Question

Which of the below represents the value of ${{\tan }^{-1}}\left( \dfrac{1}{3} \right)+{{\tan }^{-1}}\left( \dfrac{1}{7} \right)+{{\tan }^{-1}}\left( \dfrac{1}{13} \right)+......+{{\tan }^{-1}}\left( \dfrac{1}{9703} \right)=$
(1) $\dfrac{\pi }{4}$
(2) $\dfrac{\pi }{6}$
(3) $\dfrac{\pi }{3}$
(4) ${{\tan }^{-1}}\left( 0.98 \right)$

Answer 1

Hint: Here in this question we have been asked to find the value of the given expression ${{\tan }^{-1}}\left( \dfrac{1}{3} \right)+{{\tan }^{-1}}\left( \dfrac{1}{7} \right)+{{\tan }^{-1}}\left( \dfrac{1}{13} \right)+......+{{\tan }^{-1}}\left( \dfrac{1}{9703} \right)=$ . For answering this question we will use the following identity ${{\tan }^{-1}}x-{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x-y}{1+xy} \right)$ from the concepts of inverse trigonometry.

Complete step-by-step solution:
Now considering from the question we have been asked to find the value of the given expression ${{\tan }^{-1}}\left( \dfrac{1}{3} \right)+{{\tan }^{-1}}\left( \dfrac{1}{7} \right)+{{\tan }^{-1}}\left( \dfrac{1}{13} \right)+......+{{\tan }^{-1}}\left( \dfrac{1}{9703} \right)=$ .
From the basic concepts of inverse trigonometry we know the following identity given as ${{\tan }^{-1}}x-{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x-y}{1+xy} \right)$ .
Now we can write
$\begin{align}
  & {{\tan }^{-1}}\left( \dfrac{1}{3} \right)={{\tan }^{-1}}\left( \dfrac{2-1}{1+2} \right) \\
 & \Rightarrow {{\tan }^{-1}}2-{{\tan }^{-1}}1 \\
\end{align}$ .
And similarly we can write
$\begin{align}
  & {{\tan }^{-1}}\left( \dfrac{1}{7} \right)={{\tan }^{-1}}\left( \dfrac{3-2}{1+3\left( 2 \right)} \right) \\
 & \Rightarrow {{\tan }^{-1}}3-{{\tan }^{-1}}2 \\
\end{align}$.
Similarly we can also expand
$\begin{align}
  & {{\tan }^{-1}}\left( \dfrac{1}{9703} \right)={{\tan }^{-1}}\left( \dfrac{99-98}{1+99\left( 98 \right)} \right) \\
 & \Rightarrow {{\tan }^{-1}}99-{{\tan }^{-1}}98 \\
\end{align}$ .
Now by using the above simplifications for every term in the given expression we can write the given expression as
$\begin{align}
  & \Rightarrow {{\tan }^{-1}}2-{{\tan }^{-1}}1+{{\tan }^{-1}}3-{{\tan }^{-1}}2+..........+{{\tan }^{-1}}99-{{\tan }^{-1}}98 \\
 & \Rightarrow {{\tan }^{-1}}99-{{\tan }^{-1}}1 \\
\end{align}$.
Because every term except those two terms will get cancelled by each corresponding term.
By further simplify the above expression using the same identity we will get $\Rightarrow {{\tan }^{-1}}99-{{\tan }^{-1}}1={{\tan }^{-1}}\left( \dfrac{99-1}{1+99} \right)$ .
Now by simplifying this further we will get $\Rightarrow {{\tan }^{-1}}\left( \dfrac{98}{100} \right)={{\tan }^{-1}}\left( 0.98 \right)$ .
Therefore we can conclude that the value of the given expression ${{\tan }^{-1}}\left( \dfrac{1}{3} \right)+{{\tan }^{-1}}\left( \dfrac{1}{7} \right)+{{\tan }^{-1}}\left( \dfrac{1}{13} \right)+......+{{\tan }^{-1}}\left( \dfrac{1}{9703} \right)$ is ${{\tan }^{-1}}\left( 0.98 \right)$ .
Hence we will mark the option “4” as correct.

Note: This is a very easy and simple question based on just a single formula only. If we are aware of the formula surely, then we can answer the question within a short span of time. Very few mistakes are possible in questions of this type. Similar to the inverse trigonometric identity used in this question there is another identity given as ${{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)$