Question

$2{\tan }^{-1}\left(\frac{1}{3}\right)+ta{n}^{-1}\left(\frac{1}{4}\right)=$

• A. ${\tan }^{-1}\left(\frac{16}{13}\right)$
• B. ${\tan }^{-1}\left(\frac{17}{23}\right)$
• C. $\frac{\pi }{4}$
• D. 0
• E. ${\tan }^{-1}\left(\frac{11}{12}\right)$

Answer 1

Answer: (A)

${\tan }^{-1}\left(\frac{1}{3}\right)+{\tan }^{-1}\left(\frac{1}{4}\right)$
$={\tan }^{-1}\frac{\left(2\times \frac{1}{3}\right)}{1-{\left(\frac{1}{3}\right)}^2}+{\tan }^{-1}\left(\frac{1}{4}\right)$
$={\tan }^{-1}\left[\frac{\left(\frac{2}{3}\right)}{\left(1-\frac{1}{9}\right)}\right]+{\tan }^{-1}\frac{1}{4}$
$={\tan }^{-1}\left(\frac{2}{3}\times \frac{9}{8}\right)+{\tan }^{-1}\frac{1}{4}$
$={\tan }^{-1}\frac{3}{4}+{\tan }^{-1}\frac{1}{4}$
$={\tan }^{-1}\left(\frac{\frac{3}{4}+\frac{1}{4}}{1-\frac{3}{4}\times \frac{1}{4}}\right)$
$={\tan }^{-1}\frac{1}{1-\frac{3}{16}}={\tan }^{-1}\left(\frac{16}{13}\right)$