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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

A

`pi`

B

`(pi)/(2)`

C

0

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
B

We have
`(xy)/(zr)(yz)/(xr) = (y^(2))/(r^(2)) = (y^(2))/(x^(2) + y^(2) + z^(2)) lt 1`
`rArr tan^(-1) ((xy)/(zr)) + tan^(-1) ((yz)/(xr)) + tan^(-1) ((xz)/(yr))`
`= tan^(-1) (((xy)/(zr) + (yz)/(xr))/(1-(zy)/(zr) (yz)/(yz))) + tan^(-1) ((xz)/(yr))`
`= tan^(-1) (((y(x^(2) + z^(2)))/(xzy))/((r^(2) -y^(2))/(r^(2)))) + tan^(-1) ((xz)/(yr))`
`= tan^(-1) (((yr(x^(2) + z^(2)))/(xz))/((x^(2) + z^(2)))) + tan^(-1) ((xz)/(yr))`
`= tan^(-1) ((yr)/(xz)) + tan^(-1) ((xz)/(yr)) = (pi)/(2)`
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