Question Number 18184 by Tinkutara last updated on 16/Jul/17

Commented by prakash jain last updated on 16/Jul/17

Commented by Tinkutara last updated on 17/Jul/17

Answered by ajfour last updated on 16/Jul/17
![(√((m+2)^2 +(2m−1)^2 ))sin (θ+tan^(−1) ((2m−1)/(m+2)))=2m+1 θ=−tan^(−1) ((2m−1)/(m+2))+sin^(−1) ((2m+1)/( (√((m−2)^2 +(2m+1)^2 )))) =−tan^(−1) ((2m−1)/(m+2))+tan^(−1) ((2m+1)/(m−2)) tan θ=((((2m+1)/(m−2))−((2m−1)/(m+2)))/(1+((2m+1)/(m−2)).((2m−1)/(m+2)))) = (((2m+1)(m+2)−(2m−1)(m−2))/(m^2 −4+4m^2 −1)) tan θ=((10m)/(5(m^2 −1))) =((2m)/(m^2 −1)) . [ option (3) ,(2) ] for m=2 , tan θ=(4/3).](https://www.tinkutara.com/question/Q18189.png)
Commented by ajfour last updated on 16/Jul/17

Commented by Tinkutara last updated on 16/Jul/17
