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Question: Which of the below represents the value of \({{\tan }^{-1}}\left( \dfrac{1}{3} \right)+{{\tan }^{-1}...

Which of the below represents the value of tan⁡−1(13)+tan⁡−1(17)+tan⁡−1(113)+......+tan⁡−1(19703)={{\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) π4\dfrac{\pi }{4}
(2) π6\dfrac{\pi }{6}
(3) π3\dfrac{\pi }{3}
(4) tan⁡−1(0.98){{\tan }^{-1}}\left( 0.98 \right)

Explanation

Solution

Here in this question we have been asked to find the value of the given expression tan⁡−1(13)+tan⁡−1(17)+tan⁡−1(113)+......+tan⁡−1(19703)={{\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⁡−1x−tan⁡−1y=tan⁡−1(x−y1+xy){{\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(13)+tan⁡−1(17)+tan⁡−1(113)+......+tan⁡−1(19703)={{\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⁡−1x−tan⁡−1y=tan⁡−1(x−y1+xy){{\tan }^{-1}}x-{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x-y}{1+xy} \right) .
Now we can write
tan⁡−1(13)=tan⁡−1(2−11+2) ⇒tan⁡−12−tan⁡−11 \begin{aligned} & {{\tan }^{-1}}\left( \dfrac{1}{3} \right)={{\tan }^{-1}}\left( \dfrac{2-1}{1+2} \right) \\\ & \Rightarrow {{\tan }^{-1}}2-{{\tan }^{-1}}1 \\\ \end{aligned} .
And similarly we can write
tan⁡−1(17)=tan⁡−1(3−21+3(2)) ⇒tan⁡−13−tan⁡−12 \begin{aligned} & {{\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{aligned}.
Similarly we can also expand
tan⁡−1(19703)=tan⁡−1(99−981+99(98)) ⇒tan⁡−199−tan⁡−198 \begin{aligned} & {{\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{aligned} .
Now by using the above simplifications for every term in the given expression we can write the given expression as
⇒tan⁡−12−tan⁡−11+tan⁡−13−tan⁡−12+..........+tan⁡−199−tan⁡−198 ⇒tan⁡−199−tan⁡−11 \begin{aligned} & \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{aligned}.
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 ⇒tan⁡−199−tan⁡−11=tan⁡−1(99−11+99)\Rightarrow {{\tan }^{-1}}99-{{\tan }^{-1}}1={{\tan }^{-1}}\left( \dfrac{99-1}{1+99} \right) .
Now by simplifying this further we will get ⇒tan⁡−1(98100)=tan⁡−1(0.98)\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(13)+tan⁡−1(17)+tan⁡−1(113)+......+tan⁡−1(19703){{\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(0.98){{\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⁡−1x+tan⁡−1y=tan⁡−1(x+y1−xy){{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)