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If x^2+y^2+z^2=r^2,t h e ntan^(-1)((x y)...

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

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`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))`
...
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CENGAGE-INVERSE TRIGONOMETRIC FUNCTIONS-Solved Examples And Exercises
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