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

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\frac{n - 1}{\tan a_{1} . \tan a_{2}}

\frac{\tan a_{n} - \tan a_{1}}{\sin d}

\frac{\tan a}{\sin a_{1} \sin a_{2}}

\frac{\tan a_{1} - \tan a_{n}}{\sin d}

answer is B.

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Detailed Solution

Given that

$a_{1}, a_{2}, a_{3}, \dots, a_{n}$ are in A.P

$a_{2} - a_{1} = a_{3} - a_{2} = \dots = a_{n} - a_{n - 1} = d$

$\sec a_{1} \sec a_{2} + \sec a_{2} \sec a_{3} + \dots + \sec a_{n - 1} \sec a_{n}$

$= \frac{1}{\cos a_{1} \cos a_{2}} + \frac{1}{\cos a_{2} \cos a_{3}} + \dots + \frac{1}{\cos a_{n - 1} \cos a_{n}}$

$= \frac{1}{\sin d} [\frac{\sin d}{\cos a_{1} \cos a_{2}} + \frac{\sin d}{\cos a_{2} \cos a_{3}} + \dots + \frac{\sin d}{\cos a_{n - 1} \cos a_{n}}]$

$= \frac{1}{\sin d} [\frac{\sin (a_{2} - a_{1})}{\cos a_{1} \cos a_{2}} + \frac{\sin (a_{3} - a_{2})}{\cos a_{2} \cos a_{3}} + \dots + \frac{\sin (a_{n} - a_{n - 1})}{\cos a_{n - 1} \cos a_{n}}]$

$= \frac{1}{\sin d} [\frac{\sin a_{2} \cos a_{1} - \cos a_{2} \sin a_{1}}{\cos a_{1} \cos a_{2}} + \frac{\sin a_{3} \cos a_{2} - \cos a_{3} \sin a_{2}}{\cos a_{2} \cos a_{3}} + \dots + \frac{\sin a_{n} \cos a_{n - 1} - \cos a_{n} \sin a_{n - 1}}{\cos a_{n - 1} \cos a_{n}}]$