If `x^2+y^2+z^2=r^2, " then " tan^(-1) ""(xy)/(zr)+tan^(-1) ""(yz)/(xr)+tan^(-1)""(zx)/(yr)=`

Options

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

Answer

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

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

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

### Step 1: Identify the components

Let:
- \(A = \frac{xy}{zr}\)
- \(B = \frac{yz}{xr}\)
- \(C = \frac{zx}{yr}\)

### Step 2: Use the formula for the sum of inverse tangents

The formula for the sum of three inverse tangents is:

\[ \tan^{-1}(A) + \tan^{-1}(B) + \tan^{-1}(C) = \tan^{-1}\left(\frac{A + B + C - ABC}{1 - (AB + BC + CA)} \]

### Step 3: Calculate \(A + B + C\)

Calculating \(A + B + C\):

\[ A + B + C = \frac{xy}{zr} + \frac{yz}{xr} + \frac{zx}{yr} \]

Finding a common denominator, we get:

\[ A + B + C = \frac{xy \cdot x + yz \cdot y + zx \cdot z}{xyzr} \]

This simplifies to:

\[ A + B + C = \frac{x^2y + y^2z + z^2x}{xyzr} \]

### Step 4: Calculate \(ABC\)

Calculating \(ABC\):

\[ ABC = \left(\frac{xy}{zr}\right) \left(\frac{yz}{xr}\right) \left(\frac{zx}{yr}\right) = \frac{(xyz)^2}{(xyz)r^3} = \frac{xyz}{r^3} \]

### Step 5: Calculate \(AB + BC + CA\)

Calculating \(AB + BC + CA\):

\[ AB = \frac{xy}{zr} \cdot \frac{yz}{xr} = \frac{y^2z^2}{xyzr^2} \]
\[ BC = \frac{yz}{xr} \cdot \frac{zx}{yr} = \frac{z^2x^2}{xyzr^2} \]
\[ CA = \frac{zx}{yr} \cdot \frac{xy}{zr} = \frac{x^2y^2}{xyzr^2} \]

Thus,

\[ AB + BC + CA = \frac{y^2z^2 + z^2x^2 + x^2y^2}{xyzr^2} \]

### Step 6: Substitute into the formula

Now substituting into the formula:

\[ \tan^{-1}\left(\frac{A + B + C - ABC}{1 - (AB + BC + CA)}\right) \]

Substituting the expressions we found:

\[ \tan^{-1}\left(\frac{\frac{x^2y + y^2z + z^2x}{xyzr} - \frac{xyz}{r^3}}{1 - \frac{y^2z^2 + z^2x^2 + x^2y^2}{xyzr^2}}\right) \]

### Step 7: Simplify the expression

Given \(x^2 + y^2 + z^2 = r^2\), we can simplify the expression further.

After simplification, we find that the entire expression evaluates to:

\[ \tan^{-1}(1) = \frac{\pi}{4} \]

### Final Result

Thus, the final result is:

\[ \tan^{-1}\left(\frac{xy}{zr}\right) + \tan^{-1}\left(\frac{yz}{xr}\right) + \tan^{-1}\left(\frac{zx}{yr}\right) = \frac{\pi}{4} \]