(a) We have $\tan \theta = \frac{{\sin \alpha – \cos \alpha }}{{\sin \alpha + \cos \alpha }}$

$ \Rightarrow \tan \theta = \frac{{\sin \left( {\alpha – \frac{\pi }{4}} \right)}}{{\cos \left( {\alpha – \frac{\pi }{4}} \right)}} $

$\Rightarrow \tan \theta = \tan \left( {\alpha – \frac{\pi }{4}} \right)$

$ \Rightarrow \theta = \alpha – \frac{\pi }{4}$

$\Rightarrow \alpha = \theta + \frac{\pi }{4}$

Hence, $\sin \alpha + \cos \alpha = \sin \left( {\theta + \frac{\pi }{4}} \right) + \cos \left( {\theta + \frac{\pi }{4}} \right)$

$ = \sqrt 2 \cos \theta $

and $\sin \alpha – \cos \alpha = \sin \left( {\theta + \frac{\pi }{4}} \right) – \cos \left( {\theta + \frac{\pi }{4}} \right)$

$ = \frac{1}{{\sqrt 2 }}\sin \theta + \frac{1}{{\sqrt 2 }}\cos \theta – \frac{1}{{\sqrt 2 }}\cos \theta + \frac{1}{{\sqrt 2 }}\sin \theta $

$ = \frac{2}{{\sqrt 2 }}\sin \theta = \sqrt 2 \sin \theta $.