Alright, let's proceed with the given equation step-by-step to prove that it is correct.
We start with the given expression:
[tex]\[
\frac{1 - \cos(2A)}{2 \cos(A)} = \sin(A) \tan(A)
\][/tex]
First, let's simplify the left-hand side (LHS) of the equation:
[tex]\[
\frac{1 - \cos(2A)}{2 \cos(A)}
\][/tex]
To simplify this, we use a trigonometric identity for [tex]\(\cos(2A)\)[/tex]:
[tex]\[
\cos(2A) = 1 - 2 \sin^2(A)
\][/tex]
Substitute this identity into the LHS:
[tex]\[
1 - \cos(2A) = 1 - (1 - 2 \sin^2(A)) = 1 - 1 + 2 \sin^2(A) = 2 \sin^2(A)
\][/tex]
Now, substitute [tex]\(1 - \cos(2A)\)[/tex] in the LHS:
[tex]\[
\frac{2 \sin^2(A)}{2 \cos(A)}
\][/tex]
Simplify the fraction:
[tex]\[
\frac{2 \sin^2(A)}{2 \cos(A)} = \frac{\sin^2(A)}{\cos(A)}
\][/tex]
So, the simplified LHS is:
[tex]\[
\frac{\sin^2(A)}{\cos(A)}
\][/tex]
Next, let's simplify the right-hand side (RHS) of the equation:
[tex]\[
\sin(A) \tan(A)
\][/tex]
Using the definition of [tex]\(\tan(A)\)[/tex]:
[tex]\[
\tan(A) = \frac{\sin(A)}{\cos(A)}
\][/tex]
Substitute this into the RHS:
[tex]\[
\sin(A) \tan(A) = \sin(A) \left(\frac{\sin(A)}{\cos(A)}\right) = \frac{\sin^2(A)}{\cos(A)}
\][/tex]
So, the simplified RHS is:
[tex]\[
\frac{\sin^2(A)}{\cos(A)}
\][/tex]
Both the simplified LHS and RHS are:
[tex]\[
\frac{\sin^2(A)}{\cos(A)}
\][/tex]
Therefore, the original equation is verified:
[tex]\[
\frac{1 - \cos(2A)}{2 \cos(A)} = \sin(A) \tan(A)
\][/tex]
This confirms that the given equation holds true.
