Question

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

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is B

$1$

Explanation for the correct option.

Given that, ${\sin }^{-1}x+{\sin }^{-1}y+{\sin }^{-1}z=\frac{\mathrm{\pi }}{2}$.

Let $x=\sin \alpha ,y=\sin \beta ,z=\sin y$

$$\begin{gathered} \alpha +\beta +\gamma =\frac{\mathrm{\pi }}{2} \\ \Rightarrow \alpha +\beta =\frac{\mathrm{\pi }}{2}-\gamma \end{gathered}$$

$$\begin{gathered} \Rightarrow \cos (\alpha +\beta )=\cos \{\left(\frac{\pi }{2}\right)-\gamma \} \\ \Rightarrow \cos \alpha ·\cos \beta -\sin \alpha ·\sin \beta =\sin \gamma \\ \sqrt{\left(1-x^2\right)}\sqrt{\left(1-y^2\right)}-xy=z \\ \left.\left.\Rightarrow \sqrt{\left(1-x^2\right.}\right)(\sqrt{1-y^2}\right)=xy+z \\ \left.\Rightarrow \left(1-x^2\right)\sqrt{(}1-y^2\right)={\left(xy+z^2\right)}^2 \\ \Rightarrow 1-y^2-x^2+x^2{y}^2=x^2{y}^2+z^2+2xyz \\ \Rightarrow x^2+y^2+z^2+2xyz=1 \end{gathered}$$

Hence, option B is correct.