If `x=r sin theta cos phi, y=r sin theta sin phi` and `z=r cos theta` then

Options

A. `x^(2)+y^(2)+z^(2)=r^(2)`
B. `x^(2)+y^(2)-z^(2)=r^(2)`
C. `x^(2)-y^(2)+z^(2)=r^(2)`
D. `z^(2)+y^(2)-x^(2)=r^(2)`

Answer

To solve the problem, we need to verify the relationship between the variables \(x\), \(y\), and \(z\) given by the equations:

1. \(x = r \sin \theta \cos \phi\)
2. \(y = r \sin \theta \sin \phi\)
3. \(z = r \cos \theta\)

We want to determine which of the given options is true. Let's go through the steps to find the correct relationship.

### Step 1: Square Each Equation
We start by squaring each of the equations:

- \(x^2 = (r \sin \theta \cos \phi)^2 = r^2 \sin^2 \theta \cos^2 \phi\)
- \(y^2 = (r \sin \theta \sin \phi)^2 = r^2 \sin^2 \theta \sin^2 \phi\)
- \(z^2 = (r \cos \theta)^2 = r^2 \cos^2 \theta\)

### Step 2: Add the Squared Equations
Now, we add the squared equations together:

\[ x^2 + y^2 + z^2 = r^2 \sin^2 \theta \cos^2 \phi + r^2 \sin^2 \theta \sin^2 \phi + r^2 \cos^2 \theta \]

### Step 3: Factor Out Common Terms
On the right-hand side, we can factor out \(r^2\):

\[ x^2 + y^2 + z^2 = r^2 \left(\sin^2 \theta \cos^2 \phi + \sin^2 \theta \sin^2 \phi + \cos^2 \theta\right) \]

### Step 4: Simplify the Expression
Now, we can simplify the expression inside the parentheses. Notice that:

\[ \sin^2 \theta \cos^2 \phi + \sin^2 \theta \sin^2 \phi = \sin^2 \theta (\cos^2 \phi + \sin^2 \phi) \]

Using the Pythagorean identity \(\cos^2 \phi + \sin^2 \phi = 1\), we get:

\[ \sin^2 \theta (\cos^2 \phi + \sin^2 \phi) = \sin^2 \theta \cdot 1 = \sin^2 \theta \]

Thus, we can rewrite our equation as:

\[ x^2 + y^2 + z^2 = r^2 (\sin^2 \theta + \cos^2 \theta) \]

### Step 5: Apply Another Pythagorean Identity
Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\), we have:

\[ x^2 + y^2 + z^2 = r^2 \cdot 1 = r^2 \]

### Conclusion
We conclude that:

\[ x^2 + y^2 + z^2 = r^2 \]

This matches with option 1. Therefore, the correct answer is:

**Option 1: \(x^2 + y^2 + z^2 = r^2\)**.