Question

If a₁, a₂, a₃, .... a_n are in A.P. with common difference d, then the sum of the series sin d [sec a₁ sec a₂ + sec a₂ sec a₃ + .... + sec a_{n − 1} sec a_n], is

विकल्प

•  sec a₁ − sec a_n

• cosec a₁ − cosec a_n

• cot a₁ − cot a_n

• tan a_n − tan a₁

Answer 1

tan a_n − tan a₁

We have:

\[\sin d \left( \sec a_1 \sec a_2 + \sec a_2 \sec a_3 + . . . . + \sec a_{n - 1} \sec a_n \right)\]

\[ = \frac{\sin d}{\cos a_1 \cos a_2} + \frac{\sin d}{\cos a_2 \cos a_3} + . . . . . + \frac{\sin d}{\cos a_{n - 1} \cos a_n}\]

\[ = \frac{\sin ( a_2 - a_1 )}{\cos a_1 \cos a_2} + \frac{\sin ( a_3 - a_2 )}{\cos a_2 \cos a_3} + . . . . + \frac{\sin ( a_n - a_{n - 1} )}{\cos a_{n - 1} \cos a_n}\]

\[ = \frac{\sin a_2 \cos a_1 - \cos a_2 \sin a_1}{\cos a_1 \cos a_2} + \frac{\sin a_3 \cos a_2 - \cos a_3 \sin a_2}{\cos a_1 \cos a_2} + . . . . . + \frac{\sin a_2 \cos a_1 - \cos a_2 \sin a_1}{\cos a_1 \cos a_2}\]

\[ = \left( \tan a_1 - \tan a_2 \right) + \left( \tan a_2 - \tan a_3 \right) + . . . . . + \left( \tan a_{n - 1} - \tan a_n \right)\]

\[ = \tan a_1 - \tan a_n\]