - (1 - 2\sin^2 \theta)}{\sin 2\theta} = \frac{1 - 1 + 2\sin^2 \theta}{\sin 2\theta} = \frac{2\sin^2 \theta}{\sin 2\theta}
\][/tex]
Substitute \(\sin 2\theta = 2 \sin \theta \cos \theta\):
[tex]\[
\frac{2\sin^2 \theta}{2 \sin \theta \cos \theta} = \frac{\sin \theta}{\cos \theta}
\][/tex]
This reduces to:
[tex]\[
\tan \theta
\][/tex]
### Step 3: Compare Both Sides
Now we see that the left-hand side simplifies to \(\tan \theta\):
[tex]\[
\frac{1 - \cos 2\theta}{\sin 2\theta} = \tan \theta
\][/tex]
Since the right-hand side of the original equation is already \(\tan \theta\), we have shown:
[tex]\[
\tan \theta = \tan \theta
\][/tex]
### Step 4: Conclusion
Since both the simplified left-hand side and the right-hand side are equal, the equation holds true.
### Final Result
So, [tex]\(\frac{1 - \cos 2\theta}{\sin 2\theta} = \tan \theta\)[/tex] is indeed a true statement.
