If sin^{-1} x + sin^{-1} y + sin^{-1} z = fra | vedclass

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(B) Given that $\sin^{-1} x + \sin^{-1} y + \sin^{-1} z = \frac{\pi}{2}$.Let $\sin^{-1} x = \alpha$,$\sin^{-1} y = \beta$,and $\sin^{-1} z = \gamma$.T

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(B) Given that $\sin^{-1} x + \sin^{-1} y + \sin^{-1} z = \frac{\pi}{2}$.Let $\sin^{-1} x = \alpha$,$\sin^{-1} y = \beta$,and $\sin^{-1} z = \gamma$.Then $\alpha + \beta + \gamma = \frac{\pi}{2}$,which implies $\alpha + \beta = \frac{\pi}{2} - \gamma$.Taking $\cos$ on both sides: $\cos(\alpha + \beta) = \cos(\frac{\pi}{2} - \gamma)$.Using the identity $\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$ and $\cos(\frac{\pi}{2} - \gamma) = \sin \gamma$,we get:$\cos \alpha \cos \beta - \sin \alpha \sin \beta = \sin \gamma$.Since $\sin \alpha = x$,$\sin \beta = y$,and $\sin \gamma = z$,we have $\cos \alpha = \sqrt{1 - x^2}$ and $\cos \beta = \sqrt{1 - y^2}$.Substituting these values: $\sqrt{1 - x^2} \sqrt{1 - y^2} - xy = z$.Rearranging gives $\sqrt{1 - x^2} \sqrt{1 - y^2} = xy + z$.Squaring both sides: $(1 - x^2)(1 - y^2) = (xy + z)^2$.$1 - x^2 - y^2 + x^2 y^2 = x^2 y^2 + 2xyz + z^2$.$1 = x^2 + y^2 + z^2 + 2xyz$.

If $\sin^{-1} x + \sin^{-1} y + \sin^{-1} z = \frac{\pi}{2}$,then the value of $x^2 + y^2 + z^2 + 2xyz$ is equal to

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