Question

If \( a_1, a_2, \ldots, a_n \) are in Arithmetic Progression with common difference d, then the sum \( (\sin d) (\csc a_1 \csc a_2 + \csc a_2 \csc a_3 + \dots + \csc a_{n-1} \csc a_n) \) is equal to

• A. \( \cot a_1 - \cot a_n \)
• B. \( \sin a_1 - \sin a_n \)
• C. \( \csc a_1 - \csc a_n \)
• D. \( a_1 - a_n \)

Hint:

This type of problem is a classic example of a telescoping series. The trick is to manipulate the general term \( T_k \) so that it can be expressed as a difference of two consecutive terms, i.e., \( T_k = f(k) - f(k+1) \). When you sum this up, all intermediate terms cancel out, leaving only the first and last parts.

Answer 1

The Correct Option is A

Let the given sum be S. The general term in the series is \( \csc a_k \csc a_{k+1} = \frac{1}{\sin a_k \sin a_{k+1}} \). We are given an AP, so \( a_{k+1} - a_k = d \). Let's multiply and divide the general term by \( \sin d = \sin(a_{k+1} - a_k) \). \[ \frac{1}{\sin a_k \sin a_{k+1}} = \frac{1}{\sin d} \frac{\sin(a_{k+1} - a_k)}{\sin a_k \sin a_{k+1}} \] Using the identity \( \sin(A-B) = \sin A \cos B - \cos A \sin B \): \[ = \frac{1}{\sin d} \frac{\sin a_{k+1} \cos a_k - \cos a_{k+1} \sin a_k}{\sin a_k \sin a_{k+1}} \] Split the fraction: \[ = \frac{1}{\sin d} \left( \frac{\sin a_{k+1} \cos a_k}{\sin a_k \sin a_{k+1}} - \frac{\cos a_{k+1} \sin a_k}{\sin a_k \sin a_{k+1}} \right) \] \[ = \frac{1}{\sin d} \left( \frac{\cos a_k}{\sin a_k} - \frac{\cos a_{k+1}}{\sin a_{k+1}} \right) \] \[ = \frac{1}{\sin d} (\cot a_k - \cot a_{k+1}) \] Now, the original sum is: \[ S = (\sin d) \sum_{k=1}^{n-1} \csc a_k \csc a_{k+1} = (\sin d) \sum_{k=1}^{n-1} \frac{1}{\sin d} (\cot a_k - \cot a_{k+1}) \] \[ S = \sum_{k=1}^{n-1} (\cot a_k - \cot a_{k+1}) \] This is a telescoping series. Let's write out the terms: \[ S = (\cot a_1 - \cot a_2) + (\cot a_2 - \cot a_3) + \dots + (\cot a_{n-1} - \cot a_n) \] All intermediate terms cancel out. We are left with: \[ S = \cot a_1 - \cot a_n \]