Sure! Let's solve the given trigonometric equation step-by-step.
Given equation:
[tex]\[
\frac{1 - \cos(2A)}{\tan(A)} = \sin(2A)
\][/tex]
We begin by simplifying the left-hand side of the equation.
### Step 1: Simplify the Left-Hand Side
First, recall the double-angle identities for cosine and sine:
[tex]\[
\cos(2A) = 1 - 2\sin^2(A)
\][/tex]
Rewriting the numerator [tex]\(1 - \cos(2A)\)[/tex]:
[tex]\[
1 - \cos(2A) = 1 - (1 - 2\sin^2(A)) = 2\sin^2(A)
\][/tex]
Now we substitute this back into the left-hand side:
[tex]\[
\frac{1 - \cos(2A)}{\tan(A)} = \frac{2\sin^2(A)}{\tan(A)}
\][/tex]
Next, recall that [tex]\(\tan(A) = \frac{\sin(A)}{\cos(A)}\)[/tex]. Hence, [tex]\(\frac{1}{\tan(A)} = \frac{\cos(A)}{\sin(A)}\)[/tex],
So we get:
[tex]\[
\frac{2\sin^2(A)}{\tan(A)} = \frac{2\sin^2(A)}{\frac{\sin(A)}{\cos(A)}} = 2\sin^2(A) \cdot \frac{\cos(A)}{\sin(A)}
\][/tex]
This simplifies to:
[tex]\[
2\sin(A)\cos(A)
\][/tex]
### Step 2: Simplify the Right-Hand Side
Recall the double-angle identity for sine:
[tex]\[
\sin(2A) = 2\sin(A)\cos(A)
\][/tex]
### Step 3: Equating Both Sides
We found:
[tex]\[
\frac{1 - \cos(2A)}{\tan(A)} = 2\sin(A)\cos(A)
\][/tex]
And
[tex]\[
\sin(2A) = 2\sin(A)\cos(A)
\][/tex]
Thus:
[tex]\[
\frac{1 - \cos(2A)}{\tan(A)} = \sin(2A)
\][/tex]
### Step 4: Verify the Equality
Since both the simplified left-hand side and the right-hand side are [tex]\(2\sin(A)\cos(A)\)[/tex], we can clearly see that:
[tex]\[
(1 - \cos(2A)) / \tan(A) = \sin(2A)
\][/tex]
Therefore, the given equation holds true.
