Question

${\tan }^{-1}\frac{c_{1}x-y}{c_{1}y+x}+{\tan }^{-1}\frac{c_2-c_1}{1+c_2{c}_1}+{\tan }^{-1}\frac{c_3-c_2}{1+c_3{c}_2}+.....+{\tan }^{-1}\frac{1}{c_n}=k{\tan }^{-1}\left(\frac{x}{y}\right)$
Find the value of $k$

Answer 1

Consider
${\tan }^{-1}\left(\frac{c_n-c_{n-1}}{1+c_n.c_{n-1}}\right)$
$={\tan }^{-1}\left(\frac{\frac{1}{c_{n-1}}-\frac{1}{c_n}}{1+\frac{1}{c_n}.\frac{1}{c_{n-1}}}\right)$
$={\tan }^{-1}\left(\frac{1}{c_{n-1}}\right)-{\tan }^{-1}\left(\frac{1}{c_n}\right)$
Hence
${\tan }^{-1}\left(\frac{c_2-c_1}{1+c_1.c_2}\right)+{\tan }^{-1}\left(\frac{c_2-c_3}{1+c_2.c_3}\right)+....+{\tan }^{-1}\left(\frac{1}{c_n}\right)$
$={\tan }^{-1}\left(\frac{1}{c_1}\right)-{\tan }^{-1}\left(\frac{1}{c_2}\right)+{\tan }^{-1}\left(\frac{1}{c_2}\right)-{\tan }^{-1}\left(\frac{1}{c_3}\right)....+{\tan }^{-1}\left(\frac{1}{c_{n-1}}\right)-{\tan }^{-1}\left(\frac{1}{c_n}\right)+{\tan }^{-1}\left(\frac{1}{c_n}\right)$
$={\tan }^{-1}\left(\frac{1}{c_1}\right)$ ...(i)
Now
${\tan }^{-1}\frac{c_{1}x-y}{c_{1}y+x}+{\tan }^{-1}\left(\frac{1}{c_1}\right)$
$={\tan }^{-1}\left(\frac{\frac{c_{1}x-y}{c_{1}y+x}+\frac{1}{c_1}}{1-\frac{1}{c_1}.\frac{c_{1}x-y}{c_{1}y+x}}\right)$
$={\tan }^{-1}\left(\frac{c_1^{2}x-c_{1}y+c_{1}y+x}{c_1^{2}y+c_{1}x-c_{1}x+y}\right)$
$={\tan }^{-1}\left(\frac{c_1^{2}x+x}{c_1^{2}y+y}\right)$
$={\tan }^{-1}\left(\frac{x}{y}\right)$.