Explanation

(d)
$y = \tan^{- 1}{(\sec{x - \tan{x)}}}$
$\frac{dy}{dx} = \frac{d}{dx}{\ tan}^{- 1}\left( \frac{1 - \sin x}{\cos x} \right)$
$\frac{dy}{dx} = \frac{d}{dx}\tan^{- 1}\left( \frac{\cos{\left( \frac{x}{2} \right) - \sin\left( \frac{x}{2} \right)}}{\cos{\left( \frac{x}{2} \right) + \sin\left( \frac{x}{2} \right)}} \right)$
$= \frac{d}{dx}\tan^{- 1}\left( \frac{1 - \tan{x/2}}{1 + \tan{x/2}} \right)$
$= \frac{d}{dx}\tan^{- 1}\left\{ \tan\left( \frac{\pi}{4} - \frac{x}{2} \right) \right\}$
$= \frac{d}{dx}\left( \frac{\pi}{4} - \frac{x}{2} \right) = - \frac{1}{2}$