Question

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If
$\cos ^{ - 1} x + \cos { - 1} y + { \cos}^{ - 1} z = \pi$
then that
${x}^{2} + {y}^{2} + {z}^{2} + 2xy = 1$
​

Answer 1

$\mathfrak{\huge{\underline{\underline{Answer:-}}}}$

${ \cos }^{ - 1} x = a$

${ \cos}^{ - 1} y = b$

${ \cos}^{ - 1} z = c$

= A + B + C = π

= cosAcosB - sinAsinB = -cosC

$y + z = \sqrt{1 - {x}^{2} } \sqrt{1 - {y}^{2} }$

${x}^{2} + {y}^{2} + {z}^{2} + 2xyz = 1 - {x}^{2} - {y}^{2} + {x}^{2} {y}^{2}$

$= > {x}^{2} + {y}^{2} + {z}^{2} + 2xyz = 1$