• # question_answer If $\tan \theta =\frac{\sin \alpha -\cos \alpha }{\sin \alpha +\cos \alpha },$then $\sin \alpha +\cos \alpha$ and $\sin \alpha -\cos \alpha$ must be equal to [WB JEE 1971] A) $\sqrt{2}\cos \theta ,\,\,\sqrt{2}\sin \theta$ B) $\sqrt{2}\sin \theta ,\,\,\sqrt{2}\cos \theta$ C) $\sqrt{2}\sin \theta ,\,\,\sqrt{2}\sin \theta$ D) $\sqrt{2}\,\cos \theta ,\,\,\sqrt{2}\,\cos \theta$

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 $\sin \alpha -\cos \alpha =\sin \left( \theta +\frac{\pi }{4} \right)-\cos \left( \theta +\frac{\pi }{4} \right)$ $=\frac{1}{\sqrt{2}}\sin \theta +\frac{1}{\sqrt{2}}\cos \theta -\frac{1}{\sqrt{2}}\cos \theta +\frac{1}{\sqrt{2}}\sin \theta$ $=\frac{2}{\sqrt{2}}\sin \theta =\sqrt{2}\sin \theta$.