Question

If \( I(\theta) = \cos \theta_1 \cos \theta_2 \cos \theta_3 \dots \cos \theta_n \), then \[ \tan \theta_1 + \tan \theta_2 + \tan \theta_3 + \cdots + \tan \theta_n = ? \]

• A. \( \frac{I'(\theta)}{I(\theta)} \)
• B. \( \frac{I(\theta)}{I'(\theta)} \)
• C. \( \frac{I(\theta)}{I(\theta)} \)
• D. \( \frac{I'(\theta)}{I'(\theta)} \)

Hint:

To find the sum of tangents for trigonometric products, take the derivative of the product and apply the chain rule to relate it back to the original expression.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We have a product of cosine terms. When we need to find the sum of tangents (the derivatives of logs of cosines), logarithmic differentiation is the best technique.
Step 2: Key Formula or Approach:
Apply natural log (\( \ln \)) to both sides and differentiate with respect to \( \theta \).
Step 3: Detailed Explanation:
Given: \( f(\theta) = \prod_{i=1}^{n} \cos \theta_i \).
Taking log on both sides:
\( \ln f(\theta) = \ln(\cos \theta_1) + \ln(\cos \theta_2) + \dots + \ln(\cos \theta_n) \).
Differentiating with respect to \( \theta \):
\( \frac{1}{f(\theta)} \cdot f'(\theta) = \frac{-\sin \theta_1}{\cos \theta_1} + \frac{-\sin \theta_2}{\cos \theta_2} + \dots + \frac{-\sin \theta_n}{\cos \theta_n} \)
\( \frac{f'(\theta)}{f(\theta)} = -(\tan \theta_1 + \tan \theta_2 + \dots + \tan \theta_n) \).
Multiplying by -1:
\( \tan \theta_1 + \tan \theta_2 + \dots + \tan \theta_n = \frac{-f'(\theta)}{f(\theta)} \).
Step 4: Final Answer:
The sum is \( \frac{-f'(\theta)}{f(\theta)} \).