Answer
To prove the identity, cross-multiply to get \( \sin^2 \theta = (1 - \cos \theta)(1 + \cos \theta) \). Simplifying the right side using the difference of squares gives \( 1 - \cos^2 \theta \), which equals \( \sin^2 \theta \) by the Pythagorean identity. Therefore, the identity holds true.
For the second part, the expression \( \frac{\sin \theta}{14 \cos \theta} \) is undefined when the denominator is zero, i.e., when \( \cos \theta = 0 \). This occurs at \( \theta = 90^\circ \) and \( \theta = 270^\circ \) within the interval \( (0, 360) \).
Solution
Let's solve the problem step by step.
### Step 1: Prove that \( \frac{\sin \theta}{1+\cos \theta} = \frac{1-\cos \theta}{\sin \theta} \)
To prove this identity, we will start by cross-multiplying both sides:
\[
\sin^2 \theta = (1 - \cos \theta)(1 + \cos \theta)
\]
The right-hand side can be simplified using the difference of squares:
\[
(1 - \cos \theta)(1 + \cos \theta) = 1 - \cos^2 \theta
\]
Using the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we can substitute \( 1 - \cos^2 \theta \) with \( \sin^2 \theta \):
\[
\sin^2 \theta = \sin^2 \theta
\]
This shows that both sides are equal, thus proving the identity:
\[
\frac{\sin \theta}{1+\cos \theta} = \frac{1-\cos \theta}{\sin \theta}
\]
### Step 2: Find the value of \( \theta \) in the interval \( (0, 360) \) for which \( \frac{\sin \theta}{14 \cos \theta} \) is undefined.
The expression \( \frac{\sin \theta}{14 \cos \theta} \) is undefined when the denominator is zero, which occurs when:
\[
14 \cos \theta = 0
\]
This simplifies to:
\[
\cos \theta = 0
\]
The cosine function is zero at the following angles in the interval \( (0, 360) \):
\[
\theta = 90^\circ \quad \text{and} \quad \theta = 270^\circ
\]
### Final Answer
1. The identity \( \frac{\sin \theta}{1+\cos \theta} = \frac{1-\cos \theta}{\sin \theta} \) is proven.
2. The values of \( \theta \) in the interval \( (0, 360) \) for which \( \frac{\sin \theta}{14 \cos \theta} \) is undefined are \( 90^\circ \) and \( 270^\circ \).
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