If tan

Solution:

$\tan \theta=\frac{\sin \alpha-\cos \alpha}{\sin \alpha+\cos \alpha}$

Dividing numerator and denominator on the RHS by $\cos \alpha$, we get

$\tan \theta=\frac{\frac{\sin \alpha}{\cos \alpha}-1}{\frac{\sin \alpha}{\cos \alpha}+1}$

$\Rightarrow \tan \theta=\frac{\tan \alpha-\tan \frac{\pi}{4}}{1+\tan \alpha \tan \frac{\pi}{4}}$

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

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

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

Now,

$\sin \alpha+\cos \alpha$

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

$=\sin \frac{\pi}{4} \cos \theta+\cos \frac{\pi}{4} \sin \theta+\cos \frac{\pi}{4} \cos \theta-\sin \frac{\pi}{4} \sin \theta$

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

$=\frac{2}{\sqrt{2}} \cos \theta$

$=\sqrt{2} \cos \theta$

$\therefore \sin \alpha+\cos \alpha=\sqrt{2} \cos \theta$