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 given expression.
We are given the expression:
\[ I(\theta) = \cos \theta_1 \cos \theta_2 \cos \theta_3 \dots \cos \theta_n, \]
which is a product of cosines. We need to find an expression for \( \tan \theta_1 + \tan \theta_2 + \dots + \tan \theta_n \) based on this product.

Step 2: Taking the derivative of \( I(\theta) \).
To solve for the sum of the tangents, we first take the derivative of \( I(\theta) \). Using the product rule, the derivative of \( I(\theta) \) is:
\[ I'(\theta) = -\sin \theta_1 \cos \theta_2 \cos \theta_3 \dots \cos \theta_n - \cos \theta_1 \sin \theta_2 \cos \theta_3 \dots \cos \theta_n \dots. \]

Step 3: Relating \( I'(\theta) \) and \( I(\theta) \).
By applying the chain rule and further manipulating the result, we see that the sum of the tangents is related to the derivative and the original product. Specifically, we use the following identity:
\[ \tan \theta_1 + \tan \theta_2 + \dots + \tan \theta_n = \frac{I'(\theta)}{I(\theta)}. \]

Step 4: Final answer.
Thus, the sum of the tangents is:
\[ \boxed{\frac{I'(\theta)}{I(\theta)}}. \]
Final Answer:
The correct answer is:
\[ \boxed{\frac{I'(\theta)}{I(\theta)}}. \]