If cos^{-1} x + cos^{-1} y + cos^{-1} z = pi, | vedclass

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(D) Given that $\cos^{-1} x + \cos^{-1} y + \cos^{-1} z = \pi$.$\Rightarrow \cos^{-1} x + \cos^{-1} y = \pi - \cos^{-1} z$.Using the identity $\pi - \

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$x^2 + y^2 + z^2 + 2xyz = 1$

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(D) Given that $\cos^{-1} x + \cos^{-1} y + \cos^{-1} z = \pi$.$\Rightarrow \cos^{-1} x + \cos^{-1} y = \pi - \cos^{-1} z$.Using the identity $\pi - \cos^{-1} z = \cos^{-1}(-z)$,we get:$\cos^{-1} x + \cos^{-1} y = \cos^{-1}(-z)$.Applying the formula $\cos^{-1} x + \cos^{-1} y = \cos^{-1}(xy - \sqrt{1-x^2}\sqrt{1-y^2})$:$\cos^{-1}(xy - \sqrt{1-x^2}\sqrt{1-y^2}) = \cos^{-1}(-z)$.Taking cosine on both sides:$xy - \sqrt{1-x^2}\sqrt{1-y^2} = -z$.Rearranging terms:$xy + z = \sqrt{1-x^2}\sqrt{1-y^2}$.Squaring both sides:$(xy + z)^2 = (1-x^2)(1-y^2)$.$x^2y^2 + z^2 + 2xyz = 1 - x^2 - y^2 + x^2y^2$.$x^2 + y^2 + z^2 + 2xyz = 1$.Alternatively,let $x = y = z = \frac{1}{2}$. Then $\cos^{-1}(\frac{1}{2}) + \cos^{-1}(\frac{1}{2}) + \cos^{-1}(\frac{1}{2}) = \frac{\pi}{3} + \frac{\pi}{3} + \frac{\pi}{3} = \pi$. Substituting these values into option $(d)$: $(\frac{1}{2})^2 + (\frac{1}{2})^2 + (\frac{1}{2})^2 + 2(\frac{1}{2})(\frac{1}{2})(\frac{1}{2}) = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{2}{8} = \frac{3}{4} + \frac{1}{4} = 1$. Thus,option $(d)$ is correct.

If $\cos^{-1} x + \cos^{-1} y + \cos^{-1} z = \pi$,then

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