If ^ - 1 x a = ^ - 1 x b = ^ - 1 y c ; 0 < x < 1, then the value of...

So, $a + b = \dfrac{\pi}{2r}$

$cos$\left(\dfrac{\pic}{a + b}\right) = cos$\left(\dfrac{\pi\tan^{- 1}y}{\dfrac{\pi}{2r}r}\right)$

$= \cos \left(2\tan^{- 1}y\right),$ let $\tan^{- 1}y = \theta$

$= \cos \left(2\theta\right)$

$= \dfrac{1 - \tan^{2}\theta}{1 + \tan^{2}\theta} = \dfrac{1 - {y}^{2}}{1 + {y}^{2}}$