If $a_{1}, a_{2}, a_{3}, \dots \dots a_{n}$ be an A.P. with common difference d, then sec $a_{1} \sec a_{2} + \sec a_{2} \sec a_{3} + \dots \dots + \sec a_{n - 1} \sec a_{n} =$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{Tan a}{\sin a_{1} \sin a_{2}}$

$\frac{Tan a_{n} - Tan a_{1}}{\sin d}$

$\frac{n - 1}{{Tana}_{1} {Tana}_{2}}$

$\frac{Tan a_{1} - Tan a_{n}}{\sin d}$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$a_{2} - a_{1} = d = a_{3} - a_{2} = a_{4} - a_{3} . . . . . \sec a, \sec a_{2} = \frac{1}{\cos a_{1} \cdot \cos a_{2}} = \frac{1}{\sin d} [\frac{\sin d}{\cos a_{1} \cdot \cos a_{2}}] = \frac{1}{\sin d} [\frac{\sin (a_{2} - a_{1})}{\cos a_{1} \cos a_{2}}] = \frac{1}{\sin d} [\tan a_{2} - \tan a_{1}] \sec a_{2} \sec a_{3} = \frac{1}{\sin d} [\frac{\sin (a_{3} - a_{2})}{\cos a_{3} \cos a_{2}}] = \frac{1}{\sin d} [\tan a_{3} - \tan a_{2}]$

$\sec a_{1} \cdot \sec a_{2} + \sec a_{2} \cdot \sec a_{3} . . . . + \sec a_{n - 1} \sec a_{n} = \frac{1}{\sin d} [\tan a_{2} - \tan a_{1} + \tan a_{3} - \tan a_{2} + . . . . + \tan a_{n} - \tan a_{n - 1}] = \frac{1}{\sin d} [\tan a_{n} - \tan a_{1}]$