Question

Evaluate: ${\tan }^{-1}\left(\frac{1}{3}\right)+{\tan }^{-1}\left(\frac{1}{7}\right)+{\tan }^{-1}\left(\frac{1}{13}\right)+\dots {\tan }^{-1}\left(\frac{1}{9703}\right)$

• A. $\frac{\mathrm{\pi }}{4}$
• B. $\frac{\mathrm{\pi }}{6}$
• C. $\frac{\mathrm{\pi }}{3}$
• D. ${\tan }^{-1}\left(0.98\right)$

Answer 1

The correct option is D

${\tan }^{-1}\left(0.98\right)$

Explanation for the correct option:

Step 1: Evaluating the given equation

${\tan }^{-1}\left(\frac{1}{3}\right)+{\tan }^{-1}\left(\frac{1}{7}\right)+{\tan }^{-1}\left(\frac{1}{13}\right)+\dots {\tan }^{-1}\left(\frac{1}{9703}\right)$

We know the formula, ${\tan }^{-1}\left(\frac{a-b}{1+ab}\right)={\tan }^{-1}\left(a\right)-{\tan }^{-1}\left(b\right)$

By using this formula,

$$\begin{gathered} {\tan }^{-1}\left(\frac{1}{3}\right)={\tan }^{-1}\left(\frac{2-1}{1+2\times 1}\right) \\ ={\tan }^{-1}\left(2\right)-{\tan }^{-1}\left(1\right)......\left(1\right) \end{gathered}$$

$$\begin{gathered} {\tan }^{-1}\left(\frac{1}{7}\right)={\tan }^{-1}\left(\frac{3-2}{1+3\times 2}\right) \\ ={\tan }^{-1}\left(3\right)-{\tan }^{-1}\left(2\right)......\left(2\right) \end{gathered}$$

$$\begin{gathered} {\tan }^{-1}\left(\frac{1}{9703}\right)={\tan }^{-1}\left(\frac{99-98}{1+98\times 99}\right) \\ ={\tan }^{-1}\left(99\right)-{\tan }^{-1}\left(98\right)......\left(3\right) \end{gathered}$$

Step 2: Adding equations $\left(1\right),\left(2\right),\left(3\right)$

$$\begin{gathered} \mathrm{S}={\tan }^{-1}\left(2\right)-{\tan }^{-1}\left(1\right)+{\tan }^{-1}\left(3\right)-{\tan }^{-1}\left(2\right)+\dots \dots {\tan }^{-1}\left(99\right)-{\tan }^{-1}\left(98\right) \\ ={\tan }^{-1}\left(99\right)-{\tan }^{-1}\left(1\right) \\ ={\tan }^{-1}\left(\frac{99-1}{1+99\times 1}\right) \\ ={\tan }^{-1}\left(\frac{98}{100}\right) \\ ={\tan }^{-1}\left(0.98\right) \end{gathered}$$

Hence, option (D) is the correct answer.