To verify the identity:
[tex]\[
\frac{1-\cos(\alpha)}{\sin(\alpha)} = \frac{\sin(\alpha)}{1+\cos(\alpha)}
\][/tex]
We will start from the left-hand side and try to manipulate it to look like the right-hand side.
Consider the left-hand side:
[tex]\[
\frac{1-\cos(\alpha)}{\sin(\alpha)}
\][/tex]
To transform it, we will multiply the numerator and the denominator by [tex]\(1+\cos(\alpha)\)[/tex]:
[tex]\[
\frac{1-\cos(\alpha)}{\sin(\alpha)} \cdot \frac{1+\cos(\alpha)}{1+\cos(\alpha)}
\][/tex]
This can be simplified as follows:
[tex]\[
= \frac{(1-\cos(\alpha))(1+\cos(\alpha))}{\sin(\alpha)(1+\cos(\alpha))}
\][/tex]
Using the difference of squares formula in the numerator:
[tex]\[
= \frac{1 - \cos^2(\alpha)}{\sin(\alpha)(1+\cos(\alpha))}
\][/tex]
We know that [tex]\(1 - \cos^2(\alpha) = \sin^2(\alpha)\)[/tex] (from the Pythagorean identity):
[tex]\[
= \frac{\sin^2(\alpha)}{\sin(\alpha)(1+\cos(\alpha))}
\][/tex]
We can simplify the fraction further by canceling one [tex]\(\sin(\alpha)\)[/tex] from the numerator and the denominator:
[tex]\[
= \frac{\sin(\alpha)}{1+\cos(\alpha)}
\][/tex]
Thus, we have shown that:
[tex]\[
\frac{1-\cos(\alpha)}{\sin(\alpha)} = \frac{\sin(\alpha)}{1+\cos(\alpha)}
\][/tex]
Therefore, the given identity is verified.
