If y ( x ) = ^ - 1 ( 1 + x + 1 - x 1 + x - 1 - x ), x ∈ ((π)/(2),...

$y ( x ) = \cot^{- 1} \dfrac{\sqrt{1 + sin x} + \sqrt{1 - sin x}}{\sqrt{1 + sin x} - \sqrt{1 - sin x}}$

As we know that, $\sqrt{1 - sin x} = \left|\sin\dfrac{x}{2} - cos\dfrac{x}{2}\right| \text{ and } \sqrt{1 + sin x} = \left|\sin\dfrac{x}{2} + cos\dfrac{x}{2}\right|$

$y ( x ) = \cot^{- 1} \dfrac{\left|\cos \dfrac{x}{2} + sin \dfrac{x}{2}\right| + \left|\cos \dfrac{x}{2} - sin \dfrac{x}{2}\right|}{\left|\cos \dfrac{x}{2} + sin \dfrac{x}{2}\right| - \left|\cos \dfrac{x}{2} - sin \dfrac{x}{2}\right|}$

$= \cot^{- 1} \dfrac{cos \dfrac{x}{2} + sin \dfrac{x}{2} + sin \dfrac{x}{2} - cos \dfrac{x}{2}}{cos \dfrac{x}{2} + sin \dfrac{x}{2} - sin \dfrac{x}{2} + cos \dfrac{x}{2}}\text{for}x \in \left(\dfrac{\pi}{2}, \pi\right)$

$= \cot^{- 1} \left(\dfrac{sin \dfrac{x}{2}}{cos \dfrac{x}{2}}\right)$

$= \cot^{- 1} \left(tan \dfrac{x}{2}\right)$

$= \cot^{- 1} \left(\cot ( \dfrac{\pi}{2} - \dfrac{x}{2} )\right)$

$y ( x ) = \dfrac{\pi}{2} - \dfrac{x}{2}$

${y}^{′} ( x ) = - \dfrac{1}{2}$