Question

If ${a_1},{a_2},{a_3},............{a_n}$ are in AP with common difference d, then sum of series is $\sin d(\csc {a_1}\csc {a_2} + \csc {a_2}\csc {a_3} + ..... + \csc {a_{n - 1}}\csc {a_n})$ is
$
  A)\sec {a_1} - \sec {a_n} \\
  B)\cot {a_1} - \cot {a_n} \\
  C)\tan {a_1} - \tan {a_n} \\
  D)\csc {a_1} - \csc {a_n} \\
 $

Answer 1

Hint: To solve this problem you need to have knowledge about Arithmetic progression concept, which first term of AP, common difference of AP and sum of n terms of AP. We should also have basic knowledge about trigonometry as the problem includes both the concept AP and trigonometry.

Complete step-by-step answer:
Given
${a_1},{a_2},{a_3},............{a_n}$ are the n terms of AP series with common difference d.
Here common difference of AP is calculated as $d = {a_2} - {a_1} = {a_3} - {a_2} = .................{a_n} - {a_{n - 1}}$
According to the question the sum of n term of series is given as
$ \Rightarrow \sin d(\csc {a_1}\csc {a_2} + \csc {a_2}\csc {a_3} + ..... + \csc {a_{n - 1}}\csc {a_n})$
Since we know that $\csc = \dfrac{1}{{\sin }}$ on replacing the term we can rewrite the equation as
$ \Rightarrow \dfrac{{\sin d}}{{\sin {a_1}\sin {a_2}}} + \dfrac{{\sin d}}{{\sin {a_2}\sin {a_3}}} + ......... + \dfrac{{\sin d}}{{\sin {a_{n - 1}}\sin {a_n}}}$
$ \Rightarrow \dfrac{{\sin ({a_2} - {a_1})}}{{\sin {a_1}\sin {a_2}}} + \dfrac{{\sin ({a_3} - {a_1})}}{{\sin {a_2}\sin {a_3}}} + ......... + \dfrac{{\sin ({a_{n - 1}} - {a_n})}}{{\sin {a_{n - 1}}\sin {a_n}}}$
Since we know that d is the common difference of AP
On applying $\sin (a - b) = \sin a\cos b - \cos a\sin b$ formula to the numerator we will get
$ \Rightarrow \dfrac{{\sin {a_1}\cos {a_2} - \cos {a_1}\sin {a_2}}}{{\sin {a_1}\sin {a_2}}} + \dfrac{{\sin {a_3}\cos {a_2} - \cos {a_3}\sin {a_2}}}{{\sin {a_2}\sin {a_3}}} + ............ + \dfrac{{\sin {a_n}\cos {a_{n - 1}} - \cos {a_n}\sin {a_{n - 1}}}}{{\sin {a_{n - 1}}\sin {a_n}}}$On further simplification we will get
$
   \Rightarrow \cot {a_1} - \cot {a_2} + \cot {a_2} - \cot {a_3} + ........ + \cot {a_{n - 1}} - \cot {a_n} \\
   \Rightarrow \cot {a_1} - \cot {a_n} \\
 $
Hence the answer is $\cot {a_1} - \cot {a_n}$
Option B is the correct answer.

NOTE: In this problem we have converted the given trigonometric values and applied the common difference condition of AP to the numerator which gives a trigonometry formula .Later on applying the trigonometry formula we get the n values of a series in terms of trigonometry.