If ^ - 1 x + ^ - 1 y + ^ - 1 z = , then - Vidyadip

MCQ

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

A
${x^2} + {y^2} + {z^2} + xyz = 0$
B
${x^2} + {y^2} + {z^2} + 2xyz = 0$
C
${x^2} + {y^2} + {z^2} + xyz = 1$
✓
${x^2} + {y^2} + {z^2} + 2xyz = 1$

✓

Answer

Correct option: D.

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

d
(d) Given that ${\cos ^{ - 1}}x + {\cos ^{ - 1}}y + {\cos ^{ - 1}}z = \pi $

$ \Rightarrow \,\,{\cos ^{ - 1}}(x) + {\cos ^{ - 1}}(y) + {\cos ^{ - 1}}(z) = {\cos ^{ - 1}}( - 1)$

$ \Rightarrow \,\,{\cos ^{ - 1}}(x) + {\cos ^{ - 1}}(y) = {\cos ^{ - 1}}( - 1) - {\cos ^{ - 1}}(z)$

$ \Rightarrow \,\,{\cos ^{ - 1}}(xy - \sqrt {1 - {x^2}} \sqrt {1 - {y^2})} = {\cos ^{ - 1}}\,\left\{ {( - 1)\,\,(z)} \right\}$

$ \Rightarrow \,\,xy - \sqrt {(1 - {x^2})\,\,(1 - {y^2})} = - z$

$ \Rightarrow \,\,(xy + z) = \sqrt {(1 - {x^2})\,\,(1 - {y^2})} $

Squaring both sides we get ${x^2} + {y^2} + {z^2} + 2xyz = 1$.

Trick : Put $x = y = z = \frac{1}{2},$ so that

${\cos ^{ - 1}}\frac{1}{2} + {\cos ^{ - 1}}\frac{1}{2} + {\cos ^{ - 1}}\frac{1}{2} = \pi $

Obviously $ (d)$ holds for these values of $ x, y, z.$

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