Question

If $si{n}^{-1}x+si{n}^{-1}y+si{n}^{-1}z=\frac{\pi }{2},then{x}^2+y^2+z^2+2xyz=$

• A. 0.0
• B. 1
• C. -1
• D. None

Answer 1

The correct option is B 1

Given expression
$\Rightarrow co{s}^{-1}\sqrt{1-x^2}+co{s}^{-1}\sqrt{1-y^2}=\frac{\pi }{2}-si{n}^{-1}z$
$\Rightarrow$ $co{s}^{-1}[\sqrt{(1-x^2)(1-y^2)}-\sqrt{\{1-(1-x^2\left)\right\}\{1-(1-y^2\left)\right\}}]=co{s}^{-1}z$
$\Rightarrow$ $\sqrt{(1-x^2)(1-y^2)}-xy=z$
$\Rightarrow$ $(1-x^2)(1-y^2)=(xy+z{)}^2$
$\Rightarrow$ $x^2+y^2+z^2+2xyz=1$