Question

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

Answer 1

We have,

${\cos }^{-1}x+{\cos }^{-1}y+{\cos }^{-1}z=\pi ......\left(1\right)$

$\Rightarrow {\cos }^{-1}x+{\cos }^{-1}y=\pi -{\cos }^{-1}z$

We know that,

${\cos }^{-1}A+{\cos }^{-1}B={\cos }^{-1}[AB-\sqrt{1-A^2}\sqrt{1-B^2}]$

Therefore,

${\cos }^{-1}[xy-\sqrt{1-x^2}\sqrt{1-y^2}]=\pi -{\cos }^{-1}z$

$\Rightarrow [xy-\sqrt{1-x^2}\sqrt{1-y^2}]=\cos (\pi -{\cos }^{-1}z)$

$\Rightarrow [xy-\sqrt{1-x^2}\sqrt{1-y^2}]=\cos (\pi -{\cos }^{-1}z)\therefore \cos (\pi -\theta )=-\cos \theta$

$\Rightarrow [xy-\sqrt{1-x^2}\sqrt{1-y^2}]=-\cos {\cos }^{-1}z$

$\Rightarrow [xy-\sqrt{1-x^2}\sqrt{1-y^2}]=-z$

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

On squaring both side and we get$\begin{align}

${(xy+z)}^2=(1-x^2)(1-y^2)$

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

$\Rightarrow z^2+2xyz=1-x^2-y^2$

$\Rightarrow x^2+y^2+z^2+2xyz=1$

Hence proved.