JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Easy

1 of 1

Application +4 | -1

\theta_1, \theta_2, \theta_3, \ldots, \theta_n

are in A.P. whose common difference is

d

, then

\sin d\left[\sec \alpha_1 \sec \alpha_2+\sec \alpha_2 \sec \alpha_3+\ldots+\sec \alpha_{n-1} \sec \alpha_n\right]=

View answer and explanation

Correct option: D

\tan \alpha_n-\tan \alpha_1

Option D (

\tan \alpha_n - \tan \alpha_1

) is correct because the given expression simplifies to a telescoping series, resulting in the difference of tangents.

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