Question

If \( a_1, a_2, \dots, a_n \) are in A.P. with common difference \( \theta \), then the sum of the series:
\[ \sec a_1 \sec a_2 + \sec a_2 \sec a_3 + \dots + \sec a_{n-1} \sec a_n = k (\tan a_n - \tan a_1), \]
where \( k = ? \)

• A. \( \sin \theta \)
• B. \( \cos \theta \)
• C. \( \sec \theta \)
• D. \( \csc \theta \)

Hint:

For sums involving products of trigonometric functions, look for patterns or telescoping series for simplification.

Answer 1

The Correct Option is D

The general term of the A.P. is: \[ a_r = a_1 + (r - 1)\theta, \quad r = 1, 2, \ldots, n. \]

The series involves products of consecutive secants: \[ S = \sec a_1 \sec a_2 + \sec a_2 \sec a_3 + \ldots + \sec a_{n-1} \sec a_n. \]

Simplify using trigonometric identities: \[ \sec a_r \sec a_{r+1} = \frac{1}{\cos a_r \cos a_{r+1}}. \]

Summing up and simplifying using properties of tangent and secant, we find: \[ S = k (\tan a_n - \tan a_1), \quad k = \csc \theta. \]