# with common difference d, then the sum of the series sin

Question:

If a1a2a3, .... an 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 an − 1 sec an], is

(a) sec a1 − sec an

(b) cosec a1 − cosec an

(c) cot a1 − cot an

(d) tan an − tan a1

Solution:

(d) $\tan a_{n}-\tan a_{1}$

We have:

$\sin d\left(\sec a_{1} \sec a_{2}+\sec a_{2} \sec a_{3}+\ldots+\sec a_{n-1} \sec a_{n}\right)$

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

$=\frac{\sin \left(a_{2}-a_{1}\right)}{\cos a_{1} \cos a_{2}}+\frac{\sin \left(a_{3}-a_{2}\right)}{\cos a_{2} \cos a_{3}}+\ldots+\frac{\sin \left(a_{n}-a_{n-1}\right)}{\cos a_{n-1} \cos a_{n}}$

$=\frac{\sin a_{2} \cos a_{1}-\cos a_{2} \sin a_{1}}{\cos a_{1} \cos a_{2}}+\frac{\sin a_{3} \cos a_{2}-\cos a_{3} \sin a_{2}}{\cos a_{1} \cos a_{2}}+\ldots+\frac{\sin a_{2} \cos a_{1}-\cos a_{2} \sin a_{1}}{\cos a_{1} \cos a_{2}}$

$=\left(\tan a_{1}-\tan a_{2}\right)+\left(\tan a_{2}-\tan a_{3}\right)+\ldots \ldots+\left(\tan a_{n-1}-\tan a_{n}\right)$

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