b. If cos1 x  cos1 y  cos1 z   , prove that x2  y2  z2  1 2xyz

We are given the equation \(\cos^{-1} x + \cos^{-1} y + \cos^{-1} z = \pi\), and we need to prove that \(x^2 + y^2 + z^2 = 1 - 2xyz\).

### Step 1: Use the identity involving inverse cosine
We know that if \(\cos^{-1} a + \cos^{-1} b + \cos^{-1} c = \pi\), then the identity:
\[
\cos(\cos^{-1} a + \cos^{-1} b) = -\cos(\cos^{-1} c)
\]
applies. Therefore, we have:
\[
\cos(\cos^{-1} x + \cos^{-1} y) = -\cos(\cos^{-1} z)
\]

### Step 2: Apply the cosine addition formula
Using the cosine addition formula:
\[
\cos(A + B) = \cos A \cos B - \sin A \sin B
\]
we substitute \(A = \cos^{-1} x\) and \(B = \cos^{-1} y\):
\[
\cos(\cos^{-1} x + \cos^{-1} y) = \cos(\cos^{-1} x) \cos(\cos^{-1} y) - \sin(\cos^{-1} x) \sin(\cos^{-1} y)
\]
\[
\cos(\cos^{-1} x + \cos^{-1} y) = xy - \sqrt{(1-x^2)(1-y^2)}
\]

Since \(\cos(\cos^{-1} x + \cos^{-1} y) = -\cos(\cos^{-1} z) = -z\), we get:
\[
xy - \sqrt{(1-x^2)(1-y^2)} = -z
\]

### Step 3: Square both sides to eliminate the square root
Square both sides of the equation:
\[
(xy + z)^2 = (1-x^2)(1-y^2)
\]
Expanding both sides, we get:
\[
x^2y^2 + z^2 + 2xyz = (1 - x^2 - y^2 + x^2y^2)
\]

### Step 4: Simplify the equation
Bring all terms to one side:
\[
x^2y^2 + z^2 + 2xyz - 1 + x^2 + y^2 - x^2y^2 = 0
\]
The \(x^2y^2\) terms cancel out, so:
\[
x^2 + y^2 + z^2 + 2xyz = 1
\]

Rearranging the equation, we get the desired result:
\[
x^2 + y^2 + z^2 = 1 - 2xyz
\]

### Conclusion
This proves that \(x^2 + y^2 + z^2 = 1 - 2xyz\), given that \(\cos^{-1} x + \cos^{-1} y + \cos^{-1} z = \pi\).

Would you like more details on any part of this proof, or do you have any questions?

Here are 5 related questions to consider:
1. What other trigonometric identities can be derived from the inverse cosine function?
2. How does the identity \(\cos^{-1} a + \cos^{-1} b + \cos^{-1} c = \pi\) apply in geometric contexts?
3. Can the equation \(x^2 + y^2 + z^2 = 1 - 2xyz\) have a geometric interpretation?
4. How does the identity change if the sum \(\cos^{-1} x + \cos^{-1} y + \cos^{-1} z\) is not equal to \(\pi\)?
5. What is the physical significance of the equation \(x^2 + y^2 + z^2 = 1 - 2xyz\) in any practical scenario?

**Tip:** When working with inverse trigonometric functions, always consider the range of the function and how it impacts the identities you are using.

Learn how to prove the equation x^2 + y^2 + z^2 = 1 - 2xyz given the condition \(\cos^{-1} x + \cos^{-1} y + \cos^{-1} z = \pi\). This problem involves advanced concepts in inverse trigonometric functions and trigonometric identities. Suitable for advanced high school students and those preparing for competitive exams.

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