InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
If `x^2+y^2+z^2=r^2,t h e ntan^(-1)((x y)/(z r))+tan^(-1)((y z)/(x r))+tan^(-1)((x z)/(y r))`is equal to`pi`(b) `pi/2`(c) 0(d) none of these |
|
Answer» `tan^-1((xy)/(zr))+tan^-1((yz)/(xr))+tan^-1((xz)/(yr))` `=tan^-1(((xy)/(zr)+(yz)/(xr))/(1-(xy)/(zr)(yz)/(xr)))+tan^-1((xz)/(yr))` `=tan^-1(((xy)/(zr)+(yz)/(xr))/(1-(xy)/(zr)(yz)/(xr)))+tan^-1((xz)/(yr))` `=tan^-1((x^2yr+yz^2r)/(r^2zx-y^2xz))+tan^-1((xz)/(yr))` `=tan^-1((yr(x^2+z^2))/(zx(r^2-y^2)))+tan^-1((xz)/(yr))` As, `x^2+y^2 +z^2 = r^2 =>x^2+z^2 = r^2-y^2` so, it becomes, `=tan^-1((yr(x^2+z^2))/(zx(x^2+z^2)))+tan^-1((xz)/(yr))` `=tan^-1((yr)/(zx))+tan^-1((xz)/(yr))` `=tan^-1((yr)/(zx))+cot^-1((yr)/(zx))` `=pi/2` So, option `(b)` is the correct option. |
|