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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

A

`pi`

B

`(pi)/(2)`

C

0

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
B
`because"tan"^(-1)(yz)/(xr)+"tan"^(-1)(zx)/(yr)`
`rArr"tan"^(-1)(z)/(r)(((y)/(x)+(x)/(y)))/(1-(z^(2))/(r^(2)))=tan^(-1)((zr)/(xy)*(x^(2)+y^(2))/(x^(2)+y^(2)))=cot^(-1)((xy)/(zr))`
`therefore"tan"^(-1)(xy)/(zr)="cot"^(-1)(xy)/(zr)=(pi)/(2)`
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