If `x^(2)+y^(2)+z^(2)-2xyz=1`, then the value of `(dx)/(sqrt(1-x^(2)))+(dy)/(sqrt(1-y^(2)))+(dz)/(sqrt(1-z^(2)))` is equal to………..

Answer

To solve the problem, we need to find the value of

\[ \frac{dx}{\sqrt{1-x^2}} + \frac{dy}{\sqrt{1-y^2}} + \frac{dz}{\sqrt{1-z^2}} \]

given the equation

\[ x^2 + y^2 + z^2 - 2xyz = 1. \]

### Step-by-Step Solution:

1. **Differentiate the given equation**: We start by differentiating the equation \( x^2 + y^2 + z^2 - 2xyz = 1 \) with respect to \( x \).

\[ \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) + \frac{d}{dx}(z^2) - \frac{d}{dx}(2xyz) = 0 \]

This gives us:

\[ 2x + 0 + 0 - 2(yz \frac{dx}{dx} + xz \frac{dy}{dx} + xy \frac{dz}{dx}) = 0 \]

Simplifying, we have:

\[ 2x - 2(yz \frac{dy}{dx} + xz \frac{dz}{dx} + xy \frac{dy}{dx}) = 0 \]

Rearranging gives:

\[ 2x = 2(yz \frac{dy}{dx} + xz \frac{dz}{dx} + xy \frac{dy}{dx}) \]

Thus,

\[ x = yz \frac{dy}{dx} + xz \frac{dz}{dx} + xy \frac{dy}{dx} \]

2. **Partial differentiation with respect to \( y \)**: Next, we differentiate the original equation with respect to \( y \):

\[ 0 + 2y + 0 - 2(xz \frac{dx}{dy} + xy \frac{dz}{dy}) = 0 \]

This simplifies to:

\[ 2y - 2(xz \frac{dx}{dy} + xy \frac{dz}{dy}) = 0 \]

Rearranging gives:

\[ y = xz \frac{dx}{dy} + xy \frac{dz}{dy} \]

3. **Partial differentiation with respect to \( z \)**: Finally, we differentiate the original equation with respect to \( z \):

\[ 0 + 0 + 2z - 2(xy \frac{dx}{dz} + xz \frac{dy}{dz}) = 0 \]

This simplifies to:

\[ 2z - 2(xy \frac{dx}{dz} + xz \frac{dy}{dz}) = 0 \]

Rearranging gives:

\[ z = xy \frac{dx}{dz} + xz \frac{dy}{dz} \]

4. **Substituting back into the original expression**: Now we can substitute the values of \( dx, dy, dz \) back into the expression we want to evaluate:

\[ \frac{dx}{\sqrt{1-x^2}} + \frac{dy}{\sqrt{1-y^2}} + \frac{dz}{\sqrt{1-z^2}} = 0 \]

Since \( dx, dy, dz \) are all zero, the entire expression evaluates to zero.

### Final Answer:

\[ \frac{dx}{\sqrt{1-x^2}} + \frac{dy}{\sqrt{1-y^2}} + \frac{dz}{\sqrt{1-z^2}} = 0 \]