Question

If $\tan ({\tan }^{-1}\frac{1}{2}+{\tan }^{-1}\frac{1}{3}+{\tan }^{-1}\frac{1}{4}+{\tan }^{-1}\frac{1}{5})$ is expressed as a rational $\frac{a}{b}$ in lowest form then $(a+b)$ has the value equal to

• A. 19
• B. 27
• C. 38
• D. 45

Answer 1

The correct option is A 19

$tan\left(ta{n}^{-1}\right(\frac{1}{2})+ta{n}^{-1}(\frac{1}{3})+ta{n}^{-1}(\frac{1}{4})+ta{n}^{-1}(\frac{1}{5}\left)\right)$
$=tan\left(ta{n}^{-1}\right(\frac{3+2}{6-1})+ta{n}^{-1}(\frac{5+4}{20-1}\left)\right)$
$=tan\left(ta{n}^{-1}\right(1)+ta{n}^{-1}(\frac{9}{19})$
$=tan\left(ta{n}^{-1}\right(\frac{19+9}{19-9}\left)\right)$
$=\frac{28}{10}$
$=\frac{14}{5}$
$=\frac{a}{b}$
Hence
$a+b=14+5$
$=19$