Question

The sum of the infinite terms of the series ${\cot ^{ - 1}}\left( {{1^2} + \dfrac{3}{4}} \right) + {\cot ^{ - 1}}\left( {{2^2} + \dfrac{3}{4}} \right) + ...$ is equal to
A ) ${\tan ^{ - 1}}1$
B ) ${\tan ^{ - 1}}2$
C ) ${\tan ^{ - 1}}3$
D ) ${\tan ^{ - 1}}4$

Answer 1

Hint:- We will have to find out the nth term of the given series and reduce it using the inverse trigonometric identities of tan . Once that is done, we will have to use the concept of limits to further solve the question

Complete step-by-step answer:
It has been given that we have to find the sum of infinite terms of the given series ${\cot ^{ - 1}}\left( {{1^2} + \dfrac{3}{4}} \right) + {\cot ^{ - 1}}\left( {{2^2} + \dfrac{3}{4}} \right) + ...$
So, we can write the sum as
 ${S_\infty } = {\cot ^{ - 1}}\left( {{1^2} + \dfrac{3}{4}} \right) + {\cot ^{ - 1}}\left( {{2^2} + \dfrac{3}{4}} \right) + ...\infty $
Let us proceed by finding out the ${n^{th}}$ term of the given series. Let ${t_n}$ denote the ${n^{th}}$ term of the given series.
Thus, on observation we can write the ${n^{th}}$ term as
 ${t_n} = {\cot ^{ - 1}}\left( {{n^2} + \dfrac{3}{4}} \right)$
We know that ${\cot ^{ - 1}}\theta = {\tan ^{ - 1}}\left( {\dfrac{1}{\theta }} \right)$
Using this formula, we can rewrite the ${n^{th}}$ term as
${t_n} = {\tan ^{ - 1}}\left( {\dfrac{1}{{{n^2} + \dfrac{3}{4}}}} \right)$
In order to make use of the formula ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$ we will rewrite $\dfrac{3}{4}$ as $\left( {1 - \dfrac{1}{4}} \right)$ as $\dfrac{1}{4}$ is a perfect square.
Thus, we will get
${t_n} = {\tan ^{ - 1}}\left( {\dfrac{1}{{1 + \left( {{n^2} - \dfrac{1}{4}} \right)}}} \right)$
Now, using the formula ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$ , we get
${t_n} = {\tan ^{ - 1}}\left( {\dfrac{1}{{1 + \left( {n + \dfrac{1}{2}} \right)\left( {n - \dfrac{1}{2}} \right)}}} \right)$
 In order to make use of the formula ${\tan ^{ - 1}}a - {\tan ^{ - 1}}b = {\tan ^{ - 1}}\dfrac{{a - b}}{{1 + ab}}$ , we will rewrite the numerator of the ${n^{th}}$ term $1 = \left( {n + \dfrac{1}{2}} \right) - \left( {n - \dfrac{1}{2}} \right)$

Thus, we get
${t_n} = {\tan ^{ - 1}}\left( {\dfrac{{\left( {n + \dfrac{1}{2}} \right) - \left( {n - \dfrac{1}{2}} \right)}}{{1 + \left( {n + \dfrac{1}{2}} \right)\left( {n - \dfrac{1}{2}} \right)}}} \right)$
Now using the formula ${\tan ^{ - 1}}a - {\tan ^{ - 1}}b = {\tan ^{ - 1}}\dfrac{{a - b}}{{1 + ab}}$ , we get the ${n^{th}}$ term as
${t_n} = {\tan ^{ - 1}}\left( {n + \dfrac{1}{2}} \right) - {\tan ^{ - 1}}\left( {n - \dfrac{1}{2}} \right)$
Now we can write the sum of n terms as ${S_n} = {t_1} + {t_2} + {t_3} + ... + {t_n}$
Writing down the individual terms by replacing the corresponding values of n, we get
${S_n} = {\tan ^{ - 1}}\left( {\dfrac{3}{2}} \right) - {\tan ^{ - 1}}\left( {\dfrac{1}{2}} \right) + {\tan ^{ - 1}}\left( {\dfrac{5}{2}} \right) - {\tan ^{ - 1}}\left( {\dfrac{3}{2}} \right) + ... + {\tan ^{ - 1}}\left( {n + \dfrac{1}{2}} \right) - {\tan ^{ - 1}}\left( {n - \dfrac{1}{2}} \right)$
Cancelling out the terms, we get
${S_n} = {\tan ^{ - 1}}\left( {n + \dfrac{1}{2}} \right) - {\tan ^{ - 1}}\left( {\dfrac{1}{2}} \right)$
When $n \to \infty ,{S_n} \to {S_\infty }$
Thus, the sum of infinite terms of the given series can be written as
${S_\infty } = {\lim _{n \to \infty }}{S_n}$
Replacing ${S_n}$ in the above equation, we get
${S_\infty } = {\lim _{n \to \infty }}\left[ {{{\tan }^{ - 1}}\left( {n + \dfrac{1}{2}} \right) - {{\tan }^{ - 1}}\left( {\dfrac{1}{2}} \right)} \right]$
Putting $n \to \infty $ in the above equation, we get
${S_\infty } = {\tan ^{ - 1}}\left( { \to \infty } \right) - {\tan ^{ - 1}}\left( {\dfrac{1}{2}} \right)$
We know that $\tan \dfrac{\pi }{2} \to \infty $ . Using this we get
${S_\infty } = \dfrac{\pi }{2} - {\tan ^{ - 1}}\left( {\dfrac{1}{2}} \right)$
Using the identity ${\tan ^{ - 1}}\theta + {\cot ^{ - 1}}\theta = \dfrac{\pi }{2}$ in the above equation, we get
${S_\infty } = {\cot ^{ - 1}}\left( {\dfrac{1}{2}} \right)$
We know that ${\cot ^{ - 1}}\theta = {\tan ^{ - 1}}\left( {\dfrac{1}{\theta }} \right)$
Using this formula, we can rewrite the value of ${S_\infty }$ as

${S_\infty } = {\tan ^{ - 1}}2$

The correct option is (B)

Note:- In these types of questions, it is necessary to find out the nth term of the given series and break it into two terms with a minus sign in between. Then on adding, we can cancel out of terms to reduce the number of terms in the summation expression of n terms. For doing the above, we need to take help of the inverse trigonometric identities of tan.