First slide

Arithmetic progression

Question

$If a_{1}, a_{2}, a_{3}, \dots \dots . a_{n} are in A.P with common difference d \neq 0, then sum of the$
$series \sin d [\sec a_{1} \sec a_{2} + \sec a_{2} \sec a_{3} + \dots \dots \dots + \sec a_{n - 1} \sec a_{n}] is$

Moderate

A

${coseca}_{n} - \cos e c a$
B

$\cot a_{n} - \cot a$
C

$\sec a_{n} - \sec a_{1}$
D

$\tan a_{n} - \tan a_{1}$

Solution

$\begin{array}{l} As a_{1}, a_{2}, a_{3}, \dots \dots \dots \dots a_{n - 1}, a_{n} are in A.P hence \\ d = a_{2} - a_{1} = a_{3} - a_{2} = \dots \dots . . = a_{n} - a_{n - 1} \\ \sin d [\sec a_{1} \sec a_{2} + \sec a_{2} \sec a_{3} + \dots \dots \dots \dots \dots . + \sec a_{n - 1} \sec a_{n}] \\ = \frac{\sin (a_{2} - a_{1})}{\cos a_{1} \cos a_{2}} + \frac{\sin (a_{3} - a_{2})}{\cos a_{2} \cos a_{3}} + \dots \dots \dots \dots . . + \frac{\sin (a_{n} - a_{n - 1})}{\cos a_{n - 1} \cos a_{n}} \\ = (\tan a_{2} - \tan a_{1}) + (\tan a_{3} - \tan a_{2}) + \dots \dots \dots \dots + (\tan a_{n} - \tan a_{n - 1}) \\ = \tan a_{n} - \tan a_{1} \end{array}$

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