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 start with the given equation:

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

We need to find the value of the expression:

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

### Step 1: Differentiate the given equation

We will differentiate both sides of the equation with respect to a variable (let's assume it's time \(t\)):

\[ \frac{d}{dt}(x^2 + y^2 + z^2 - 2xyz) = \frac{d}{dt}(1) \]

Using the chain rule, we differentiate each term:

\[ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} + 2z \frac{dz}{dt} - 2\left( y z \frac{dx}{dt} + x z \frac{dy}{dt} + x y \frac{dz}{dt} \right) = 0 \]

### Step 2: Rearranging the differentiated equation

Rearranging the equation gives:

\[ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} + 2z \frac{dz}{dt} = 2\left( y z \frac{dx}{dt} + x z \frac{dy}{dt} + x y \frac{dz}{dt} \right) \]

Dividing through by 2 simplifies it to:

\[ x \frac{dx}{dt} + y \frac{dy}{dt} + z \frac{dz}{dt} = y z \frac{dx}{dt} + x z \frac{dy}{dt} + x y \frac{dz}{dt} \]

### Step 3: Isolate the differentials

Rearranging gives:

\[ \left( x - yz \right) \frac{dx}{dt} + \left( y - xz \right) \frac{dy}{dt} + \left( z - xy \right) \frac{dz}{dt} = 0 \]

### Step 4: Solve for the expression

Now, we can express the differentials in terms of the original variables. We can use the identity \( \sqrt{1 - x^2} \) in our expression:

From the given equation, we know that \( x^2 + y^2 + z^2 = 1 + 2xyz \), which means that \( 1 - x^2 = y^2 + z^2 - 2xyz \).

Thus, we can substitute back into our expression:

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

### Final Result

Thus, the value of the expression is:

\[ \boxed{2} \]