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

Options

A. `pi`
B. `(pi)/(2)`
C. 0
D. none of these

Answer

To solve the problem, we need to evaluate the expression:

\[ \tan^{-1}\left(\frac{xy}{zr}\right) + \tan^{-1}\left(\frac{yz}{xr}\right) + \tan^{-1}\left(\frac{xz}{yr}\right) \]

given the condition \(x^2 + y^2 + z^2 = r^2\).

### Step-by-Step Solution:

1. **Use the Formula for the Sum of Inverse Tangents**:
We can use the identity for the sum of inverse tangents:
\[ \tan^{-1}(a) + \tan^{-1}(b) + \tan^{-1}(c) = \tan^{-1}\left(\frac{a + b + c - abc}{1 - ab - ac - bc}\right) \]
where \(a = \frac{xy}{zr}\), \(b = \frac{yz}{xr}\), and \(c = \frac{xz}{yr}\).

2. **Calculate \(a + b + c\)**:
\[ a + b + c = \frac{xy}{zr} + \frac{yz}{xr} + \frac{xz}{yr} \]
To combine these fractions, we find a common denominator, which is \(xyzr\):
\[ a + b + c = \frac{xy^2 + y^2z + x^2z}{xyzr} \]

3. **Calculate \(abc\)**:
\[ abc = \left(\frac{xy}{zr}\right) \left(\frac{yz}{xr}\right) \left(\frac{xz}{yr}\right) = \frac{xyz \cdot xyz}{(zr)(xr)(yr)} = \frac{x^2y^2z^2}{r^3xyz} = \frac{xyz}{r^3} \]

4. **Calculate \(1 - ab - ac - bc\)**:
We need to compute \(ab\), \(ac\), and \(bc\):
\[ ab = \frac{xy}{zr} \cdot \frac{yz}{xr} = \frac{y^2z}{r^2} \]
\[ ac = \frac{xy}{zr} \cdot \frac{xz}{yr} = \frac{x^2z}{r^2} \]
\[ bc = \frac{yz}{xr} \cdot \frac{xz}{yr} = \frac{y^2x}{r^2} \]
Therefore,
\[ ab + ac + bc = \frac{y^2z + x^2z + y^2x}{r^2} \]
Thus,
\[ 1 - ab - ac - bc = 1 - \frac{y^2z + x^2z + y^2x}{r^2} \]

5. **Substituting into the Inverse Tangent Formula**:
Now we substitute \(a + b + c\) and \(abc\) into the formula:
\[ \tan^{-1}\left(\frac{\frac{xy^2 + y^2z + x^2z}{xyzr} - \frac{xyz}{r^3}}{1 - \frac{y^2z + x^2z + y^2x}{r^2}}\right) \]

6. **Using the Given Condition**:
Since \(x^2 + y^2 + z^2 = r^2\), we can conclude that:
\[ 1 - \frac{y^2z + x^2z + y^2x}{r^2} = 0 \]
This means we are dividing by zero, leading to:
\[ \tan^{-1}(\text{undefined}) = \tan^{-1}(\infty) \]

7. **Final Result**:
The value of \(\tan^{-1}(\infty)\) is \(\frac{\pi}{2}\).

### Conclusion:
Thus, the answer to the given problem is:

\[ \frac{\pi}{2} \]