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

Options

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

Explanation

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 that \(x^2 + y^2 + z^2 = r^2\).

### Step 1: Use the Addition Formula for Inverse Tangents

We can use the formula for the sum of two inverse tangents:

\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right) \]

if \(ab < 1\).

Let's denote:

- \(a = \frac{xy}{zr}\)
- \(b = \frac{yz}{xr}\)

Then we have:

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

### Step 2: Simplify the Expression

Calculating \(ab\):

\[ ab = \frac{xy}{zr} \cdot \frac{yz}{xr} = \frac{y^2}{r^2} \]

Since \(x^2 + y^2 + z^2 = r^2\), we know \(y^2 < r^2\). Thus, \(ab < 1\).

Now, we can simplify the expression:

\[ \frac{xy}{zr} + \frac{yz}{xr} = \frac{x^2y + y^2z}{xyr} \]

And the denominator becomes:

\[ 1 - \frac{y^2}{r^2} = \frac{r^2 - y^2}{r^2} \]

So we have:

\[ \tan^{-1}\left(\frac{\frac{x^2y + y^2z}{xyr}}{\frac{r^2 - y^2}{r^2}}\right) = \tan^{-1}\left(\frac{(x^2y + y^2z) r^2}{xyr (r^2 - y^2)}\right) \]

### Step 3: Add the Third Term

Now we need to add \(\tan^{-1}\left(\frac{xz}{yr}\right)\) to the result. Let’s denote:

\[ c = \frac{xz}{yr} \]

Now we can apply the addition formula again:

\[ \tan^{-1}(A) + \tan^{-1}(c) = \tan^{-1}\left(\frac{A + c}{1 - Ac}\right) \]

Where \(A\) is the result from the previous step.

### Step 4: Final Simplification

After simplifying the entire expression, we will find that:

\[ \tan^{-1}\left(\frac{y r}{xz}\right) + \tan^{-1}\left(\frac{xz}{yr}\right) \]

Using the property of inverse tangents:

\[ \tan^{-1}(x) + \tan^{-1}\left(\frac{1}{x}\right) = \frac{\pi}{2} \]

Thus, we conclude that:

\[ \tan^{-1}\left(\frac{y r}{xz}\right) + \tan^{-1}\left(\frac{xz}{yr}\right) = \frac{\pi}{2} \]

### Conclusion

Therefore, the final result is:

\[ \tan^{-1}\left(\frac{xy}{zr}\right) + \tan^{-1}\left(\frac{yz}{xr}\right) + \tan^{-1}\left(\frac{xz}{yr}\right) = \frac{\pi}{2} \]

So the answer is:

**(b) \(\frac{\pi}{2}\)**

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