Question

Find the value of the series \[{\cot ^{ - 1}}({2.1^2}) + {\cot ^{ - 1}}({2.2^2}) + {\cot ^{ - 1}}({2.3^2}) + ...\infty \]
A.\[\dfrac{\pi }{4}\]
B.\[\dfrac{\pi }{3}\]
C.\[\dfrac{\pi }{2}\]
D.\[\dfrac{\pi }{5}\]

Answer 1

Hint First write the nth term of the given series. Then convert the nth term to tangent function. Then set the obtained expression in the form \[{\tan ^{ - 1}}(\dfrac{{A + B}}{{1 - AB}})\] so that we can write it as \[{\tan ^{ - 1}}A + {\tan ^{ - 1}}B\] and then substitute the values for \[n = 1,2,3,...\] and calculate to obtain the required result.

Formula Used:
1.\[{\cot ^{ - 1}}x = {\tan ^{ - 1}}\dfrac{1}{x}\]
2. \[{\tan ^{ - 1}}A + {\tan ^{ - 1}}B = {\tan ^{ - 1}}\left( {\dfrac{{A + B}}{{1 - AB}}} \right)\]
3.\[{\tan ^{ - 1}}1 = \dfrac{\pi }{4}\]
4.\[{\tan ^{ - 1}}\infty = \dfrac{\pi }{2}\]

Complete step by step solution
The given series is,
\[{\cot ^{ - 1}}({2.1^2}) + {\cot ^{ - 1}}({2.2^2}) + {\cot ^{ - 1}}({2.3^2}) + ...\infty \]
\[ = \sum\limits_{n = 1}^\infty {{{\cot }^{ - 1}}\left( {2.{n^2}} \right)} \]
Use the formula \[{\cot ^{ - 1}}x = {\tan ^{ - 1}}\dfrac{1}{x}\] to convert the expression into a tangent function.
\[ = \sum\limits_{n = 1}^\infty {{{\tan }^{ - 1}}\left( {\dfrac{1}{{2{n^2}}}} \right)} \]
\[ = \sum\limits_{n = 1}^\infty {{{\tan }^{ - 1}}\left( {\dfrac{{(1 + 2n) + (1 - 2n)}}{{1 - (1 + 2n)(1 - 2n)}}} \right)} \]
Use the formula \[{\tan ^{ - 1}}\left( {\dfrac{{A + B}}{{1 - AB}}} \right) = {\tan ^{ - 1}}A + {\tan ^{ - 1}}B\]for further calculation.
\[ = \sum\limits_{n = 1}^\infty {[{{\tan }^{ - 1}}\left( {1 + 2n} \right)} + {\tan ^{ - 1}}\left( {1 - 2n} \right)]\]
Substitute the values for \[n = 1,2,3,...\]in the expression \[\sum\limits_{n = 1}^\infty {[{{\tan }^{ - 1}}\left( {1 + 2n} \right)} + {\tan ^{ - 1}}\left( {1 - 2n} \right)]\] to obtain the required answer.
\[ = {\tan ^{ - 1}}3 - {\tan ^{ - 1}}1 + {\tan ^{ - 1}}5 - {\tan ^{ - 1}}3 + {\tan ^{ - 1}}7 - {\tan ^{ - 1}}5 + .... + {\tan ^{ - 1}}\infty \]
\[ = - {\tan ^{ - 1}}1 + {\tan ^{ - 1}}\infty \]
Put \[{\tan ^{ - 1}}1 = \dfrac{\pi }{4}\] and \[{\tan ^{ - 1}}\infty = \dfrac{\pi }{2}\] and calculate.
\[ = - \dfrac{\pi }{4} + \dfrac{\pi }{2}\]
\[ = \dfrac{\pi }{4}\]
The correct option is A.

Note When we are solving this type of problems we use a lot of formulas in one question, so to solve this type of problems students must have the knowledge of these formulas otherwise they will not be unable to solve these problems.