Question

If $\sum _{n=1}^5\sum _{m=1}^5{\tan }^{-1}\frac{m}{n}=k\pi ,k\in R,$ then $\left[k\right]=$
where ($[.]$ denotes greatest integer function)

Answer 1

Given: $\sum _{n=1}^5\sum _{m=1}^5{\tan }^{-1}\frac{m}{n}=k\pi$
on expanding
L.H.S.$=({\tan }^{-1}\frac{1}{1}+{\tan }^{-1}\frac{1}{2}+\cdots +{\tan }^{-1}\frac{1}{5})+({\tan }^{-1}\frac{2}{1}+{\tan }^{-1}\frac{2}{2}+\cdots +{\tan }^{-1}\frac{2}{5})⋮+({\tan }^{-1}\frac{5}{1}+{\tan }^{-1}\frac{5}{2}+\cdots +{\tan }^{-1}\frac{5}{5})$
$=({\tan }^{-1}\frac{1}{1}+{\tan }^{-1}\frac{2}{2}+\cdots +{\tan }^{-1}\frac{5}{5})+({\tan }^{-1}\frac{1}{2}+{\tan }^{-1}\frac{2}{1}+{\tan }^{-1}\frac{1}{3}+{\tan }^{-1}\frac{3}{1}\cdots +{\tan }^{-1}\frac{1}{5}+{\tan }^{-1}\frac{5}{1})+({\tan }^{-1}\frac{2}{3}+{\tan }^{-1}\frac{3}{2}+{\tan }^{-1}\frac{2}{4}+{\tan }^{-1}\frac{4}{2}+{\tan }^{-1}\frac{2}{5}+{\tan }^{-1}\frac{5}{2})+({\tan }^{-1}\frac{3}{4}+{\tan }^{-1}\frac{4}{3}+{\tan }^{-1}\frac{3}{5}+{\tan }^{-1}\frac{5}{3})+({\tan }^{-1}\frac{4}{5}+{\tan }^{-1}\frac{5}{4})$
$(\because {\tan }^{-1}x+{\tan }^{-1}\frac{1}{x}=\frac{\pi }{2};x>0)$
$=5\times \frac{\pi }{4}+10\times \frac{\pi }{2}$
$=\frac{25\pi }{4}$
$\therefore \left[k\right]=\left[\frac{25}{4}\right]=6$

Alternate solution :
Let $f\left(x\right)=\sum _{n=1}^5\sum _{m=1}^5{\tan }^{-1}\frac{m}{n}=k\pi$
and $g\left(x\right)=\sum _{n=1}^5\sum _{m=1}^5{\tan }^{-1}\frac{n}{m}=k\pi$
$(\text{where}x=\frac{m}{n})$

Adding both, $f\left(x\right)+g\left(x\right)=2k\pi$
$\Rightarrow \sum _{n=1}^5\sum _{m=1}^5({\tan }^{-1}\frac{m}{n}+{\tan }^{-1}\frac{n}{m})=2k\pi \cdots \left(1\right)$
$\because \frac{m}{n}>0\Rightarrow {\tan }^{-1}\frac{m}{n}={\cot }^{-1}\frac{n}{m}$
So, equation $\left(1\right)$ becomes,
$\sum _{n=1}^5\sum _{m=1}^5\frac{\pi }{2}=2k\pi$
$\Rightarrow 25\times \frac{\pi }{2}=2k\pi$
$\Rightarrow k=\frac{25}{4}$
$\therefore \left[k\right]=\left[\frac{25}{4}\right]=6$