Question

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

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

Answer 1

Answer: (A)

$\tan^{-1}\left(\frac{16}{13}\right)$

Answer 2

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