Solution:
Given, $ \displaystyle \sin^{-1} \cos \left ( \sin^{-1} \text{x}\right ) + \cos^{-1} \sin \left ( \cos^{-1} \text{x} \right )$
$ =\displaystyle \sin^{-1} \cos \left ( \cos^{-1} \sqrt{1-\text{x}^2}\right ) + \cos^{-1} \sin \left ( \sin^{-1} \sqrt{1-\text{x}^2} \right ) $
$ =\displaystyle \sin^{-1} \left(\sqrt{1-\text{x}^2}\right ) + \cos^{-1}\left ( \sqrt{1-\text{x}^2} \right ) =\frac{\pi}{2}$
Since $ \sin^{-1}\text{x}+\cos^{-1}\text{x} = \frac{\pi}{2} \forall |\text{x}|\leq $
Solution:
$ \sin^{-1}(\text{x})+\sin^{-1}(1-\text{x})=\cos^{-1}(\text{x})$
Now, x ϵ [-1, 1]
and -1 ≤ (1 - x) ≤ 1
x ϵ [0, 2]
Hence, x ϵ [0, 1]
Now we get two solutions of the above equation as 0 and $ \frac{1}{2}$
Thus, the two roots are
x = 0 and $ \text{x}=\frac{1}{2}$
$ α+β=\frac{2}{1}$
and $ \alpha\beta = 0$
$\cos^{-1}\bigg(\cos\frac{7\pi}{6}\bigg)=\cos^{-1}\bigg[\cos\bigg(2\pi-\frac{7\pi}{6}\bigg)\bigg]$
$\left[\because\cos\left(2\pi-\theta\right)=\cos\theta\right]$
$=2\pi-\frac{7\pi}{6}=\frac{12\pi-7\pi}{6}=\frac{5\pi}{6}$