Question

The value of ${\tan }^{-1}\left(\frac{1}{3}\right)+{\tan }^{-1}\left(\frac{2}{9}\right)+{\tan }^{-1}\left(\frac{4}{33}\right)+{\tan }^{-1}\left(\frac{8}{129}\right)+$.......$n$ terms is

• A. ${\tan }^{-1}{2}^n-\frac{\pi }{4}$
• B. ${\tan }^{-1}{2}^n$
• C. ${\cot }^{-1}{2}^n$
• D. $\frac{{\sin }^{-1}{2}^n}{{\cos }^{-1}{2}^n}$

Answer 1

Answer: (A)

${\tan }^{-1}{2}^n-\frac{\pi }{4}$

${\tan }^{-1}\left(\frac{1}{3}\right)+{\tan }^{-1}\left(\frac{2}{9}\right)+{\tan }^{-1}\left(\frac{4}{33}\right)+{\tan }^{-1}\left(\frac{8}{129}\right)+....n$ terms.

$={\tan }^{-1}\left(\frac{2-1}{1+2.1}\right)+{\tan }^{-1}\left(\frac{4-2}{1+4.2}\right)+{\tan }^{-1}\left(\frac{8-4}{1+8.4}\right)+{\tan }^{-1}\left(\frac{16-8}{1+16\times 8}\right)+...+{\tan }^{-1}\left(\frac{2^n-2^{n-1}}{1+2^n\times 2^{n-1}}\right)$

we know that ${\tan }^{-1}x-{\tan }^{-1}y={\tan }^{-1}\frac{x-y}{1+xy}$

$={\tan }^{-1}2-{\tan }^{-1}1+{\tan }^{-1}4-{\tan }^{-1}2+{\tan }^{-1}8-{\tan }^{-1}4+{\tan }^{-1}16-{\tan }^{-1}8+...+{\tan }^{-1}{2}^n-{\tan }^{-1}{2}^{n-1}$

$={\tan }^{-1}{2}^n-{\tan }^{-1}1$

$={\tan }^{-1}{2}^n-\frac{\pi }{4}$