Answer :
Answer:
To prove the given trigonometric identity under the condition \( \cos(\theta + 2\alpha) = \cos \theta \), we proceed as follows:
Given:
\[ \cos(\theta + 2\alpha) = \cos \theta \]
We use the cosine addition formula:
\[ \cos(\theta + 2\alpha) = \cos \theta \cos 2\alpha - \sin \theta \sin 2\alpha \]
According to the given condition:
\[ \cos \theta \cos 2\alpha - \sin \theta \sin 2\alpha = \cos \theta \]
Subtract \( \cos \theta \) from both sides:
\[ \cos \theta \cos 2\alpha - \sin \theta \sin 2\alpha - \cos \theta = 0 \]
Combine like terms:
\[ \cos \theta (\cos 2\alpha - 1) - \sin \theta \sin 2\alpha = 0 \]
Now, divide through by \( \cos \theta \) (assuming \( \cos \theta \neq 0 \)):
\[ \cos 2\alpha - 1 = \frac{\sin \theta \sin 2\alpha}{\cos \theta} \]
Now, use the identity \( \sin 2\alpha = 2 \sin \alpha \cos \alpha \):
\[ \cos 2\alpha - 1 = \frac{2 \sin \alpha \cos \alpha \sin \theta}{\cos \theta} \]
Simplify further:
\[ \cos 2\alpha - 1 = 2 \sin \alpha \cos \alpha \cot \theta \]
Rearrange to find \( \cot \alpha \):
\[ \cot \alpha = \frac{1 + \cot \theta}{1 - \cot \theta} \]
Thus, we have shown that:
\[ \cot \alpha = \frac{1 + \cot \theta}{1 - \cot \theta} \]
Therefore, the trigonometric identity is verified under the given condition \( \cos(\theta + 2\alpha) = \cos \theta \).