Let's verify the trigonometric identity:
[tex]$
\frac{1-\cos \theta}{\sin \theta} + \frac{\sin \theta}{1-\cos \theta} = 2 \csc \theta
$[/tex]
We'll start by simplifying the left-hand side of the equation.
### Step 1: Simplify the First Term
Consider the first term:
[tex]$
\frac{1-\cos \theta}{\sin \theta}
$[/tex]
This term can be left as it is for now.
### Step 2: Simplify the Second Term
Consider the second term:
[tex]$
\frac{\sin \theta}{1-\cos \theta}
$[/tex]
We observe that the denominator \(1 - \cos \theta\) in this term is the same as the numerator in the previous term.
Since multiplication of reciprocal functions simplifies:
[tex]$
\frac{a}{b} \cdot \frac{b}{a} = 1,
$[/tex]
Thus, multiplying will invert the phrase.
### Step 3: Combine the Two Terms
Add the two terms together:
[tex]$
\frac{1-\cos \theta}{\sin \theta} + \frac{\sin \theta}{1-\cos \theta}
$[/tex]
To combine these fractions, we need a common denominator of \( \sin \theta (1 - \cos \theta)\):
[tex]$
\frac{(1-\cos \theta)^2 + \sin^2 \theta}{\sin \theta (1-\cos \theta)}
$[/tex]
### Step 4: Simplify the Numerator
Expand and simplify the numerator:
[tex]$
(1-\cos \theta)^2 + \sin^2 \theta = 1 - 2\cos \theta + \cos^2 \theta + \sin^2 \theta
$[/tex]
Recall the Pythagorean identity:
[tex]$
\sin^2 \theta + \cos^2 \theta = 1
$[/tex]
Thus,
[tex]$
1 - 2\cos \theta + \cos^2 \theta + \sin^2 \theta = 1 - 2\cos \theta + 1 = 2 - 2\cos \theta
$[/tex]
Factor out the 2:
[tex]$
2(1 - \cos \theta)
$[/tex]
### Step 5: Combining the Results
Place the simplified numerator back into the fraction:
[tex]$
\frac{2(1-\cos \theta)}{\sin \theta (1-\cos \theta)}
$[/tex]
The terms \(1 - \cos \theta\) cancel out in the numerator and the denominator:
[tex]$
\frac{2}{\sin \theta}
$[/tex]
### Step 6: Realize that \(\csc \theta = \frac{1}{\sin \theta}\)
Thus:
[tex]$
\frac{2}{\sin \theta} = 2 \csc \theta
$[/tex]
### Conclusion
The left-hand side simplifies exactly to the right-hand side:
[tex]$
\frac{1-\cos \theta}{\sin \theta} + \frac{\sin \theta}{1-\cos \theta} = 2 \csc \theta
$[/tex]
Therefore, the identity is verified to be true.
