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

A) $x^{2} + y^{2} = \mathcal{z}^{2}$
B) $x^{2} + y^{2} + \mathcal{z}^{2} = 0$
C) $x^{2} + y^{2} + \mathcal{z}^{2} = 1 - 2xy\mathcal{z}$
D) None of the above
Explanation

(c)
$\cos^{- 1}{x + \cos^{- 1}{y + \cos^{- 1}{\mathcal{z =}\pi}}}$
$\Rightarrow \cos^{- 1}{\left( xy - \sqrt{1 - y^{2}}\ \sqrt{1 - x^{2}} \right) = \pi - \cos^{- 1}\mathcal{z}}$
$\Rightarrow xy - \sqrt{1 - x^{2}\ }\sqrt{1 - y^{2} =}cos(\pi - \cos^{- 1}\mathcal{z)}$
$\Rightarrow xy - \sqrt{1 - x^{2}}$ $\sqrt{1 - y^{2}}\mathcal{= - z}$
$\Rightarrow xy + \mathcal{z =}\sqrt{1 - x^{2}}\ \sqrt{1 - y^{2}}$
On squaring both sides, we get
$x^{2}y^{2} + \mathcal{z}^{2} + 2xy\mathcal{z -}1 - x^{2} - y^{2} + x^{2}y^{2}$
$\Rightarrow x^{2} + y^{2} + \mathcal{z}^{2} = 1 - 2xy\mathcal{z}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.