Discussion Forum

The value of $\tan^{-1}\frac{1}{2}+\tan^{-1}\frac{1}{3}+\tan^{-1}\frac{7}{8}$ is

Answer:

Explanation:

$\tan^{-1}\frac{1}{2}+\tan^{-1}\frac{1}{3}+\tan^{-1}\frac{7}{8}$

$=\left[\frac{\frac{1}{2}+\frac{1}{3}+\frac{7}{8}-\frac{1}{2}*\frac{1}{3}*\frac{7}{8}}{1-\frac{1}{2}*\frac{1}{3}-\frac{1}{3}*\frac{7}{8}-\frac{7}{8}*\frac{1}{2}}\right]$

$[\therefore\tan^{-1}x+\tan^{-1}y+\tan^{-1}z=\tan^{-1}(\frac{x+y+z-xyz}{1-xy-yz-zx})]$

$=\tan^{-1}[\frac{\frac{41}{24}-\frac{7}{48}}{1-\frac{1}{6}-\frac{7}{24}-\frac{7}{16}}]$

$=\tan^{-1}[\frac{\frac{75}{48}}{1-\frac{43}{48}}]=\tan^{-1}(\frac{75}{48-43})$

$=\tan^{-1}[\frac{75}{5}]=\tan^{-1}15$