<p>If \(\cos^{-1}x + \cos^{-1}y + \cos^{-1}z = \pi\), then \(x^2 + y^2 + z^2 + 2xyz\) equals</p>
<p>(a) \(x^2 + y^2 = z^2\)</p>
<p>(b) \(x^2 + y^2 + z^2 = 0\)</p>
<p>(c) \(x^2 + y^2 + z^2 = 1 - 2xyz\)</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: Use the constraint $\cos^{-1}x + \cos^{-1}y + \cos^{-1}z = \pi$ to derive a relationship. Set $\cos^{-1}x = \alpha$, $\cos^{-1}y = \beta$, $\cos^{-1}z = \gamma$ where $\alpha + \beta + \gamma = \pi$, then use trigonometric identities to relate the cosines and their products.
<p><strong>Step 1: Set up the constraint</strong></p><p>Let $\cos^{-1}x = \alpha$, $\cos^{-1}y = \beta$, $\cos^{-1}z = \gamma$.</p><p>Given: $\alpha + \beta + \gamma = \pi$</p><p>Therefore: $\gamma = \pi - (\alpha + \beta)$</p><p>This means: $x = \cos\alpha$, $y = \cos\beta$, $z = \cos\gamma$</p></p><p><strong>Step 2: Apply cosine to the constraint</strong></p><p>Since $\gamma = \pi - (\alpha + \beta)$:</p><p>$$z = \cos\gamma = \cos[\pi - (\alpha + \beta)] = -\cos(\alpha + \beta)$$</p><p>$$z = -(\cos\alpha\cos\beta - \sin\alpha\sin\beta)$$</p><p>$$z = -\cos\alpha\cos\beta + \sin\alpha\sin\beta$$</p><p>$$z = -xy + \sin\alpha\sin\beta$$</p></p><p><strong>Step 3: Find $\sin\alpha$ and $\sin\beta$</strong></p><p>Since $\alpha, \beta \in [0, \pi]$ (range of $\cos^{-1}$), we have $\sin\alpha \geq 0$ and $\sin\beta \geq 0$.</p><p>$$\sin\alpha = \sqrt{1 - \cos^2\alpha} = \sqrt{1 - x^2}$$</p><p>$$\sin\beta = \sqrt{1 - \cos^2\beta} = \sqrt{1 - y^2}$$</p></p><p><strong>Step 4: Substitute back into the equation</strong></p><p>$$z = -xy + \sqrt{1-x^2}\sqrt{1-y^2}$$</p><p>$$z + xy = \sqrt{(1-x^2)(1-y^2)}$$</p></p><p><strong>Step 5: Square both sides</strong></p><p>$$(z + xy)^2 = (1-x^2)(1-y^2)$$</p><p>$$z^2 + 2xyz + x^2y^2 = 1 - x^2 - y^2 + x^2y^2$$</p><p>$$z^2 + 2xyz = 1 - x^2 - y^2$$</p><p>$$x^2 + y^2 + z^2 + 2xyz = 1$$</p><p>$$x^2 + y^2 + z^2 = 1 - 2xyz$$</p></p><p><strong>∴ Answer: C</strong></p>
Correct Answer: C