2022/013) How do you prove that $\tan^{-1}\frac{1}{5}+ \tan^{-1} \frac{1}{7} + \tan^{-1} \frac{1}{3} + \tan^{-1} \frac{1}{8} = \frac{\pi}{4}$

Let us have $\tan^{-1}\frac{1}{5} = \alpha$

$\tan^{-1}\frac{1}{7} = \beta$

$\tan^{-1}\frac{1}{3} = \gamma$

$\tan^{-1}\frac{1}{8} = \delta$

we need to find $\alpha + \beta + \gamma + \delta$

Let us add 2 at a time $\alpha + \beta$ , $\gamma +\delta$ and then $\alpha + \beta + \gamma + \delta$

We have 

$\tan \alpha = \frac{1}{5}\cdots(1)$

and  $\tan \beta = \frac{1}{7}\cdots(2)$

So $\tan (\alpha + \beta) = \frac{\tan (\alpha) + \tan (\beta)}{1 - \tan (\alpha) \tan (\beta)}$

$= \frac{\frac{1}{5} + \frac{1}{7}}{1 - \frac{1}{5} \frac{1}{7}}$

$ = \frac{\frac{12}{35}}{ \frac{34}{35}} = \frac{12}{34} = \frac{6}{17}$ 

We have 

$\tan \gamma = \frac{1}{3}\cdots(3)$

and  $\tan \delta  = \frac{1}{8}\cdots(4)$

So $\tan (\gamma + \delta) = \frac{\tan (\gamma) + \tan (\delta)}{1 - \tan (\gamma) \tan (\delta)}$

$= \frac{\frac{1}{3} + \frac{1}{8}}{1 - \frac{1}{3} \frac{1}{8}}$

$ = \frac{\frac{11}{24}}{ \frac{23}{24}} = \frac{11}{23}$ 

now $\tan (\alpha + \beta + \gamma + \delta)$

$=\frac{\tan (\alpha + \beta)  + \tan (\gamma + \delta)}{1 - \tan (\alpha + beta) \tan (\gamma + \delta)}$

$= \frac{\frac{6}{17} + \frac{11}{23}}{1 - \frac{6}{17} \frac{17}{23}}$

$ = \frac{\frac{6 * 23 + 11 * 17}{17 * 23}}{1-  \frac{66}{17 * 23}}$

$ = \frac{6 * 23 + 11 * 17}{17 * 23 - 66}$

$=\frac{ 325}{325} = 1$

so $\tan (\alpha + \beta + \gamma + \delta) = \tan \frac{\pi}{4}$

but we need to show that $\alpha + \beta + \gamma + \delta = \frac{\pi}{4}$

as it could be $\frac{5\pi}{4}$

but as  $\alpha,\beta,\gamma, \delta$ each is less than $\frac{\pi}{4}$ so sum is less than $\pi$ so it is $\frac{\pi}{4}$