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

Question:

If a1a2a3, .... an are in A.P. with common difference d, then the sum of the series sin d [cosec a1 cosec a2 + cosec a1 cosec a3 + .... + cosec an − 1 cosec an] is

(a) sec a1 − sec an

(b) cosec a1 − cosec an

(c) cot a1 − cot an

(d) tan a1 − tan an

Solution:

(c) $\cot a_{1}-\cot a_{n}$

We have:

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

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

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

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

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

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

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