Question

Find the value of $S = {\tan ^{ - 1}}\left( {\dfrac{1}{{{n^2} + n + 1}}} \right) + {\tan ^{ - 1}}\left( {\dfrac{1}{{{n^2} + 3n + 3}}} \right) + ... + {\tan ^{ - 1}}\left( {\dfrac{1}{{1 + (n + 19)(n + 20)}}} \right)$ ,then .$\tan S$. is equal to :
    (a)$\dfrac{{20}}{{401 + 20n}}$ (b)$\dfrac{n}{{{n^2} + 20n + 1}}$

    (c)$\dfrac{{20}}{{{n^2} + 20n + 1}}$ (d)$\dfrac{n}{{401 + 20n}}$

Answer 1

Hint: Here, we will be using the inverse tangent formula for this. The formula is
${\tan ^{ - 1}}x - {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\dfrac{{x - y}}{{1 + xy}}} \right)$ .

Complete step-by-step answer:
$\eqalign{
  & S = {\tan ^{ - 1}}\left( {\dfrac{1}{{{n^2} + n + 1}}} \right) + {\tan ^{ - 1}}\left( {\dfrac{1}{{{n^2} + 3n + 3}}} \right) + ... + {\tan ^{ - 1}}\left( {\dfrac{1}{{1 + (n + 19)(n + 20)}}} \right) \cr
  & = {\tan ^{ - 1}}\left( {\dfrac{1}{{1 + n(n + 1)}}} \right) + {\tan ^{ - 1}}\left( {\dfrac{1}{{1 + (n + 1)(n + 2)}}} \right) + ... + {\tan ^{ - 1}}\left( {\dfrac{1}{{1 + (n + 19)(n + 20)}}} \right) \cr
  & = {\tan ^{ - 1}}\left( {\dfrac{{(n + 1) - n}}{{1 + n(n + 1)}}} \right) + {\tan ^{ - 1}}\left( {\dfrac{{(n + 2) - (n + 1)}}{{1 + (n + 1)(n + 2)}}} \right) + ... + {\tan ^{ - 1}}\left( {\dfrac{{(n + 20) - (n + 19)}}{{1 + (n + 19)(n + 20)}}} \right) \cr
  & = {\tan ^{ - 1}}(n + 1) - {\tan ^{ - 1}}(n) + {\tan ^{ - 1}}(n + 2) - {\tan ^{ - 1}}(n + 1) + ... + {\tan ^{ - 1}}(n + 20) - {\tan ^{ - 1}}(n + 19) \cr
  & = {\tan ^{ - 1}}(n + 20) - {\tan ^{ - 1}}(n)\,\,\,\,\,\,\,\,[as\,\,other\,\,terms\,\,are\,\,vanished] \cr
  & = {\tan ^{ - 1}}\left( {\dfrac{{(n + 20) - n}}{{1 + n(n + 20)}}} \right) \cr
  & = {\tan ^{ - 1}}\left( {\dfrac{{20}}{{{n^2} + 20n + 1}}} \right) \cr} $
Hence,$\tan \,S = \dfrac{{20}}{{{n^2} + 20n + 1}}.$
So, option (c) is correct here.

Note: For solving,such type of problems inverse trigonometric rule is necessary here. These inverse functions in trigonometry are used to get the angle with any of the trigonometry ratios. In expansion we will left out with only 2 values other values will be cancelled.