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Verify the identity:

[tex]\ \textless \ br/\ \textgreater \ \begin{array}{l} \ \textless \ br/\ \textgreater \ \frac{1-\cos (\alpha)}{\sin (\alpha)}=\frac{\sin (\alpha)}{1+\cos (\alpha)} \\\ \textless \ br/\ \textgreater \ \frac{1-\cos (\alpha)}{\sin (\alpha)}=\frac{1-\cos (\alpha)}{\sin (\alpha)} \cdot \frac{1+\cos (\alpha)}{1+\cos (\alpha)} \\\ \textless \ br/\ \textgreater \ =\frac{1-\cos^2 (\alpha)}{(\sin (\alpha))(1+\cos (\alpha))} \\\ \textless \ br/\ \textgreater \ =\frac{\sin^2 (\alpha)}{(\sin (\alpha))(1+\cos (\alpha))} \\\ \textless \ br/\ \textgreater \ =\frac{\sin (\alpha)}{1+\cos (\alpha)}\ \textless \ br/\ \textgreater \ \end{array}\ \textless \ br/\ \textgreater \ [/tex]



Answer :

GinnyAnswer
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.

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