Question

The sum of the infinite terms of the series ${\cot }^{-1}\left(1^2+{\displaystyle \frac{3}{4}}\right)+{\cot }^{-1}\left(2^2+{\displaystyle \frac{3}{4}}\right)+...$ 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 ${\cot }^{-1}\left(1^2+{\displaystyle \frac{3}{4}}\right)+{\cot }^{-1}\left(2^2+{\displaystyle \frac{3}{4}}\right)+...$
So, we can write the sum as
 $S_{\mathrm{\infty }}={\cot }^{-1}\left(1^2+{\displaystyle \frac{3}{4}}\right)+{\cot }^{-1}\left(2^2+{\displaystyle \frac{3}{4}}\right)+...\mathrm{\infty }$
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
 $t_n={\cot }^{-1}\left(n^2+{\displaystyle \frac{3}{4}}\right)$
We know that ${\cot }^{-1}\theta ={\tan }^{-1}\left({\displaystyle \frac{1}{\theta }}\right)$
Using this formula, we can rewrite the $n^{th}$ term as
$t_n={\tan }^{-1}\left({\displaystyle \frac{1}{n^2+{\displaystyle \frac{3}{4}}}}\right)$
In order to make use of the formula $a^2-b^2=\left(a-b\right)\left(a+b\right)$ we will rewrite ${\displaystyle \frac{3}{4}}$ as $\left(1-{\displaystyle \frac{1}{4}}\right)$ as ${\displaystyle \frac{1}{4}}$ is a perfect square.
Thus, we will get
$t_n={\tan }^{-1}\left({\displaystyle \frac{1}{1+\left(n^2-{\displaystyle \frac{1}{4}}\right)}}\right)$
Now, using the formula $a^2-b^2=\left(a-b\right)\left(a+b\right)$ , we get
$t_n={\tan }^{-1}\left({\displaystyle \frac{1}{1+\left(n+{\displaystyle \frac{1}{2}}\right)\left(n-{\displaystyle \frac{1}{2}}\right)}}\right)$
 In order to make use of the formula ${\tan }^{-1}a-{\tan }^{-1}b={\tan }^{-1}{\displaystyle \frac{a-b}{1+ab}}$ , we will rewrite the numerator of the $n^{th}$ term $1=\left(n+{\displaystyle \frac{1}{2}}\right)-\left(n-{\displaystyle \frac{1}{2}}\right)$

Thus, we get
$t_n={\tan }^{-1}\left({\displaystyle \frac{\left(n+{\displaystyle \frac{1}{2}}\right)-\left(n-{\displaystyle \frac{1}{2}}\right)}{1+\left(n+{\displaystyle \frac{1}{2}}\right)\left(n-{\displaystyle \frac{1}{2}}\right)}}\right)$
Now using the formula ${\tan }^{-1}a-{\tan }^{-1}b={\tan }^{-1}{\displaystyle \frac{a-b}{1+ab}}$ , we get the $n^{th}$ term as
$t_n={\tan }^{-1}\left(n+{\displaystyle \frac{1}{2}}\right)-{\tan }^{-1}\left(n-{\displaystyle \frac{1}{2}}\right)$
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
$S_n={\tan }^{-1}\left({\displaystyle \frac{3}{2}}\right)-{\tan }^{-1}\left({\displaystyle \frac{1}{2}}\right)+{\tan }^{-1}\left({\displaystyle \frac{5}{2}}\right)-{\tan }^{-1}\left({\displaystyle \frac{3}{2}}\right)+...+{\tan }^{-1}\left(n+{\displaystyle \frac{1}{2}}\right)-{\tan }^{-1}\left(n-{\displaystyle \frac{1}{2}}\right)$
Cancelling out the terms, we get
$S_n={\tan }^{-1}\left(n+{\displaystyle \frac{1}{2}}\right)-{\tan }^{-1}\left({\displaystyle \frac{1}{2}}\right)$
When $n\to \mathrm{\infty },S_n\to S_{\mathrm{\infty }}$
Thus, the sum of infinite terms of the given series can be written as
$S_{\mathrm{\infty }}=\underset{n\to \mathrm{\infty }}{lim}{S}_n$
Replacing $S_n$ in the above equation, we get
$S_{\mathrm{\infty }}=\underset{n\to \mathrm{\infty }}{lim}\left[{\tan }^{-1}\left(n+{\displaystyle \frac{1}{2}}\right)-{\tan }^{-1}\left({\displaystyle \frac{1}{2}}\right)\right]$
Putting $n\to \mathrm{\infty }$ in the above equation, we get
$S_{\mathrm{\infty }}={\tan }^{-1}\left(\to \mathrm{\infty }\right)-{\tan }^{-1}\left({\displaystyle \frac{1}{2}}\right)$
We know that $\tan {\displaystyle \frac{\pi }{2}}\to \mathrm{\infty }$ . Using this we get
$S_{\mathrm{\infty }}={\displaystyle \frac{\pi }{2}}-{\tan }^{-1}\left({\displaystyle \frac{1}{2}}\right)$
Using the identity ${\tan }^{-1}\theta +{\cot }^{-1}\theta ={\displaystyle \frac{\pi }{2}}$ in the above equation, we get
$S_{\mathrm{\infty }}={\cot }^{-1}\left({\displaystyle \frac{1}{2}}\right)$
We know that ${\cot }^{-1}\theta ={\tan }^{-1}\left({\displaystyle \frac{1}{\theta }}\right)$
Using this formula, we can rewrite the value of $S_{\mathrm{\infty }}$ as

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

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.