If `c_j >0` for `i=1,2,....., n ` prove that `tan^(-1)((c_1x-y)/(c_1y+x))+tan^(-1)((c_2-c_1)/(1+c_2c_1))+tan^(-1)((c_3-c_2)/(1+c_3c_2))+.........+tan^(-1)1/(c_n)=tan^(-1)(x/y)`

Explanation

`tan^(-1)((c_1x-y)/(c_1y+x))
=>tan^(-1)((x/y-y/c_1)/(1+x/(yc_1)))`

`=>tan^(-1)(x/y)-tan^(-1)(1/(c_1))`

Similarly,

`=tan^(−1)(​1/(c_1​))−tan^(−1)(1/(c_2)​)`

Summing we get
,
`tan^(-1)((c_1x-y)/(c_1y+x))+tan^(-1)((c_2-c_1)/(1+c_2c_1))+tan^(-1)((c_3-c_2)/(1+c_3c_2))+.........+tan^(-1)1/(c_n)=tan^(-1)(x/y)`