Explanation

(b)
$\frac{\sin{2 - 1}}{\cos 2} = - \frac{1 - \sin 2}{\cos 2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ = - \frac{\left( \cos{1 - \sin 1} \right)^{2}}{\left( \cos{1 + \sin 1} \right)(\cos{1 - \sin 1})}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ = - \frac{\cos{1 - \sin 1}}{\cos{1 + \sin 1}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ = - \frac{1 - \tan 1}{1 + \tan 1}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ = - \tan\left( \frac{\pi}{4} - 1 \right)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ = \tan\left( 1 - \frac{\pi}{4} \right)$
$\Rightarrow \ \ \ \tan^{- 1}\left( \frac{\sin{2 - 1}}{\cos 2} \right)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ = \tan^{- 1}\left\lbrack \tan\left( 1 - \frac{\pi}{4} \right) \right\rbrack$
$\ \ \ \ \ \ \ \ \ \ \ \ \ = 1 - \frac{\pi}{4}$