Prove that $\;$ $\cos^{-1} \left(x\right) + \cos^{-1} \left(y\right) = \cos^{-1} \left[x \cdot y - \sqrt{1 - x^2} \cdot \sqrt{1 - y^2}\right]$
Let $\;$ $\cos^{-1} \left(x\right) = \theta_1$, $\;$ $\cos^{-1} \left(y\right) = \theta_2$
Then, $\;$ $\cos \theta_1 = x$, $\;$ $\cos \theta_2 = y$
$\begin{aligned}
\text{Now, } \cos \left(\theta_1 + \theta_2\right) & = \cos \theta_1 \cos \cdot \theta_2 - \sin \theta_1 \cdot \sin \theta_2 \\\\
& = \cos \theta_1 \cdot \cos \theta_2 - \sqrt{1 - \cos^2 \theta_1} \cdot \sqrt{1 - \cos^2 \theta_2} \\\\
& = x \cdot y - \sqrt{1 - x^2} \cdot \sqrt{1 - y^2}
\end{aligned}$
$\therefore \;$ $\theta_1 + \theta_2 = \cos^{-1} \left[x \cdot y - \sqrt{1 - x^2} \cdot \sqrt{1 - y^2}\right]$
i.e. $\;$ $\cos^{-1} \left(x\right) + \cos^{-1} \left(y\right) = \cos^{-1} \left[x \cdot y - \sqrt{1 - x^2} \cdot \sqrt{1 - y^2}\right]$
Hence proved.