To prove the given trigonometric identity:
[tex]\[
\frac{1 - \cos y}{\sin y} + \frac{\sin y}{1 - \cos y} = 2 \csc y
\][/tex]
We'll start by simplifying the left-hand side (LHS) of the equation and show that it equals the right-hand side (RHS).
### Step-by-Step Solution:
1. Write down the left-hand side (LHS):
[tex]\[
\text{LHS} = \frac{1 - \cos y}{\sin y} + \frac{\sin y}{1 - \cos y}
\][/tex]
2. Find a common denominator for the fractions:
To simplify the LHS, we get a common denominator, which would be [tex]\( \sin y (1 - \cos y) \)[/tex]:
[tex]\[
\text{LHS} = \frac{(1 - \cos y)^2 + \sin^2 y}{\sin y (1 - \cos y)}
\][/tex]
3. Simplify the numerator:
Notice that we can rewrite [tex]\( (1 - \cos y)^2 \)[/tex] as [tex]\( 1 - 2\cos y + \cos^2 y \)[/tex]:
[tex]\[
(1 - \cos y)^2 = 1 - 2 \cos y + \cos^2 y
\][/tex]
Also, recall the Pythagorean identity:
[tex]\[
\sin^2 y + \cos^2 y = 1
\][/tex]
Using this identity, substitute [tex]\( \cos^2 y \)[/tex]:
[tex]\[
\sin^2 y + \cos^2 y = 1 \quad \Rightarrow \quad 1 - \sin^2 y = \cos^2 y
\][/tex]
Substituting these identities back into the numerator:
[tex]\[
(1 - \cos y)^2 + \sin^2 y = (1 - 2 \cos y + \cos^2 y) + \sin^2 y = 1 - 2 \cos y + \cos^2 y + \sin^2 y = 1 - 2 \cos y + 1 = 2 - 2 \cos y
\][/tex]
4. Factor out the common term in the numerator:
[tex]\[
\text{Above Proposition} = 2 (1 - \cos y)
\][/tex]
Thus, the numerator simplifies to:
[tex]\[
\frac{2 (1 - \cos y)}{\sin y (1 - \cos y)}
\][/tex]
5. Cancel the common term [tex]\( (1 - \cos y) \)[/tex] in the numerator and the denominator:
[tex]\[
\frac{2 (1 - \cos y)}{\sin y (1 - \cos y)} = \frac{2}{\sin y} = 2 \csc y
\][/tex]
We see that the left-hand side (LHS) simplifies to:
[tex]\[
\text{LHS} = 2 \csc y
\][/tex]
6. Compare it with the right-hand side (RHS):
[tex]\[
\text{RHS} = 2 \csc y
\][/tex]
7. Conclusion:
Since the simplified LHS is equal to the RHS, we have proven the given identity:
[tex]\[
\frac{1 - \cos y}{\sin y} + \frac{\sin y}{1 - \cos y} = 2 \csc y
\][/tex]
